Fall 2025 Math 147: Solutions to Final Exam Practice Problems

Below are problems to practice for the final exam. The problems below, together with the problems from the two midterm exams, are a good representation of what to expect on the final exam. There will also be a few short answer and true-false questions on the final exam. The final exam is comprehensive. The final exam is Thursday, December 11 from 4:30–7:00pm and will take place in Snow Hall 302.

Problem 1

1. Let \( f(x,y) = \begin{cases} x^2+y^2, & \text{if } x^2+y^2 < 1 \\ 1, & \text{if } x^2+y^2 \geq 1. \end{cases} \) Determine at which points \(f(x,y)\) is continuous.

Solution. Note that \(g(x,y) = x^2+y^2\) and \(h(x,y) = 1\) are both continuous on all of \(\mathbb{R}^2\). Thus, if we let \(D\) denote the unit disk \(0 \le x^2+y^2 \le 1\), then \(g(x,y)\) is continuous on the interior of \(D\) and \(h(x,y)\) is continuous on \(\mathbb{R}^2\setminus D\), and thus \(f(x,y)\) is continuous on both the interior of \(D\) and \(\mathbb{R}^2\setminus D\). For points \((a,b)\) on the boundary of \(D\), \(a^2+b^2=1\), and we can consider \(\lim_{(x,y)\to(a,b)} f(x,y)\). Let \(\epsilon>0\). Since \(g(x,y)\), as a function on \(\mathbb{R}^2\), is continuous at \((a,b)\) there exists \(\delta>0\) such that \(||(x,y)-(a,b)||<\delta\) implies \(|g(x,y)-g(a,b)| = |g(x,y)-1|<\epsilon\). Taking the same \(\delta\), if \(||(x,y)-(a,b)||<\delta\) and \((x,y)\in D\), then \(g(x,y)=f(x,y)\), which gives \(|f(x,y)-f(a,b)| = |g(x,y)-1|<\epsilon\). If \((x,y)\not\in D\), then \(f(x,y)-f(a,b) = 1-1=0\), so \(|f(x,y)-f(a,b)|<\epsilon\). Thus \(f(x,y)\) is continuous at \((a,b)\).

Problem 2

2. Show that the function \( f(x,y) = \begin{cases} \dfrac{(2^x-1)\sin(y)}{xy}, & \text{if } xy \ne 0 \\ \ln(2), & \text{if } xy = 0 \end{cases} \) is continuous at \((0,0)\).

Solution. Set \(g(x) = \begin{cases} \dfrac{2^x-1}{x}, & \text{if } x\ne 0\\ \ln(2), & \text{if } x=0 \end{cases}\) and \(h(y) = \begin{cases} \dfrac{\sin(y)}{y}, & \text{if } y \ne 0 \\ 1, & \text{if } y = 0 \end{cases}\). Then \(f(x,y) = g(x)h(y)\). Thus, \(\lim_{(x,y)\to(0,0)} f(x,y) = \{\lim_{x\to 0} g(x)\}\cdot \{\lim_{y\to 0} h(y)\}\). By L'Hospital's Rule, \(\lim_{x\to 0} g(x) = \ln(2)\) and \(\lim_{y\to 0} h(y) = 1\). Therefore, \(\lim_{(x,y)\to(0,0)} f(x,y) = \ln(2)\).

Problem 3

3. Use the limit definition to show that \(f(x,y) = 5x+4y^2\) is differentiable at \((2,1)\).

Solution. \(\dfrac{\partial f}{\partial x}(2,1) = 5\), \(\dfrac{\partial f}{\partial y}(2,1) = 8\), and \(f(2,1)=14\), so \(L(x,y) = 5(x-2)+8(y-1)+14\). Therefore,
\[ \begin{align*} f(x,y)-L(x,y) &= 5x+4y^2-(5x-10+8y-8+14) \\ &= 4y^2-8y+4 \\ &= 4(y-1)^2. \end{align*} \]
Thus, \(\dfrac{f(x,y)-L(x,y)}{\sqrt{(x-2)^2+(y-1)^2}} = \dfrac{4(y-1)^2}{\sqrt{(x-2)^2+(y-1)^2}} \le \dfrac{4(y-1)^2}{\sqrt{(y-1)^2}} = 4|y-1|\). Therefore,
\[ \lim_{(x,y)\to(2,1)} \frac{f(x,y)-L(x,y)}{\sqrt{(x-2)^2+(y-1)^2}} \le \lim_{y\to 1} 4|y-1| = 0, \]
which shows that \(f(x,y)\) is differentiable at \((2,1)\).

