Fall 2025: Math 147 Daily Update

We continued our discussion of the change of variables theorem for double integration. After recalling the translation transformation, we discussed at length linear transformations:

Important Properties of linear transformations.

1. (i) \(T(P_1+P_2) = T(P_1)+T(P_2)\), for points \(P_1, P_2\) in the \(uv\)-plane.
2. (ii) \(T(\lambda P) = \lambda T(P)\), for all points \(P\) in the \(uv\)-plane and \(\lambda \in \mathbb{R}\).
3. (iii) \(T\) is one-to-one if and only if the Jacobian \(ad-bc \not = 0\)

Part (ii) shows that \(T\) takes lines through the origin in the \(uv\)-plane to lines through the origin in the \(xy\)-plane. In general, \(T = (au+bv, cu+vd))\) transforms the unit square in the \(uv\)-plane to the parallelogram in the \(xy\)-plane spanned by the vectors \((a,c)\) and \((b,d)\). We noted that this can be seen using the same reasoning as in Example 1 from the previous lecture.

We also considered a second type of transformation,

For example: If \(u^2+v^2= 2^2\), and \(G(u,v) = (u+a,v+b)\), then \(u = x-a\) and \(v = y-b\), so

In other words, \(G\) translates the circle (and the disk) of radius \(2\) in the \(uv\)-plane centered at (0,0) to the circle (and disk) of radius \(2\) in the \(xy\)-plane, centered at \((a,b)\). We illustrated this by calculating \(\int\int_R\sqrt{(x-2)^2+(y-3)^2}\ dA\) where \(R\) is the disk of radius two centered at (2,3). By setting \(G(u,v) = (u+2, v+3)\), the given integral becomes \(\int\int_{R_0} \sqrt{u^2+v^2}\ dA\), where \(R_0\) is the disk of radius two centered at the origin in the \(uv\)-plane. This latter integral can be solved using polar coordinates, which itself is another transformation. Alternately, one can do one transformation by setting \(G = (u\cos(v)+2, u\sin(v)+3)\). We also used similar ideas to calculate the area of the ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2} = 1\).

Here is an example combining both a linear transformation with a translation.

in the \(xy\)-plane.

Solution: We see that this parallelogram is the translation of one similar to it with lower left corner at the origin. If we move the vertex (1,2) to the origin,we get a new parallelogram with vertices (0,0), (2,2), (5,3), (3,1), moving counterclockwise along the perimeter. This new parallelogram is spanned by the vectors (3,1) and (2,2), so that by the previous lecture, \(T(u,v) = (3u+2v, u+2v)\) takes the unit square in the \(uv\)-plane to the new parallelogram. If we add (1,2) to the new coordinates, we get

and this transformation takes the unit square in the \(uv\)-plane to the original parallelogram in the \(xy\)-plane.

Notice that \(G\) takes the vertices (0,0), (1,0), (1,1), (0,1), in the \(uv\)-plane to the vertices (1,2), (4,3), (6,5), (3,4) in the \(xy\)-plane. Moreover,

.

We ended class with a discussion of the transformations needed to calculate the volume of the ellipsoid \(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2} = 1\).

Solution: If we let \(D\) be the elliptic disk \(0\leq \frac{x^2}{a^2}+\frac{y^2}{b^2} \leq 1\), then:

We can do this two different ways using the change of variables formula. First, we can use the transformation from the end of last lecture, \(G(u,v) = (au\cos(v), bu\sin(v))\), with \(|\frac{\partial(x,y)}{\partial(u,v)} | = abu\), which takes the rectangle \(0\leq u\leq 1\), \(0\leq v\leq 2\pi\) to \(D\). \ We then get:

Alternately, we can use the linear transformation \(T(u,v) = (au,bv)\), with \(|\frac{\partial(x,y)}{\partial(u,v)} | = ab\). This transformation stretches the plane \(a\) units horizontally and \(b\) units vertically. It takes the unit disk \(D': 0\leq u^2+v^2 \leq 1\) to \(D\). Thus:

Now, we can either use polar coordinates to evaluate the last double integral, or in this case, recognize it as the volume of the top half of the unit sphere, which is \(\frac{2}{3}\pi\). Thus, \(vol(E) = \frac{4}{3}\pi abc\).

We began with an example which shows that a change of variables can be used when the integrand has no (obvious) antiderivative.

We also have

. Thus, \(|\mathrm{Jac} (G)| = \frac{1}{2}\). Therefore,

the last step being a standard Calculus 2 problem that can be solved using integration by parts.

We then discussed how the previous problem is related to the notion of an inverse for \(G(u,v)\) We knew how to express \(u,v\) in terms of \(x,y\), but really wanted to express \(x,y\) in terms of \(u,v\). What this means, is that we were given a function \(F(x,y) = (u(x,y), v(x,y))\) that writes \(u,v\) in terms of \(x,y\), and we want to “unravel” \(F(x,y)\) so that we can express \(x,y\) in terms of \(u,v\), via the function \(G(u,v)\). This involves realizing \(G(u,v)\) as the inverse of \(F(x,y)\), or equivalently, regarding \(F(x,y)\) as the inverse of \(G(u,v)\). This lead to the:

It is important that both these equations hold, otherwise \(F\) is not the inverse of \(G\). By symmetry, it follows that \(G\) is the inverse of \(F\). While it may not always be possible to find \(F\) given \(G\), the idea is that if we express \(x\) and \(y\) in terms of \(u\) and \(v\), we try to solve for \(u\) and \(v\) in terms of \(x\) and \(y\) to find \(F\). Conversely, if we are given equations expressing \(u\) and \(v\) in terms of \(x\) and \(y\), we regard this as \(F\), and if we can solve for \(x\) and \(y\) in terms of \(u\) and \(v\) this gives \(G\). We checked that in the example above \(F(x,y) = (x+y, x-y)\) is the inverse of \(G(u,v) = (\frac{u+v}{2}, \frac{u-v}{2})\). We also noted that if \(D\) is the original domain of integration in the \(xy\)-plane, and one knows the inverse \(F(x,y)\), then \(F(D) = D_0\) is the domain of integration in the \(uv\)-plane.

We followed this with a discussion explaining how the Jacobian comes into the change of variables formula for double integrals. In other words, why do we use \(dA = |\frac{\partial(x,y)}{\partial(u,v)} | dudv \)? To see this, we started with the transformation \(G(u,v) = (x(u,v), y(u,v))\) and the following two facts:

1. (i) If \(A = a\vec{i}+b\vec{j} \) and \(B = c\vec{i}+d\vec{j}\), then the area of the parallelogram spanned by \(A\) and \(B\) is the absolute value of
   \[ det \begin{pmatrix} a & c\\b & d\end{pmatrix} = |ad-bc| \]
   .
2. (ii) For a function \(f(u,v)\) whose partial derivatives exist:
   \[ f(u+\Delta u,v)-f(u,v) \approx \Delta u \frac{\partial f}{\partial u}\ \ and\ \ f(u,v+\Delta v)-f(u,v) \approx \Delta v \frac{\partial f}{\partial v}, \]
   when \(\Delta u\) and \(\Delta v\) are small. This follows, since for example, \(\frac{\partial f}{\partial u} \approx \frac{f(u+\Delta u, v)-f(u,v)}{\Delta u}\).

Now, \(G\) transforms the rectangle with area \(\Delta u \Delta v\) to the curvilinear rectangle shown below:

In the Riemann sums of the double integral in \(x\) and \(y\) over the region \(G(\mathcal{R})\), we may use the parallelogram \(P\) spanned by spanned by the vectors \(\mathbf{A}\) and \(\mathbf{B}\) as small portions of area \(dA\). Note that

Similarly: \(\mathbf{B} \approx \Delta v \frac{\partial x}{\partial v}\ \vec{i}+\Delta v \frac{\partial y}{\partial v}\ \vec{j}.\) Therefore:

Thus, the Riemann sum: \(\sum_i\sum_j f(x_i,y_j)\ dA\) in \(xy\)-coordinates is approximately the Riemann sum

in \(uv\)-coordinates. Passing to the limit as the units of area tend to zero gives the change of variables formula.

We began class by discussing improper double integrals, first recalling familiar cases for single improper integrals. As in the single variable case, we noted improper integrals occur either when the integrand is not defined on boundary points of the domain of integration, is unbounded on the domain of integration, or when the domain of integration is infinite. We illustrated these ideas with the following examples.

We can use polar coordinates:

Taking the limit as \(a \to 1\), we get \(2\pi\). Thus:

In other words, \(\int\int_D \frac{1}{\sqrt{1-x^2-y^2}}\ dA\) converges to \(2\pi\).

Solution: In this case we can proceed as one might expect:

Here's one way to evaluate the iterated integral:

Calculating these integrals separately:

Similarly: \(\int_0^a xe^{-x^2}\ dx = \frac{1}{2}(-e^{-a^2}+1)\). Thus:

Passing to the limit at \(a\to \infty\) and \(b\to \infty\), we get:

We then calculated the improper integral \(\int_{-\infty}^{\infty} e^{-x^2}\ dx\) by calculating its square as a double integral over \(\mathbb{R} ^2\). We noted that the answer then yields \(\frac{1}{\sqrt{\pi}} \int_{-\infty}^{\infty} e^{-x^2}\ dx = 1\), a formula used when discussing normal distributions.