Problem 4

4. From class, we saw that if the first order partial derivatives of \(f(x,y)\) are continuous in a neighborhood of \((a,b)\), then \(f(x,y)\) is differentiable at \((a,b)\). This problem shows why those conditions are necessary. Let \( g(x,y) = \begin{cases} \dfrac{2xy(x+y)}{x^2+y^2}, & \text{if } (x,y)\ne(0,0) \\ 0, & \text{if } (x,y)=(0,0) \end{cases} \). Show that:

  1. (i) \(g(x,y)\) is continuous at \((0,0)\).
  2. (ii) Use the limit definitions to show that \(g_x(0,0)\) and \(g_y(0,0)\) exist and are equal to 0.
  3. (iii) Conclude that \(L(x,y) = 0\).
  4. (iv) Show that \(g(x,y)\) is not differentiable at \((0,0)\).
  5. (v) Show that \(g_x(x,y)\) is not continuous at \((0,0)\).

Solution. (i) Taking limits, we have
\[ \begin{align*} \lim_{(x,y)\to(0,0)} g(x,y) &= \lim_{r\to 0} \frac{2r^2\cos(\theta)\sin(\theta)(r\cos(\theta)+r\sin(\theta))}{r^2} \\ &= \lim_{r\to 0} r\cdot\{2\cos(\theta)\sin(\theta)(\cos(\theta)+\sin(\theta))\} \\ &= 0 \\ &= g(0,0), \end{align*} \]
so \(g(x,y)\) is continuous at \((0,0)\).

(ii) \(\dfrac{\partial g}{\partial x}(0,0) = \lim_{h\to 0} \dfrac{g(0+h,0)-g(0,0)}{h} = \lim_{h\to 0} \dfrac{0}{h} = 0\). \(\dfrac{\partial g}{\partial y}(0,0) = \lim_{h\to 0} \dfrac{g(0,0+h)-g(0,0)}{h} = \lim_{h\to 0} \dfrac{0}{h} = 0\).

(iii) \(L(x,y) = 0(x-0)+0(y-0) = 0\).

(iv) Thus, \(g(x,y)-L(x,y) = g(x,y)\). If \(g(x,y)\) were differentiable at \((0,0)\), then the limit (when \((x,y)\ne(0,0)\))

\[ \lim_{(x,y)\to(0,0)} \frac{g(x,y)}{\sqrt{x^2+y^2}} = \lim_{(x,y)\to(0,0)} \frac{2xy(x+y)}{(x^2+y^2)^{3/2}} \]
should equal zero. If we take the limit along the line \(y=x\), then \(\dfrac{g(x,y)}{\sqrt{x^2+y^2}} = \dfrac{4x^3}{\sqrt{2}\,x^3} = \sqrt{2}\), so the limit along the line \(y=x\) as \(x\to 0\) is not 0. Thus, \(g(x,y)\) is not differentiable at \((0,0)\).

(v) Differentiating the non-zero part of \(g(x,y)\) gives \(g_x(x,y) = \dfrac{-2x^2y^2+4xy^3+2y^4}{(x^2+y^2)^2}\). If we take the limit along the line \(y=0\), then \(\lim_{(x,y)\to(0,0)} g_x(x,y) = 0\), while if we take the limit along the line \(x=0\), the limit becomes 2. Thus, \(\lim_{(x,y)\to(0,0)} g_x(x,y)\) does not exist, so that \(g_x(x,y)\) is not continuous at \((0,0)\).

Problem 5

5. Find and classify the critical points for \(f(x,y) = x^4-4xy+2y^2\).

Solution. To find critical points we solve
\[ \begin{align*} f_x &= 4x^3-4y = 0 \\ f_y &= -4x+4y = 0. \end{align*} \]
From the second equation, we get \(x=y\). Using this in the first equation gives \(4x^3-4x=0\), so that \(x=0,-1,1\). Thus, the critical points are \((0,0)\), \((-1,-1)\), and \((1,1)\). For the discriminant, we have
\[ D = f_{xx}f_{yy}-f_{xy}^2 = (12x^2)4-(-4)^2 = 48x^2-16. \]
For \((0,0)\): \(D(0,0)=-16\), so that \(f(x,y)\) has a saddle point at \((0,0)\).

For \((-1,-1)\): \(D(-1,-1) = 32 > 0\) and \(f_{xx}(-1,-1) = 12 > 0\). Thus, \(f(x,y)\) has a relative minimum value (of \(-3\)) at \((-1,-1)\).

For \((1,1)\): \(D(1,1) = 32 > 0\) and \(f_{xx}(1,1) = 12 > 0\). Thus, \(f(x,y)\) has a relative minimum value (of \(-3\)) at \((1,1)\).