We then began a discussion of triple integrals.

A triple integral is an integral of the form \(\iiint_B f(x,y,z)\ dV\), where \(B\) is a solid region contained in \(\mathbb{R}^3\). The underlying idea for the definition of \(\iiint_B f(x,y,z)\ dV\) is the same as we discussed for double integrals:

First, partition the domain of integration \(B\) into small subregions - in this case solids - of a similar type.

Second, select a point from each small subregion and evaluate the function \(f(x,y,z)\) at that point.

Third, multiply the value obtained in the second step by the size of the subregion the point was chosen from. In this case we are multiplying the function value by a small unit of volume.

Fourth, add up the values from the previous step, thereby getting a Riemann sum.

Fifth, take a limit of the Riemann sums as the maximum volumes of the subregions in the partition go to zero.

Sixth, if the limit exists, we denote it \(\iiint_B f(x,y,z)\ dV\).

As in previous discussions concerning double integrals, \(\iiint_B f(x,y,z)\ dV\) is a quantity that depends upon \(f(x,y,z)\) and the geometry of \(B\) and does not depend upon the coordinate system used to describe \(B\) or used to calculate \(\iiint_B f(x,y,z)\ dV\).

we began class noting that we expect various versions Fubini's Theorem for triple integrals. If \(B\) is a rectangular box, and we use rectangular coordinates, our Riemann sums look something like this, which justifies Fubini's Theorem:

Solution: Applying Fubini's Theorem,

Integrating in a different order we get:

Fundamental Fact: \(\iiint_B \ dV = vol(B).\) This follows, since in the Riemann sums approximating \(\iiint_B \ dV\), we are just adding volumes of small solids covering \(B\), so that each Reimann sum provides a better approximation to the volume of \(B\) and by passing to the limit, we obtain the volume of \(B\).

What about more general domains of integration? Suppose we have a region \(\mathcal{W} \subseteq \mathbb{R}^3\) defined as all points \((x,y,z)\) in \(\mathbb{R}^3\) such that \((x,y) \in D\) and \(z_1(x,y)\leq z\leq z_2(x,y)\), as pictured below.

We have a corresponding version of Fubini's Theorem:

Note that if we calculate the integral in the brackets by integrating with respect to \(z\), and then evaluating \(z\) at \(z_2(x,y)\) and \(z_1(x,y)\) and subtracting, we then have a double integral of a function in \(x\) and \(y\) only over the domain \(D\subseteq \mathbb{R}^2\). In this version of Fubini's Theorem, we can reduce a triple to a double integral. In the case above, we call \(W\) a \(z\)-simple region.

{Solution.} Regarding \(B\) as a \(z\)-simple region, we have that \(0\leq z\leq 1-x-y\) and \(D\) is the triangle \(0\leq x\leq 1\), \(0\leq y\leq 1-x\). Thus:

We then noted the following. Suppose \(B\) is a \(z\)-simple region defined over the domain \(D\subseteq \mathbb{R}^2\). If \(D\) is a region of Type 1, say \(c(x)\leq y\leq d(x)\), \(a\leq x\leq b\), then

while if \(D\) is a region of Type 3 described as: \(a(y)\leq x\leq b(y)\) and \(c\leq y\leq d\), then

We also noted that the one may have \(x\)-simple and \(y\)-simple regions, where in the first case, \(B\) is bounded in the front and back by functions \(x = a(y,z)\) and \(x = b(y,z)\), while in the second case, \(B\) is bounded to the left and right by functions \(y = c(x,z)\) and \(y = d(x,z)\), with the usual orientation of the \(x,y,z\)-axes. In our MW text, the \(z\)-simple, \(x\)-simple, and \(y\)-simple regions are called regions of Type I, II, or III, respectively.

To elaborate, nn \(x\)-simple solid has the form \(x_1(y,z)\leq x\leq x_2(y,z)\), \((y,z)\in D\), with \(D\) in the \(yz\)-plane.

By Fubini's Theorem:

If \(D\) is described as: \(a(z)\leq y\leq b(z)\) and \(c\leq z\leq d\), then

While if \(D\) is described as: \(c(y)\leq z\leq d(y)\) and \(a\leq y\leq b\), then

A \(y\)-simple solid \(E\) has the form \(u_1(x,z)\leq y\leq u_2(x,z)\), \((y,z)\in D\), with \(D\) in the \(xz\)-plane.

By Fubini's Theorem:

which can take the form

We then calculated the following example in two ways, first as a \(z\)-simple region, then as an \(x\)-simple region.

Solution

To calculate Calculate \(\iiint_B x\ dV\) for \(B\) above, viewed as a \(x\)-simple surface, we first notice that if we move inside \(B\) in the direction of the \(x\)-axis, we get \(0\leq x\leq y^2\). The projection of \(B\) onto the \(yz\)-plane is the area in the first quadrant below the line \(z = 1-y\). Thus, we have

We began class by stating the two facts:

1. (i) \(\iiint_B \ dV = volume(B)\).
2. (ii) The average value of \(f(x,y,z)\) over \(B\) is: \(\frac{1}{vol(B)} \iiint_B f(x,y,z)\ dV\).

We verified the volme formula by using a triple integral to show that the volme of a cylinder with radis \(r\) and height \(h\) is \(\pi r^2h\).

We then worked the following example:

Regarding \(E\) as a \(z\)-simple region, with domain of integration \(A\), we have

which is difficult to integrate

On the other hand, regarding \(E\) as a \(y\)-simple region with domain of integration , we tegrate with respect to \(y\) first, to obtain

We ended class by considering the integral \(\iiint_B \sqrt{x^2+y^2+z^2} \ dV\), where \(B\) is the solid sphere of radius one centered at the origin. We noted that this requires a three dimensional version of polar coordinates, and suggested that a substitution using spherical coordinates should be the analogous procedure.Using the substitutions \(x = \rho \cos(\theta)\sin(\phi), y = \rho\sin(\theta)\sin(\phi), z = \rho\cos(\phi), dV = \rho^2\sin(\phi)\), with limits of integration \(0\leq \rho\leq 1, 0\leq \phi\leq \pi, 0\leq \theta\leq 2\pi\), we saw that

This led to the interesting conclusion that the average distance from the origin for points in \(B\) is \(\frac{3}{4}\).

We began class by reviewing the definition of spherical coordinates. One begins by observing that every point in \(\mathbb{R}^3\) lies on a sphere of radius \(\rho\) centered at (0,0,0) and thus can be expressed in terms of spherical coordinates, \(\rho, \phi, \theta\). Here is typical point \(P\) using spherical coordinates

Note: \(P = (\rho,\phi,\theta)\), with \(0\leq \phi\leq \pi\) and \(0\leq \theta\leq 2\pi\). If we set each of the spherical coordinate equal to a constant, we get:

We then showed the relation between the spherical coordinates and the rectangular coordinates of \(P\).

Note:

It follows from these equations that the expression \(x^2+y^2+z^2\), in spherical coordinates, becomes \(\rho^2\). We will use this fact often. Writing spherical coordinates in terms of rectangular coordinates.

1. () \(\rho = \sqrt{x^2+y^2+z^2}\)
2. () \(\tan (\theta) = \frac{y}{x}\), so \(\theta = \tan^{-1}(\frac{y}{x})\).
3. () \(\cos(\phi) = \frac{z}{\rho}\), so \(\phi = \cos^{-1}(\frac{z}{\rho})\).

We then stated

Thus, \(\tan(\phi) = \frac{1}{\sqrt{3}}\), and thus, \(\phi = \frac{\pi}{6}\). Therefore,

which can easily be worked out.

We then briefly discussed why we use \(dV = \rho^2\sin(\phi)\ d\rho d\phi d\theta\) in the conversion to spherical coordinates by looking at the diagram below.

Recall that \(r = \rho \sin(\phi)\). The top face of the spherical cube above has dimensions \((\rho\sin(\phi) \Delta \theta)\times \Delta \rho\). The remaining edge (subtending an angle \(\Delta \phi\)) of the spherical cube is approximately \(\rho \Delta \phi\). The volume of the spherical cube is approximately: Since once can use small spherical wedges like above as a partition of \(B\) in the Riemann sum for \(\iiint_B f(x,y,z)\ dV\), it follows that the Riemann sum in spherical coordinates takes the form

so that taking the limit as the volumes tend to zero gives the formula for \(\iiint_B f(x,y,z)\ dV\) in spherical coordinates.

We then noted that every point in \(\mathbb{R}^3\) lies on the top edge of a cylinder, which enables us to describe points in \(\mathbb{R}^3\) in terms of cylindrical coordinates, noting that cylindrical coordinates are essentially like polar coordinates, though with the extra variable \(z\).

In cylindrical coordinates, \(P = (r,\theta, z)\).

To transform a triple integral into cylindrical coordinates, we set:

We can easily guess what the volume of a cylindrical wedge should be.

It should be approximately the area of the corresponding polar rectangle times \(\Delta z\), the change in the \(z\) direction, i.e., the height of the cylindrical wedge. Thus, in cylindrical coordinates, small units of volume \(dV\) are approximately \((r\Delta r\Delta \theta)\cdot \Delta z = r\ \Delta z\ \Delta r\ \Delta \theta\). Using expressions of this type in the Riemann sum for \(\iiint_B f(x,y,z)\ dV\) in cylindrical coordinates, we get the following version of Fubini's theorem:

We then worked some examples using cylindrical coordinates.

Solution: In cylindrical coordinates \(B: 0\leq r\leq 2, 0\leq \theta\leq 2\pi, 1\leq z\leq 5\). Thus,

To describe \(W\) is cylindrical coordinates: \(0\leq z\leq y\), so \(0\leq z\leq r\sin(\theta).\) Since \(y\geq 0\), the projection of \(W\) onto the \(xy\)-plane is the semi-circle \(D\). Thus, \(0\leq r\leq 2\) and \(0\leq \theta \leq \pi\).

The class worked on practice problems for Exam 2.

The class turned in their Math Diaries and worked on practice problems for Exam 2.

We briefly discussed the change of variables theorem for triple integrals. For this we needed to define the determinant of a \(3\times 3\) matrix. We noted that there are several ways to do this. With this in hand, we stated

Our next goal is to integrate along a curve in \(\mathbb{R}^3\). For this, we began a discussion of vector valued functions.

with \(t\) belonging to a subset of \(\mathbb{R}\).

If the values of \(\mathbf{r} (t)\) lie in the \(xy\)-plane, we write The variable \(t\) is called the parameter. It is often convenient to think of \(t\) as time. The set of points traced out by \(\mathbf{r}(t)\) is a curve. How the curve is traced out is called a path. For example, for the curve \(C\), the circle of radius one, with center \((0,0,1)\)

one can have several different paths:

\(\mathbf{r}_1(t) = (\cos(t), \sin(t), 1)\), with \(0\leq t\leq 2\pi\) traces the curve once.

\(\mathbf{r}_2(t) = (\cos(2t), \sin(2t), 1)\), with \(0\leq t\leq \pi\) traces once, but twice as fast.

\(\mathbf{r}_3(t) = (\cos(t), \sin(t), 1)\), with \(0\leq t\leq 4\pi\), traces the curve twice, but at the same speed at \(\mathbf{r}_1(t)\).

\(\mathbf{r}_4(t) = (\cos(2\pi-t), \sin(2\pi-t), 1)\), traces the curve once, in the opposite direction of \(\mathbf{r} _1(t)\).

The different paths \(\mathbf{r} _i(t)\) above are referred to as different parametrizations of \(C\). We then gave two more examples of curves in \(\mathbb{R}^3\).

We then noted that limits and continuity for vector valued functions are defined in a similar way as for functions we have previously encountered.

Limits. For a vector valued function \(\mathbf{r}(t)\) and a fixed vector \(\vec{u} = u_1i+u_2j+u_3k\), we write:

Note that this means the vectors \(\mathbf{r} (t)\) get closer to the vector \(\vec{u}\) as \(t\) approaches \(t_0\).

This is equivalent to

To explain why this holds, we noted

by continuity of the square root and square functions. If then the limits under the radical are zero. Thus, \(\lim_{t\to t_0} || \mathbf{r}(t)-\vec{u} || = 0\). i.e., \(\lim_{t\to t_0} \mathbf{r} (t) = \vec{u}\).

Continuity. \(\mathbf{r}(t)\) is continuous at \(t_0\) if \(\lim_{t\to t_0} \mathbf{r}(t) = \mathbf{r}(t_0)\). From the discussion of limits, we observed that this is equivalent to:

We then noted that differentiabiility is defined as expected.

This follows, since we may take limits coordinate-wise (and because the scalar \(\frac{1}{h}\) can be moved inside of an ordered triple):

We finished class by noting that the vector \(\mathbf{r}'(t_0)\) is tangent to the curve \(\mathbf{r}(t)\) at the point \(P_0 = \mathbf{r}(t_0)\), as illustrated in the diagram below.

We then noted that we have versions of the familiar rules of differentiation for vector valued functions.

We continued our discussion of curves in \(\mathbb{R}^3\) or \(\mathbb{R}^2\) by considering the length of a path or curve.

The parameter \(s\) is called arc length. We can keep track of the arc length as we move along the path by considering the function:

for \(a\leq t\leq b\). We noted that \(s\) gives the length of the path \(\mathbf{r}(t)\) for the range \(a\leq t\leq b\), and that this is also the length of the corresponding curve, under an additional hypothesis.

(b) The length of the curve traced out by \(\mathbf{r}(t)\) is independent of the parametrization.

We expect the lengths of (i), (i), (iv) to be \(2\pi\) and the length of the path (iii) to be \(4\pi\).

For \(\mathbf{r}_1(t):\) \(|| \mathbf{r}'_1(t)|| = \sqrt{(-\sin(t))^2+(\cos(t))^2+0} = 1\). Thus:

Note that \(\mathbf{r}_1(t)\) is 1-1 on \([0,2\pi]\), so this gives the expected length of the curve.

For \(\mathbf{r}_2(t):\) \(|| \mathbf{r}'_2(t)|| = \sqrt{(-2\sin(2t))^2+(2\cos(2t))^2+0} = 2\). Thus:

Note that \(\mathbf{r}_2(t)\) is also 1-1, so we get that the length of the path equals the length of the curve. Note also that the previous two parametrizations are the different, but yield the same arc length.

For \(\mathbf{r}_3(t):\) \(|| \mathbf{r}'_3(t)|| = \sqrt{(-\sin(t))^2+(\cos(t))^2+0} = 1\). Thus

Note here that the path traces the curve twice, so the length of the path is \(4\pi\), while the length of the curve is \(2\pi\). In this case, \(\mathbf{r}_3(t)\) is NOT 1-1 on the interval \([0,4\pi]\), which explains why the length of the path differs from the length of the curve.

For \(\mathbf{r} _4(t): ||\mathbf{r} _4'(t)|| = \sqrt{(\cos(2\pi-t))^2+(-\sin(2\pi-t))^2+0^2} = 1\). Thus

as expected.

We then noted that if we think of \(t\) as time, then \(s(t) = \int_a^t ||\mathbf{r}'(t) ||\ dt\) gives the distance traveled in time \(t\) along the path described by \(\mathbf{r}(t)\), so that \(s'(t) = ||\mathbf{r}'(t)||\) denotes the speed. Thus, the velocity vector \(\mathbf{r}'(t)\) points in the direction of a point traveling along the curve (since it is tangent to the curve) and the length of the velocity vector gives the speed at time \(t\).

We then gave a heuristic description accounting for the formula for the arc length along a segment of a curve, say \(C\) is given by \(\mathbf{r}(t)\), with \(a\leq t\leq b\). Partition the interval \([a,b]: a = t_1 < t_1\cdots < t_n = b\) so that \(t_{i+1}-t_i = \Delta t\) is small.

Create a polygonal path whose endpoints are the \(\mathbf{r}(t_i)\). The length of each line segment in the path is \( || \mathbf{r}(t_{i+1})-\mathbf{r}(t_i)||\). We have:

For small values of \(\Delta t\):

Thus:

Summing these expressions we get an approximation of the arc length on the one hand, and a Riemann sum for \(\int_a^b||\mathbf{r}'(t)||\ dt\) on the other hand. Passing to the limit as \(\Delta t\to 0\) gives the arc length formula.

We then turned to the definition of \(\int_C f(x,y,z)\ ds\), the line integral of the scalar function \(f(x,y, z)\) along the curve \(C\). We proceed as before, as with all previous types of integration.

Step 1. Subdivide \(C\) into finitely many smaller curves \(C_i\) of the same length \(\Delta s\).

Step 2. Choose a point \((x_i,y_i,z_i)\) from the component \(C_i\).

Step 3. Multiply \(f(x_i,y_,z_i)\) by the size of each \(C_i\) to get \(f(x_i,y_i,z_i)\ \Delta s\).

Step 4. Add the products in Step 3 to get the Riemann sum: \(\Sigma _i f(x_i,y_i,z_i)\Delta s\).

Step 5. Take the limit of the Riemann sums as \(\Delta s \to 0\), to get:

the line (or path) integral over \(f(x,y,z)\) over \(C\). This integral is sometimes called the the line integral of \(f(x,y,z)\) with respect to arc length.

We must use the parametrization \(\mathbf{r}(t)\) of \(C\) to calculate \(\int_C f(x,y,z)\ ds\). From our discussion of arc length, we have that small portions of length \(\Delta s\) along the curve are approximated by \(||\mathbf{r}'(t_i)|| \Delta t\), where \((x_i,y_i,z_i) = \mathbf{r}(t_i)\). Now use

in the Riemann sum (Step 4) above. Taking the limit as \(t\to 0\) we get:

A word of caution: Depending upon the context, one will either calculate a path integral, if the parametrization is given or one may have to choose a parametrization to calculate a line integral. Moreover, if the path \(\mathbf{r}(t)\) doubles back on itself, or repeats portions of the curve with non-zero length, this will be reflected in \(\int_a^bf(\mathbf{r}(t))\ ||\mathbf{r}'(t)||\ dt\). Note that:

is the arc length of \(C\), which is what we expect.