Problem 6

6. Find the absolute maximum and absolute minimum values of \(f(x,y) = x^2y\) on the closed and bounded set \(D: 0 \le 4x^2+9y^2 \le 36\).

Solution. Solving
\[ \begin{align*} f_x &= 2xy = 0 \\ f_y &= x^2 = 0 \end{align*} \]
we see that \(x=0\), and \(y\) can be any real number. Thus, critical points in the interior of \(D\) are of the form \((0,y)\) with \(0 \le 9y^2 \le 36\). However, \(f(0,y)=0\) for all such points. On the boundary of \(D\), we must maximize and minimize \(f(x,y)\) subject to the constraint \(4x^2+9y^2=36\). Calling this equation \(g(x,y)\), we set \(\nabla f = \lambda \nabla g\) and solve the resulting system of equations
\[ \begin{align*} 2xy &= \lambda 8x \\ x^2 &= \lambda 18y \\ 4x^2+9y^2 &= 36. \end{align*} \]
Notice that if \(x\) or \(y\) equal 0, then \(f(x,y)=0\). So we can assume neither \(x\) nor \(y\) is zero. Dividing the first equation by \(2x\) gives \(y=4\lambda\). Using this in the second equation gives \(x^2 = 72\lambda^2\). Putting both of these into the constraint equation gives \(4(72\lambda^2)+9(4\lambda)^2=36\), so that
\[ 3\lambda^2 = \frac{36}{144}, \]
so that \(\lambda = \pm\sqrt{\dfrac{1}{12}}\). Thus, \(y = \pm\sqrt{\dfrac{4}{12}}\) and \(x=\pm\sqrt{6}\). Thus, substituting these into \(x^2y\) gives \(\pm\dfrac{\sqrt{24}}{12} = \pm 4\sqrt{3}\). Thus, on the domain \(D\), the maximum value of \(f(x,y)\) is \(4\sqrt{3}\) and the minimum value is \(-4\sqrt{3}\). Note that the critical points \((0,y)\) do not determine a minimum or maximum value of \(f(x,y)\) on \(D\).

Problem 7

7. Let \(S\) be the surface parametrized by \(G(u,v) = (2u\sin(\tfrac{v}{2}), 2u\cos(\tfrac{v}{2}), 3v)\), with \(0 \le u \le 1\) and \(0 \le v \le 2\pi\).

  1. (i) Find the tangent plane to \(S\) at the point \(P = G(1, \tfrac{\pi}{3})\).
  2. (ii) Find the surface area of \(S\).

Solution.
\[ \mathbf{T}_u \times \mathbf{T}_v = \begin{vmatrix} \vi & \vj & \vk \\ 2\sin(v/2) & 2\cos(v/2) & 0 \\ u\cos(v/2) & -u\sin(v/2) & 3 \end{vmatrix} = (6\cos(v/2),\, -6\sin(v/2),\, -2u). \]
Therefore, \(\mathbf{T}_u \times \mathbf{T}_v(1,\tfrac{\pi}{3}) = (3\sqrt{3}, -3, -2)\), since \(G(1,\tfrac{\pi}{3}) = (1,\sqrt{3},\pi)\), so for the tangent plane we have:
\[ 3\sqrt{3}(x-1) - 3(y-\sqrt{3}) - 2(z-\pi) = 0. \]
For the surface area, we have \(||\mathbf{T}_u \times \mathbf{T}_v|| = \sqrt{36+4u^2} = 2\sqrt{9+u^2}\).

Surface area \(\displaystyle = \int\!\!\int_D ||\mathbf{T}_u \times \mathbf{T}_v||\, dA = \int_0^{2\pi}\!\int_0^1 2\sqrt{9+u^2}\, du\, dv = 4\pi \int_0^1 \sqrt{9+u^2}\, du\)

\[ = 4\pi\left\{ \frac{u}{2}\sqrt{9+u^2} + \frac{9}{2}\ln\left|u+\sqrt{9+u^2}\right| \right\}\Bigg|_0^1 \quad \text{(using a table of integrals)} \]
\[ = 4\pi\left\{ \frac{\sqrt{10}}{2} + \frac{9}{2}\ln(1+\sqrt{10}) - \frac{9}{2}\ln(3) \right\} = 4\pi\left\{ \frac{\sqrt{10}}{2} + \frac{9}{2}\ln\!\left(\frac{1+\sqrt{10}}{3}\right) \right\}. \]

Problem 8

8. Let \(C\) be a curve from the point \(P\) to the point \(Q\) in the \(xy\)-plane. Let \(R\) be the region enclosed by \(C\) and the two radial lines from the origin to \(P\) and \(Q\). (See the figure below.) Use Green's Theorem to show that \(\int_C \mathbf{F}\cdot d\mathbf{r}\) gives the area of \(R\), for \(\mathbf{F} = -\dfrac{y}{2}\vi + \dfrac{x}{2}\vj\).