We then worked the following example

Solution. \(\mathbf{r}'(t) = (-s\in )t_, \cos (t), 1)\), so \(||\mathbf{r}'(t)|| = \sqrt{2}\).

Note that \(\mathbf{r}'(t) = (\sin (t), \cos (t), 1)\), so that \(||\mathbf{r}'(t)|| = \sqrt{2}\)

%We then discussed the following properties of line integrals.

We began class by discussing the following properties of line integrals.

(iv) \(\int_C f(x,y,z)\ ds\) is independent of the parametrization, if the parametrization is 1-1 with continuous derivative.

The parametrizations for these curves, respectively, are:

It is easy to check that \(||\mathbf{r}_i(t)|| = 8\), for all \(i\). Thus,

Therefore, so that For the \(\mathrm{parametrization}\) \(\mathbf{s}(t)\) we have so that, Using \(u\)-substitution with \(u = \cos(\pi-2t)\), we have

which agrees with the calculation above.

We then stated the following applications of a line integral.

Applications.

1. (i) \(\frac{1}{length(C)} \int_c f(x,y,z)\ ds\) give the average value of \(f(x,y,z)\) over \(C\).
2. (ii) If \(C\subseteq \mathbb{R}^2\) and \(f(x,y) \geq 0\), for \((x,y)\in C\), then \(\int_C f(x,y)\ ds\) represents the area under the surface \(z = f(x,y)\) above the curve \(C\).
3. (iii) If \(C\) represents a wire and \(f(x,y,z)\) is the density of the wire at the point \((x,y,z)\), then the total mass of the wire is \(M = \int_C f(x,y,z)\ ds\).
4. (iv) The center of mass of the wire is the point \((\overline{x}, \overline{y}, \overline{z})\), where
   \[ \overline{x} = \frac{\int_C xf(x,y,z)\ ds}{M},\quad \overline{y} = \frac{\int_C y f(x,y,z)\ ds}{M},\quad \overline{z} = \frac{\int_C zf(x,y,z)\ ds}{M}. \]

We ended our initial discussion of line integrals by showing that if we are given two 1-1 parameterizations of the curve \(C\), \(\mathbf{r}(t)\), \(a\leq t\leq b\) and \(\mathbf{s}(t)\), \(c\leq t\leq d\), with continuous derivatives, then

This was accomplished by viewing \(\mathbf{s}(t)\) as a re-parameterization of \(\mathbf{r}(t)\), i.e., by writing \(\mathbf{s}(t) = \mathbf{r}(\theta(t))\), for \(\theta :[c, d]\to [a, b]\), where \(\theta\) is 1-1, with a continuous derivative, and \(\theta(c) = a\) and \(\theta (d) = b\).

We then began our discussion of surface integrals whose integrands are scalar functions. We follow the same process used in all previous forms of integration. To integrate \(f(x,y,z)\) over the surface \(S\)

Step 1. Subdivide \(S\) into finitely many smaller surfaces \(S_i\) of the same area \(\Delta S\). We are using \(\Delta S\) for a small element of surface area.

Step 2. Choose a point \((x_i,y_i,z_i)\) from the component \(S_i\).

Step 3. Multiply \(f(x_i,y_i,z_i)\) by the size of each \(S_i\) to get \(f(x_i,y_i,z_i)\ \Delta S\).

Step 4. Add the products in Step 3 to get the Riemann sum: \(\Sigma _i f(x_i,y_i,z_i)\Delta S\).

Step 5. Take the limit of the Riemann sums as \(\Delta S \to 0\), to get:

the surface integral of \(f(x,y,)\) over \(S\) with respect to surface area. Note that we write a double integral, since our domain of integration is two-dimensional.

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Following an analogy with curves, to calculate \(\int\int_S f(x,y,z)\ dS\), we will need:

1. (i) A way to describe or parametrize a surface as a function of two variables.
2. (ii) A way to calculate surface area.

such that \(S = G(D)\) for some domain \(D\) in the \(uv\)-plane.

As usual, we assume that all first order partials exists and are continuous, at least on the interior of \(D\).

Here we can write \(G(u,v) = (u,v, f(u,v))\) or \(G(x,y) = (x,y, f(x,y))\). For example, if \(S\) is that portion of the paraboloid \(z = x^2+y^2\) lying over \(D: 0\leq x^2+y^2 \leq 9\), then \(G(u,v) = (u,v,u^2+v^2)\), with \(0\leq u^2+v^2\leq 9\) is a parametrization that takes the disk of radius 3 in the \(uv\)-plane to \(S\).

Though the top half of the cone can be expressed as \(z = \sqrt{x^2+y^2}\) and we could parametrize it by \(G(u,v) = (u,v,\sqrt{u^2+v^2} )\) with \(0\leq x^2+y^2\leq 4\), the whole surface cannot be expressed as the graph of a function of \(x\) and \(y\).

Better: \(G(u,v) = (u\cos(v), u\sin(v) u)\), with \(-2\leq u\leq 2\) and \(0\leq v\leq 2\pi\). Note

so the points \(G(u,v)\) lie on the cone. More over, if we hold \(u\) fixed at \(u_0\) and let \(v\) vary between \(0\) and \(2\pi\), this vertical line segment in the \(uv\)-plane is taken by \(G\) to the circle \(G(u_0,v) = (u_0\cos(v), u_0\sin(v), u_0)\), i.e., the circle of radius \(u_0\) centered at \((0,0, u_0)\). As \(u_0\) varies between 2 and -2, these circles sweep out the cone.

Spherical and cylindrical coordinates tell us how to parametrize spheres and cylinders.

with \(0\leq \phi \leq \pi, 0\leq \theta < 2\pi\).

As we have seen in our earlier discussion concerning spherical coordinates, this transformation take the vertical line segment \(\phi = \phi_0\), with \(0\leq \theta\leq 2\pi\) in the \(\phi\theta\)-plane to the circle \((R\sin(\phi_0)\cos(\theta), R\sin(\phi_0)\cos(\theta), R\cos(\phi_0))\) of radius \(R\sin(\phi_0)\) centered at \(0,0, R\cos(\phi_0)\). As \(\phi_0\) varies from \(0\) to \(\pi\) these circles sweep our the sphere.

Now that we have a description of a surface in terms of parameters \(u,v\), we can use this parameterization to find the plane tangent to the surface at \(P = (x_0,y_0, z_0) = G(u_0,v_0)\). Since the equation of any plane is determined by a point on the plane and a vector normal to the plane, to find the tangent plane to the surface \(S\) at a point \(P\), we find two tangent vectors, and take their cross product to find a normal vector.

Suppose \(P = G(u_0,v_0).\) How do we get tangent vectors to \(S\) at \(P\)? Hold \(v\) fixed at \(v_0\) and let \(u\) vary. Then \(G(u,v_0)\) gives a curve \(C_1\) that moves in the \(u\)-direction on \(S\) passing through \(P\).

Thus, \(\mathbf{T}_u(u_0,v_0) = \frac{\partial G}{\partial u} (u_0,v_0)\) is a vector tangent to \(C_1\), and hence \(S\), at \(P\).

We get a second tangent vector to \(S\) by taking a tangent to the curve obtained by holding \(u\) fixed at \(u_0\).

The normal vector to \(S\) at \(P\) is

Thus, the equation of the plane tangent to \(S\) at the point \(P = G(u_0,v_0) = (x_0,y_0,z_0)\) is given by

After a brief reminder about parameterized surfaces, we considered

where the double integral is a standard double integral in the \(uv\)-plane.

Solution. We use

with \(0\leq \phi \leq \pi, 0\leq \theta < 2\pi\), which defines \(D\).

Thus,

We have:

since \((\sin(\phi)\cos(\theta),\sin(\phi)\sin(\theta),\cos(\phi))\) lies on the sphere of radius 1. Note that \(R^2\sin(\phi)\) is non-negative since \(0\leq \phi \leq \pi\). Thus:

Question: Do you see a connection between the formula for the volume of a sphere of radius \(R\) and its surface area?

Answer: The surface area is the derivative of the volume.

In order to understand our formulas for calculating surface area and surface integrals, we need to understand, where does the formula for surface area come from? For this we will use the fact from Calculus 2 that if \(\mathbf{v}_1 = ai+bj+ck\) and \(\mathbf{v}_2 = di+cj+ek\), then the area of the parallelogram spanned by \(\mathbf{v}_1\) and \(\mathbf{v}_2\) is \(||\mathbf{v}_1\times \mathbf{v}_2||\).

We first subdivide \(S\) into small portions, \(S_i\) with surface area \(\Delta S\). We approximate the small portions of surface area \(\Delta S\) with small approximating tangent parallelograms. We start with the parametrization \(G(u,v)\) of \(S\). We use a tangent parallelogram to estimate the surface of the curved parallelogram on the surface.

Note that the vector from \(P\) to \(Q\) is \(G(u_0+\Delta u,v_0)-G(u_0,v_0)\). This vector is approximated by the tangent vector in red, \(\Delta u\cdot \mathbf{T}_u (u_0,v_0)\). Similarly, the other tangent vector in red \(\Delta v\cdot \mathbf{T}_v(u_0,v_0)\) approximates the vector \(G(u_0,v_0+\Delta v)-G(u_0,v_0)\). Thus, our small approximating tangent parallelogram is spanned by the vectors \(\Delta u\cdot \mathbf{T}_u(u_0,v_0)\) and \(\Delta v\cdot \mathbf{T}_v(u_0,v_0)\). The area of the approximating rectangle is

Adding these areas for each component of the subdivision and taking the limit as \(\Delta u, \Delta v \to 0\), gives a total surface area of \(\int\int_D ||\mathbf{T}_u\times \mathbf{T}_v||\ dA\).

By retracing our steps in defining a surface integral, using a parameterization leads to a formula for calculating \(\int\int_S f(x,y,z)\ dS\).

Step 1. Subdivide \(S\) into finitely many smaller surfaces \(S_i\) of area \(\Delta S \approx ||\mathbf{T}_u\times \mathbf{T}_v||\ dA\).

Step 2. Choose a point \(G(u_i,v_i)\) from the component \(S_i\).

Step 3. Multiply \(f(G(u_i,v_i))\) the area of \(S_i\) to get

Step 4. Add the products in Step 3 to get the Riemann sum :

Step 5. Take the limit of the Riemann sums as \(\Delta S \to 0\), to get:

We then calculated the following examples.

Solution. We take \(G(u,v) = (u,v,u^2+v^2)\), with \(0\leq u^2+v^2\leq 4\). We need to calculate \(||\mathbf{T}_u\times \mathbf{T}_v||\).

Thus \(||\mathbf{T}_u\times \mathbf{T}_v|| = \sqrt{(-2u)^2+(-2v)^2+1^2} = \sqrt{4u^2+4v^2+1}\). Using this in the formula for surface integrals, we have

Solution. From last lecture, \(G(u,v) = (u\cos(v), u\sin(v), u)\), with \(0\leq u\leq 2\), \(0\leq v\leq 2\pi\), is a parametrization of \(S\).

\(\mathbf{T}_u = (\cos(v),\sin(v),1)\) and \(\mathbf{T}_v = (-u\sin(v),u\cos(v),0)\).

\(||\mathbf{T}_u\times \mathbf{T}_v|| = \sqrt{(-u\cos(v))^2+(-u\cos(v))^2+u^2} = \sqrt{2u^2} = \sqrt{2}|u|\).

since \(u\) is non-negative on \(D\).

We then recorded the following familiar looking properties for surface integrals of continuous functions defined on a surface \(S\):

1. (i) \(\int\int_S f+g\ dS = \int\int_S f\ dS + \int\int_S g\ dS\).
2. (ii) \(\int\int_S \lambda f\ dS = \lambda \int\int_S f\ dS\), for \(\lambda \in \mathbb{R}\).
3. (iii) If \(S = S_1\cup S_2\), then \(\int\int_S f\ dS = \int\int_{S_1} f] dS + \int\int_{S_2} f\ dS\), as long as \(S_1\) and \(S_2\) only intersect along their boundaries.
4. (iv) \(Surface area(S) = \int\int_S\ dS\).

Important comments about calculating surface area and surface integrals.

1. The quantity \(||\mathbf{T}_u\times \mathbf{T}_v||\) should always be non-negative (and almost never zero). If you you calculate this expression and it involves variables from the parametrization, this function should be positive on the domain of integration. For example, if you get \(||\mathbf{T}_u\times \mathbf{T}_v|| = 2uv\), then you have to make sure that the product \(uv\) is positive on the domain of integration. If not, you may have dropped an absolute value somewhere.

2. Aside from having component functions with continuous first order partial derivatives, \(G(u,v)\) should be 1-1 on the interior of the domain \(D\). Otherwise, one may be double counting portions of the surface area.

3. For calculating surface area and surface integrals of scalar functions, the orientation of the normal vector does not matter, because these quantities use the length of the normal vector. However, it will matter when we integrate vector valued functions over a surface. After all, we care whether or not a fluid flows into or out of a chamber. Or as Hagrid might say: Better out than in.

We began a discussion of integrating vector fields over curves and surfaces.

We will make our usual assumptions that the first order partial derivatives of all component functions exist and are continuous in suitable regions contained in \(\mathbb{R}^3\). Some other pictures of vector fields:

Tangent vector fields and normals vector fields play a central role in what we want to do next.

Case 1. Let \(C\) be a smooth curve. Then at each point \(P = (x,y,z) \) of the curve, we can assign a tangent vector \(\mathbf{F} (x,y,z)\) to \(C\) at the point \(P\). This gives a vector field along the curve \(C\). Note that if \(\mathbf{r}(t)\) is a parametrization of \(C\), then \(\mathbf{r}'(t)\) is a tangent vector at each point along the curve. Recall that \(\mathbf{r}'(t)\) points in the direction of a point traveling along the curve and the length of \(\mathbf{r}'(t)\) gives the speed of the point at time \(t\).

If we want to just keep track of the direction of a point moving along the curve, we can consider the unit tangent vector \(\mathbf{T}(x,y,z)\) along the curve. If \(\mathbf{r}(t)\) is a parametrization of \(C\) and \(\mathbf{r}'(t) \not = 0\), then

How do we use the unit tangent vectors? If we have a vector field \(\mathbf{F}\) pushing a point along a curve, the dot product \(\mathbf{F}(P)\cdot \mathbf{T}(P)\) gives us the component of \(\mathbf{F}\) in the direction of the curve since:

where \(\theta\) is the angle between \(\mathbf{F}(P)\) and \(\mathbf{T}(P)\).

Informally, we can think of \(\mathbf{F}\cdot \mathbf{T}\) as “how much of \(\mathbf{F}\) acts in the direction of \(\mathbf{T}\)”. We will see that when we want to integrate a vector field \(\mathbf{F}\) along the curve \(C\), we will really be integrating the scalar function \(\mathbf{F}\cdot \mathbf{T}\) along the \(C\).

Case 2. If we have a smooth surface \(S\), then the normal vector \(\mathbf{N}\) at each point of \(S\) defines a vector field on \(S\).

Let \(S\) be a smooth surface in \(\mathbb{R}^3\). We obtain a vector field \(\mathbf{F}(x,y,z)\) by assigning to each point \(P\) on the surface a vector that is normal to \(S\) at \(P\). At each point there are two normal vectors, each the negative of the other. The parametrization of the surface may not always yield the desired vector. If \(G(u,v)\) is a parametrization of \(S\), then \(\mathbf{N}(u,v) = \mathbf{T}_u\times \mathbf{T}_v\) gives a normal vector and

gives a unit normal to the surface. If we have a vector field \(\mathbf{F}\) passing through a surface - think of a fluid passing through a membrane - then \(\mathbf{F}(P)\cdot \mathbf{n}(P)\) gives the component of \(\mathbf{F}(P)\) in the direction of \(\mathbf{n}(P)\). In fluid mechanics, one would call this the flux of the fluid across the boundary \(S\) at the point \(P\).

We can now define line and surface integrals of vector fields.

where the quantity on the right is the line integral along \(C\) of the scalar function \(\mathbf{F}\cdot \mathbf{T}\). In terms of calculating the line integral, given the parametrization \(\mathbf{r}(t)\) we have

since \(\frac{1}{||\mathbf{r}'(t)||}\) cancels with \(||\mathbf{r}'(t)||\). Thus, we do not have to calculate \(|| \mathbf{r}'(t)||\) even though it is implicit in the definition of \(\int_C\mathbf{F}\cdot d\mathbf{r}\). Note that to calculate \(\mathbf{F}(\mathbf{r}(t)) = \mathbf{F}(x(t), y(t), z(t))\), everywhere \(x, y, \) or \(z\) appear in the formula for \(\mathbf{F}\) we replace these by \(x(t), y(t), z(t)\) given in the parametrization of \(C\).

Solution. \(\mathbf{F}(\mathbf{r}(t)) = t^2i + (e^t)^2j+ (t+1)k\) and \(\mathbf{r}'(t) = i + e^tj + 2tk\). Thus,

Therefore,

In other words, by definition,

We calculate \(\int\int_S\mathbf{F}\cdot d\mathbf{S}\) using a parametrization. If

is a parametrization of \(S\), the surface integral of \(\mathbf{F}\) over \(S\) is given by

Thus, in a similar vein to the line integral above, in calculating \(\iint_S\mathbf{F}\cdot d\mathbf{S}\), we do not need to calculate \(||\mathbf{T}_u\times\mathbf{T}_v||\), even though this quantity is implicit in the definition of \(\iint_S\mathbf{F}\cdot d\mathbf{S}\). Note also, that ultimately, the calculation of \(\iint_S\mathbf{F}\cdot d\mathbf{S}\) reduces to the calculation of a double integral of a function of \(u\) and \(v\) over the flat region \(D\) in the \(uv\)-plane. Moreover, \(\mathbf{F} (G(u,v))\) is obtained by replacing each occurrence of \(x,y,z\) in the definition of \(\mathbf{F}\) by \(x(u,v), y(u,v), z(u,v)\) respectively.

Solution. We have

with \(0\leq \phi \leq \frac{\pi}{2}, 0\leq \theta < 2\pi\), which defines \(D\).

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Thus,

On the other hand,

Thus, Therefore,

We began class by recalling and discussing the formulas presented in the last lecture for integrating vectors fields along a curve or surface.

We then looked at the following interesting example.

For \(C_2\) we have

We noted that this example illustrated the concept of path independence enjoyed by certain vector fields \(\mathbf{F}\), the so-called conservative vector fields. We noted that in this case \(\mathbf{F} = \nabla f\), for \(f(x,y) = 3xy+x\), and recorded, but did not discuss at length, the formula \({\int_C \nabla f}\cdot d\mathbf{r} = f(Q)-f(P)\), for the smooth curve \(C\) starting at the point \(P\) and ending at the point \(Q\).

We then began a discussion of the concept of the flux of vector field.

If we imagine that the vector field \(\mathbf{F}\) represents a fluid passing through a membrane (e.g., blood through capillaries or an electromagnetic field passing though a shield) then, by definition, the flux is the quantity that passes through the membrane or shield. We want to see that the flux can be calculated as a surface integral.

Case 1. First consider a constant vector field \mathbf{F} passing through a rectangle \(R\) perpendicular to the vectors in \mathbf{F}. How much of \(\mathbf{F}\) passes through \(R\)? Answer:

Note when \(\mathbf{F}\) is parallel to \(R\), the flux is zero. In this case \(\mathbf{F}\cdot \mathbf{n} = 0 = \) flux.

Case 2. \(\mathbf{F}\) is constant, passing through \(R\), neither perpendicular or parallel to \(R\).

The amount of \(\mathbf{F}\) flowing though \(R\) is \((\mathbf{F}\cdot \mathbf{n} )\cdot area(R)\).

General case. \mathbf{F} is not constant, and the surface \(S\) is not a rectangle. Subdivide \(S\) into very small sections \(S_i\) of size \(dS\). For \(P_i\in S_i\), \mathbf{F} on \(S_i\) is approximately \(\mathbf{F}(P_i)\). The flux across each \(S_i\) is approximately: \(\mathbf{F}(P_i)\cdot \mathbf{n}(P_i)\ dS\). Adding over the surface gives a Riemann sum:

Taking the limit as \(dS\to 0\) gives \(\int\int_S \mathbf{F}\cdot \mathbf{n}\ dS = \int\int_S \mathbf{F}\cdot d\mathbf{S}\).

Therefore, we get the Flux of \(\mathbf{F}\) passing through \(S\) equals \(\int\int_S \mathbf{F}\cdot \mathbf{n}\ dS\).

We then did two examples of calculating flux.

We began a discussion of the divergence of a vector field and the Divergence Theorem.

Suppose we wish to interpret how the flow of a vector field \mathbf{F} behaves near a point P. For example, a point source of a radiating electromagnetic field or a sink attracting a fluid in motion. Ideally, this should tell us whether or not, on the whole, the field \mathbf{F} is flowing towards or away from the point \(P\), and with what magnitude. A first attempt might be to consider a small closed sphere \(S_{\epsilon}\) of radius \(\epsilon\) about \(P\), and then take the limit of the flux of \(\mathbf{F}\) through \(S_{\epsilon}\) as \(\epsilon \to 0\),

The problem with this is that since the surface area of \(S_{\epsilon}\) tends to zero as \(\epsilon \to 0\), this limit will always be zero. If we think of \(S_{\epsilon}\) as being full of a liquid, and we empty this liquid as \(\mathbf{F}\) moves from \(P\), this all passes through the boundary of \(S_{\epsilon}\) as \(\epsilon \to 0\). In other words, what we want is the flux per unit volume of \(\mathbf{F}\) over \(S_{\epsilon}\), as \(\epsilon \to 0\), which would be

This limit is called the Divergence of \mathbf{F} at the point \(P\). Letting \(P\) vary, we have a new scalar function, called the divergence of \mathbf{F}, which we denote by \(\mathrm{div}\,\mathbf{F}\).

Important facts about the Divergence of \mathbf{F}.

1. (i) The divergence does not depend upon the coordinate system used.
2. (ii) The divergence of \mathbf{F} at \(P\) can be calculated as
   \[ \lim_{vol(S)\to 0} \frac{1}{vol(S)} \iint_S\mathbf{F}\cdot d\mathbf{S} \]

   where the limit is taken over any sequence of (sufficiently smooth) closed surfaces containing \(P\), whose volumes tend to 0.

3. (iii) \(\mathrm{div}\,\mathbf{F}\) is a scalar function of \(x,y,z\).
4. (iv) \(div(\mathbf{F} +\mathbf{G}) = \mathrm{div}\,\mathbf{F} + \mathrm{div}\,\mathbf{G}\).
5. (v) \(\lambda\cdot \mathrm{div}\,\mathbf{F} = div (\lambda \mathbf{F})\), for any \(\lambda \in \mathbb{R}\).

Solution. Let \(C_s\) denote the cube centered at \(P\) having sides of length \(s\). Then,

We first evaluate the surface integral of \(\mathbf{F}\) over the front and back faces of \(C_s\). For the front face \(S_1\), we note that \(\mathbf{n} = \vec{\imath}\) and that the \(x\) coordinate of every point on the front fact is \(x_0+\frac{s}{2}\). Since \(\mathbf{F} \cdot \mathbf{n} = x^3\), it follows that on the from fact, \(\mathbf{F}\cdot \mathbf{n} = (x_0+\frac{s}{2})^3\). Noting that \(x_0\) and \(s\) are constants (relative to \(S_1\)), it follows that The calculation for \(S_2\) is similar. In this case, \(\mathbf{n} = -\vec{\imath}\), \(\mathbf{F}\cdot \mathbf{n} = -x^3\) and \(-x^3\) evaluate on \(S_2\) is the constant \(-(x_0-\frac{s}{2})^3\). Thus, Therefore,

We proceedED in similar fashion for the next two faces, the right and left faces of \(C_s\). Suppose \(S_3\) is the right face of the cube \(C_s\). Then \(\mathbf{n} = \vec{\jmath}\), \(\mathbf{F}\cdot \mathbf{n} = 4y\), and \(4y\) evaluated on \(S_3\) equals \(4(y_0 +\frac{s}{2})\), a constant on \(S_3\). Thus,

For the left face \(S_4\), \(\mathbf{n} = -\vec{\jmath}\), \(\mathbf{F}\cdot \mathbf{n} = -4y\) and \(-4y\) evaluated on \(S_3\) equals \(-4(y_0 -\frac{s}{2})\), a constant on \(S_3\). Thus, Thus, Finally, for the top face \(S_5\), \(\mathbf{n} = \vec{k}\), \(\mathbf{F}\cdot \mathbf{n} = 2z^2\) and \(2z^2\) on \(S_5\) equals \(2(z_0+\frac{s}{2})^2\). Thus, as before, Similarly, if \(S_6\) denotes the bottom face of \(C_s\), \(\mathbf{n} = -\vec{k}\), \(\mathbf{F}\cdot \mathbf{n} = 2z^2\), which equals \(2(z_0-\frac{s}{2})^2\) on \(S_6\). Thus, Therefore, Adding these six integrals, we now have Therefore,

We noted that one might guess from the preceding example the following formula for the divergence in rectangular coordinate:

In cylindrical coordinates: \(\mathbf{F} = (F_r,F_{\theta},F_z)\) (not \(F_ri+F_{\theta}j+F_zk\)), and

In spherical coordinates: \(\mathbf{F} = (F_r{\rho} F_{\phi}, F_{\theta})\), and

We then stated one of the main theorems of our course:

The Divergence Theorem. Let \(S\) be a closed (piece-wise smooth) surface that bounds the solid \(B\) in \(\mathbb{R}^3\). If the first order partial derivatives of the component functions of \(\mathbf{F}\) are continuous on \(B\), then

We noted that the integral on the right is a standard triple integral of a scalar function over a solid in \(\mathbb{R}^3\), while the integral on the left is a surface integral of the the vector field \({\mathbf{F}(x,y,z)}\) with respect to the outward pointing normal vectors. We also noted that the theorem makes intuitive sense using our coordinate free definition of the divergence of a vector field. Indeed, on the one hand, the left hand side of the equation above is the flux of \(\mathbf{F}\) across the boundary of \(B\), so that if we think of \(S\) as a chamber, full of fluid, this is the amount of fluid leaving the chamber. On the other hand, the divergence of \(\mathbf{F}\) at each point \(P\in B\) measures how much fluid flows from \(P\). Summing this over all points in \(B\) should give the right hand side of the equation in the theorem.

We then showed how to use the Divergence theorem.

Solution. We can set up the surface integral by regarding it as \(\int\int_S\mathbf{F}\cdot \mathbf{n}\ dS\). For a point \(P = (x,y,z)\) on \(S\), the unit normal at \(P\) is \(\frac{1}{R}(x,y,z)\). Thus \(\mathbf{F}\cdot \mathbf{n}\) on \(S\) equals

Therefore,

Which is doable, but not pleasant. Better: Use the Divergence Theorem. \(\mathrm{div}\,\mathbf{F} = z^2++x^2+y^2\). For \(B\) the solid sphere,

which can easily be calculated using spherical coordinates.

Therefore, \(\iint_S\mathbf{F}\cdot d\mathbf{S} = \frac{4}{5}\pi R^5\).

We then had a heuristic discussion about why the Divergence Theorem works? We want to compare \(\iiint_B \mathrm{div}\,\mathbf{F}\ dV\) and \(\iint_S\mathbf{F}\cdot d\mathbf{S}\).

Step 1. Working in rectangular coordinates, we subdivide \(B\) into small solids solids \(B_i\) that are approximately cubes, each with volume \(\Delta V\).

Step 2. We take a point \(P_i\) in each \(B_i\).

Step 3. \(\mathrm{div}\,\mathbf{F} (P_i)\approx \frac{1}{vol(B_i)} \cdot \int\int_{S_i} \mathbf{F}\cdot \mathbf{n}\ dS\), where \(S_i\) is the boundary of \(B_i\).

Step 4. \(\mathrm{div}\,\mathbf{F} (P_i) \cdot vol(B_i)\approx \int\int_{S_i} \mathbf{F}\cdot \mathbf{n}\ dS.\)

Summing the left hand side we get:

a Reimann sum. Passing to the limit as \(\Delta V \to 0\) we get:

Something interesting happens when we sum the terms belonging to the right hand side of Step 4.

When we integrate \(\mathbf{F}\) over the boundaries of adjacent \(V_i\), the surface integrals over common faces cancel.

Why: We integrate twice over the common face, once for \(S_1\) and again for \(S_2\), but once with a normal vector \(\mathbf{n}\) and again with \(-\mathbf{n}\). For the common face \(S_0\):

\(\int\int_{S_0}\mathbf{F}\cdot \mathbf{n_1}\ dS = \int\int_{S_0}\mathbf{F}\cdot -\mathbf{n_2}\ dS\). Thus when we add

the components over \(S_0\) cancel and the sum becomes \(\int\int_{\tilde{S}} \mathbf{F}\cdot \mathbf{n}\ dS\), where \(\tilde{S}\) is the combined outer shell of \(S_1\) and \(S_2\).

Thus, the sum of the terms in the RHS in Step 4 approximates \(\iint_S\mathbf{F}\cdot d\mathbf{S}\), and equals it in the limit.

We ended class by mentioning that Gauss's Law is a physics version of the Divergence Theorem We did not provide the details below in class. .

Gauss' Law (Physics Version). The net electric flux through any hypothetical closed surface is equal to \(\frac{1}{\epsilon}_0\) times the net electric charge \(q\) within that closed surface, where \(\epsilon_0\) is the electric constant.

Gauss' Law (Math Version). Let \(M\) be a solid in \(\mathbb{R}^3\) with a smooth boundary \(\partial M\). Assume (0,0,0) is not on the boundary \(\partial M\). Set \(\mathbf{r} = xi+yj+zk\), and \(r = ||\mathbf{r}|| = \sqrt{x^2+y^2+z^2}\). Then:

In the second case, the vector field \(\frac{1}{r^3}\mathbf{r}\) is defined throughout \(M\), so that if we show its divergence is zero, then by the Divergence Theorem,

The \(i\) component of \(\frac{1}{r^3}\mathbf{r}\) is \(\frac{x}{(x^2+y^2+z^2)^{\frac{3}{2}}}\). Differentiating with respect to \(x\) we get:

Similarly, the corresponding partials of the \(j\) and \(k\) components are:

Adding these three terms shows \(div (\frac{1}{r^3} \mathbf{r}) = 0\).

Now, suppose \((0,0,0) \in M\). Let \(N\) be a sphere of radius \(\epsilon\) centered at (0,0,0) contained in \(M\). Let \(W\) be the complement of \(N\) in \(M\). The boundary of \(W\) is \(\partial N \cup \partial M\), i.e., the union of the boundaries of \(M\) and \(N\).

Since \(div \frac{1}{r^3}\mathbf{r} = 0\) over \(W\), \(\int\int_{\partial N\cup \partial M} \frac{1}{r^3}\mathbf{r}\cdot \mathbf{n}\ dS = 0\). Therefore,

Note that the outward normal for \(\partial N\) as part of the boundary of \(B\) is the inward normal for \(N\). Thus, where now \(\mathbf{n}\) is the outward normal for the sphere \(N\).

On the sphere \(\partial N\), the sphere of radius \(\epsilon\) centered at the origin, \(\mathbf{n} = \frac{1}{\epsilon} \mathbf{r}\) and \(r = \epsilon\). Then:

Thus:

Comments not stated in class. In rectangular coordinates, we often write \(\nabla\cdot \mathbf{F}\) for \(\mathrm{div}\,\mathbf{F}\), since we think of \(\nabla\) as \(\frac{\partial }{\partial x}\ \vec{\imath} + \frac{\partial }{\partial y}\ \vec{\jmath}+\frac{\partial }{\partial z}\ \vec{k}\), which is a differential operator. Thus,

The use of \(\nabla\) in this way is consistent with our prior use, if we recall that the gradient of a scalar function \(f(x,y,z)\) is:

If we use the symbol \(\partial B\) to denote the boundary of the solid in the Divergence Theorem, and the notation \(\nabla\cdot \mathbf{F}\) for \(\mathrm{div}\,\mathbf{F}\), then the Divergence Theorem becomes:

Note how the differential operator on the domain of integration on the left hand side of the equation, moves up to the integrand on the right hand side of the equation.

We began class with the discussion of Gauss's Law presented in the Daily Update from Monday, November 17. In particular, we worked through the calculation of the math version of Gaiss's law. We then turned to a dsicssuion of the curl of a vector field in \(\mathbb{R}^2\).

We discussed the curl of a vector field in \(\mathbb{R}^2\) and Green's Theorem.

We start with a vector field \(\mathbf{F}= F_1(x,y)\vec{\imath}+F_2(x,y)\vec{\jmath}\) in \(\mathbb{R}^2\). If \(C\) is a closed path, the circulation of F along \(C\) is the line integral

where we assume \(C\) is oriented in a counter-clockwise direction.

Throughout we assume that the first order partial derivatives of the component functions of \(\mathbf{F}\) exist and are continuous. Later we will need to assume that all second order partial derivatives of the components functions exist and are continuous.

where \(D\) is a region in \(\mathbb{R}^2\) whose boundary \(\partial D\) is a simple closed curve. A simple closed curve is a (piecewise) smooth curve with no self intersections. We have the following properties of the curl of \(\mathbf{F}\):

1. (i) Curl\ \mathbf{F} is a scalar function.
2. (ii) We can calculate (Curl \mathbf{F})(\(P\)) using shrinking regions of our choice.
3. (iii) The value of (Curl \mathbf{F})(\(P\)) does not depend upon the regions we choose.
4. (iv) (Curl \mathbf{F} )(\(P\)) is also independent of the coordinate system.

How do we calculate the curl using rectangular coordinates? Answer:

Checking the formula with Example 1: \(\mathbf{F} = -y\vec{\imath}+xj\). \(\frac{\partial F_2}{\partial x} = \frac{\partial x}{\partial x} = 1.\) \(\frac{\partial F_1}{\partial y} = \frac{\partial -y}{\partial y} = -1.\)

After recalling the limit definition of \(Curl\ \mathbf{F}\), we can an informal explanation of why in rectangular coordinates one has

We start with a \((\Delta x)\times (\Delta y)\) square centered at \((x_0,y_0)\). We travel counter-clockwise around the square.

Along the base \(C_B\), the unit tangent is \(\vec{\imath}\), so that \(\mathbf{F}\cdot \mathbf{t}\) is \(F_1\) along \(C_B\). Since \(y = y_0-\frac{\Delta y}{2}\) along the base, \(\mathbf{F}\cdot \mathbf{t} = F_1 (x, y_0-\frac{\Delta y}{2})\). Moreover, if \(\Delta x\) is small, and \(x_0\) is the center of \(C_B\), then the average value of \(F_1\) along \(C_B\) is approximately \(F_1(x_0, y_0-\frac{\Delta y}{2})\). Thus,

Along the top \(C_T\), the unit tangent is \(-\vec{\imath}\), so that \(\mathbf{F}\cdot \mathbf{t}\) is -\(F_1\) along \(C_T\). Thus, similarly, we have

Adding these terms we have

Dividing by the area \(\Delta x\Delta y\), we get:

Taking the limit as \(\Delta x, \Delta y \to 0\) gives

A similar analysis yields

Adding we get

We then discussed at length another major theorem from vector calculus.

This was followed by:

When then gave a heuristic proof of Green's Theorem. We start with \(\int\int_D \text{Curl}\ \mathbf{F}\ dA\).

Step 1. Subdivide \(D\) into small rectangular regions \(D_i\), and select \(P_i\in D_i\).

Step 2. \(\int\int_D \text{Curl}\ \mathbf{F}\ dA \approx \sum_i \text{Curl}\ \mathbf{F}(P_i) \text{Area}(D_i)\).

Step 3. \(\sum_i \text{Curl}\ \mathbf{F}(P_i)\cdot \text{Area}(D_i) \approx \sum_i (\frac{1}{\text{area}(D_i)}\int_{\partial D_i} \mathbf{F}\cdot d\mathbf{r})\cdot \text{area}(D_i)\approx \sum_i \int_{\partial D_i} \mathbf{F}\cdot d\mathbf{r}.\)

Step 4. The line integrals along common boundaries cancel, since the tangent vectors point in opposite directions,

to get \(\int_{\partial D} \mathbf{F}\cdot d\mathbf{r}\).

which can actually be done with a substitution. Better: Use Green's Theorem. Curl\ \mathbf{F} = 2-3 = -1, therefore,

Comments on Green's Theorem.

1. (i) Other notation for \(\int_C\mathbf{F}\cdot d\mathbf{r}\) is \(\int_C F_1(x,y)\ dx + F_2(x,y)\ dy\), so that
   \[ \int\int_D \text{Curl}\ \mathbf{F}\ dA = \int_C F_1(x,y)\ dx + \int_C F_2(x,y)\ dy. \]
2. (ii) The area of the domain \(D\) can be calculated as a line integral. Set \(\mathbf{F} = -\frac{y}{2}i+\frac{x}{2}j\). So, \(\text{Curl}\ \mathbf{F} = 1.\)
   \[ \begin{align*} \text{area}(D) &= \int\int_D 1\ dA\\ &= \int\int_D \text{Curl}\ \mathbf{F}\ dA\\ &= \int_C\mathbf{F}\cdot d\mathbf{r}. \end{align*} \]

Over the spiral we have:

Thus:

For the line segment: \(\mathbf{L}(t) = ((1-t)2\pi, 0)\), with \(0\leq t\leq 1\). Note that \(\mathbf{L}'(t) = (-1,0)\).

Therefore:

Consequently:

We began class with a discussion of path independence of line integral as an application of Green's Theorem, starting with the following example.

When does this occur?

If

Then

This suggests \(\frac{dF_2}{dx}-\frac{\partial F_1}{\partial y} = 0\), leading to the following theorem.

Moreover, in this case, if \(C\) is any smooth curve from \(P\) to \(Q\) in \(\mathbb{R}^2\),

which is a generalization of the fundamental theorem of calculus. We also noted that the condition of second order partial guarantees equality of mixed second order partial derivatives, from with it easily follows that if \(\mathbf{F} = \nabla f\), then Curl \mathbf{F} = 0.

The notion of the curl of a vector field in \(\mathbb{R}^3\) can be defined similarly, as in the previous lecture, however in this case, Curl \mathbf{F} is a vector field. The definition gives the values of the component of the curl in a given direction.

Comments. Given a vector field, \(\mathbf{F}\):

1. (i) This definition is independent of the coordinate system.
2. (ii) If we choose a coordinate system, say rectangular coordinates, then taking \(\mathbf{n}\) to be \(\vec{\imath}\), then \(\vec{\jmath}\), then \(\vec{k}\), we get components of the vector field \(\mathbf{Curl}\,\mathbf{F}\) at the point \(P\). Thus, \(\mathbf{Curl}\,\mathbf{F}\) is a vector field derived from the field \(\mathbf{F}\).
3. (iii) The limit
   \[ \mathbf{Curl}\,\mathbf{F} (P)\cdot \mathbf{n} = \lim_{\Delta S\to 0} \frac{1}{\Delta S} \int_C \mathbf{F}\cdot d\mathbf{r} \]

   gives the circulation of \mathbf{F} at \(P\) per unit area about an axis in the direction of \(\mathbf{n}\).

4. (iv) If we take \(\mathbf{n} = \vec{k}\), and assume \(P\) lies in the \(xy\)-plane, we get the curl of F at P as defined in the previous lecture.

A similar, though more complicated, argument as the one given in the previous lecture gives

Formula for Curl in \(\mathbb{R}^3\):} \(\mathbf{F} = F_1i+F_2j+F_3k\), then

We noted that one can also use the notation \(\nabla \times \mathbf{F}\) to denote \(\mathbf{Curl}\,\mathbf{F}\).

In cylindrical coordinates: \(\mathbf{F} = (F_r,F_{\theta},F_z)\) and

In spherical coordinates: \(\mathbf{F} = (F_{\rho}, F_{\phi}, F_{\theta})\), and

We then began a discussion of the last major theorem of the semester, one of the most important theorems in vector calculus.

Stoke's Theorem. Let \(S \subseteq \mathbb{R}^3\) be a smooth oriented surface with whose boundary \(\partial{S}\) is a simple closed curve, oriented according to the right hand thumb rule. If \(\mathbf{F}\) is a vector field whose components have continuous first order and second order partial derivatives, then

Moreover: If \(S\) is a closed surface, \(\int\int_S \nabla \times \mathbf{F}\cdot d\mathbf{S} = 0\).

We first noted that we could derive a heuristic proof of Stokes theorem along the lines of the argument given for Green's theorem given in the previous lecture by using the intrinsic definition of the curl and the fact that one can partition \(S\) so that interior line integrals along common boundaries in the partition cancel. We also gave three explanations as to why \(\int\int_S \nabla \times \mathbf{F}\cdot d\mathbf{S} = 0\), when \(S\) is a closed surface: (i) \(S\) has no boundary, so using Stoke's theorem, the line integral is zero; (ii) Using the divergence theorem, \(\int\int_S \nabla \times \mathbf{F})\cdot d\mathbf{S} = \int\int\int_B \nabla \cdot (\nabla \times \mathbf{F})\ dV\), which is zero, since \(\nabla \cdot (\nabla \times \mathbf{F}) = 0\), under suitable differentiability conditions; (iii) Thinking of \(S\) as the union of two open surfaces \(S = S_1\cup S_2\) joined along a common boundary, and noting that \(\partial S_2 = -\partial S_1\), so that

We ended class by working problem 2 on the worksheet from November 18, asking to verify Stokes theorem, obtaining \(18\pi\) for each expression in the theorem.

We began class by restating Stoke's Theorem and discussed the nabla notation introduced in the Daily Update of November 17, \(\nabla = \frac{\partial }{\partial x}\ \vec{\imath} + \frac{\partial }{\partial y}\ \vec{\jmath}+\frac{\partial }{\partial z}\ \vec{k}\), which is a differential operator, noting that the Divergence Theorem with this notation becomes \(\int\int\int_B \nabla \cdot \mathbf{F}\ dV = \int\int_{\partial{B}} \mathbf{F}\cdot d\mathbf{S}\), where \(\partial{B}\) is the closed boundary of the solid \(B\) and Stokees Theorem becomes \(\int\int_{S}\nabla \times \mathbf{F}\cdot d\mathbf{S} = \int_{\partial{S}}\mathbf{F}\cdot d\mathbf{r}\), under the previously stated conditions for each theorem.

This led to a brief discussion of the following theorem, which can be considered a three variable analog of the theorem characterizing conservative vector fields.

Thus, (iii) is a surface integral analogue of the path independence of line integrals of conservative vector fields that we saw above.

We then illustrated the latter point further by considering \(\int\int_S(\nabla\times\mathbf{F})\cdot d\mathbf{S}\) where \(S\) is the surface represented by the graph of \(z = e^{-(x^2+y^2)}\), with \(z\geq \frac{1}{e}\) and \(\mathbf{F} = (e^{y+z}-2y)\vec{\imath} + (xe^{y+z}+y)\vec{\jmath} + e^{x+y}\vec{k}\). We saw that computing \(\int\int_S(\nabla\times\mathbf{F})\cdot d\mathbf{S}\) and \(\int_{\partial S} \mathbf{F}\cdot d\mathbf{r}\) directly lead to unmanageable double and single integrals. However, since \(\int\int_S(\nabla\times\mathbf{F})\cdot d\mathbf{S}\) is independent of surfaces sharing the same boundary, we saw that the calculation is much easier if we take \(S'\) the disk of radius 1 at the \(z = \frac{1}{e}\) level, so that \(\partial S = \partial S'\), and thus, \(\int\int_S(\nabla\times\mathbf{F})\cdot d\mathbf{S} = \int\int_{S'}(\nabla\times\mathbf{F})\cdot d\mathbf{S} = 2\pi\). (Note: When \(z = \frac{1}{e}\), we get \(e^{-1} = e^{-(x^2+y^2)}\), so \(1 = x^2+y^2\).)

Though not stated in class, we can end the collection of Daily Updates with the following two Observations.

1. Using the symbol \(\nabla\), the Divergence Theorem and Stoke's Theorem become:

\begin{align* \int\int_{\partial B}\mathbf{F}\cdot d\mathbf{S} &= \int\int\int_B \nabla \cdot \mathbf{F}\ dV\\ \int_{\partial S} \mathbf{F}\cdot dr &= \int\int_S (\nabla\times\mathbf{F})\cdot d\mathbf{S}. \end{align*} } Note how the differential operator on the domain of integration on the left hand side of the equations, moves up to the integrand on the right hand side of the equations. There are more general theorem of this type when one studies a much more theoretical course on the calculus of several variables in arbitrary dimension.

2. Just like areas can be calculated using line integrals via Green's Theorem, we can calculate volumes as surface integrals, using the divergence theorem. Taking \(B\) to be a closed and bounded solid in \(\mathbb{R}^3\) whose boundary \(\partial{B}\) is a smooth closed surface, if we set \(\mathbf{F} := \frac{1}{3}\cdot (x\vec{\imath}+y\vec{\jmath}+z\vec{k})\), then

Jake presented review slides for the final exam.

The class worked on practice problems for the final exam.