Diagram of the region R in the xy-plane, bounded by the curve C from point P to point Q and the two radial line segments from the origin to P and to Q, with angles alpha and beta marked at the origin.
Solution. Let \(D\) denote the oriented closed curve forming the boundary of \(R\), namely, the line segment from the origin to \(P\), followed by \(C\), followed by the line segment from \(Q\) to the origin. We take \(\mathbf{F} = (-\tfrac{1}{2}y, \tfrac{1}{2}x)\), so that by Green's Theorem, \(\int_D \mathbf{F}\cdot d\mathbf{r}\) equals the area enclosed by the closed curve \(D\). Let \(C_1\) denote the line segment from \((0,0)\) to \(P=(a,b)\), \(C_2\) denote the line segment from \(Q=(c,d)\) to \((0,0)\), and \(C\) the curve given in the illustration. Let \(D = C_1 \cup C \cup C_2\) so that
\[ \text{area}(R) = \int_D \mathbf{F}\cdot d\mathbf{r} = \int_{C_1} \mathbf{F}\cdot d\mathbf{r} + \int_C \mathbf{F}\cdot d\mathbf{r} + \int_{C_2} \mathbf{F}\cdot d\mathbf{r}. \]
We have to show \(\int_{C_1} \mathbf{F}\cdot d\mathbf{r} = \int_{C_2} \mathbf{F}\cdot d\mathbf{r} = 0\).

Take \(\mathbf{r}_1(t) = (at,bt)\), \(0 \le t \le 1\) for the parametrization of \(C_1\). Then \(\mathbf{F}(\mathbf{r}(t)) = (-\tfrac{1}{2}bt, \tfrac{1}{2}at)\) and \(\mathbf{r}'(t) = (a,b)\). Therefore \(\mathbf{F}(\mathbf{r}(t))\cdot \mathbf{r}'(t) = -\tfrac{1}{2}abt+\tfrac{1}{2}abt = 0\). So \(\int_{C_1} \mathbf{F}\cdot d\mathbf{r} = 0\). The calculation showing \(\int_{C_2} \mathbf{F}\cdot d\mathbf{r} = 0\) is similar.

Problem 9

9. Let \(C\) be the triangle with vertices \((1,0,0)\), \((0,2,0)\), \((0,0,1)\). Compute \(\int_C \mathbf{F}\cdot d\mathbf{r}\), for the vector field \(\mathbf{F} = (x^2+yz,\, x+y,\, y-z^2)\).

Solution. We have to integrate along each side of the triangle.

\(C_1: \mathbf{r}(t) = (1-t,2t,0)\), with \(0\le t\le 1\). \(\mathbf{r}'(t)=(-1,2,0)\). \(\mathbf{F}(\mathbf{r}(t)) = ((1-t)^2, 1+t, 2t)\).

\[ \begin{align*} \int_{C_1} \mathbf{F}\cdot d\mathbf{r} &= \int_0^1 \mathbf{F}(\mathbf{r}(t))\cdot \mathbf{r}'(t)\, dt \\ &= \int_0^1 -(1-t)^2+2+2t\, dt \\ &= \int_0^1 -1+2t-t^2+2+2t\, dt \\ &= \int_0^1 1+4t-t^2\, dt \\ &= 1+2-\frac{1}{3} = \frac{8}{3}. \end{align*} \]

\(C_2: \mathbf{r}(t) = (0,2-2t,t)\), \(0\le t\le 1\), \(\mathbf{r}'(t)=(0,-2,1)\), \(\mathbf{F}(\mathbf{r}(t)) = (2t-2t^2, 2-2t, 2-2t-t^2)\).

\[ \begin{align*} \int_{C_2} \mathbf{F}\cdot d\mathbf{r} &= \int_0^1 \mathbf{F}(\mathbf{r}(t))\cdot \mathbf{r}'(t)\, dt \\ &= \int_0^1 -4+4t+2-2t-t^2\, dt \\ &= \int_0^1 -2+2t-t^2\, dt \\ &= -2+1-\frac{1}{3} = -\frac{4}{3}. \end{align*} \]

\(C_3: \mathbf{r}(t) = (t,0,1-t)\), \(\mathbf{r}'(t)=(1,0,-1)\), \(\mathbf{F}(\mathbf{r}(t)) = (t^2, t, -(1-t)^2)\).

\[ \begin{align*} \int_{C_3} \mathbf{F}\cdot d\mathbf{r} &= \int_0^1 \mathbf{F}(\mathbf{r}(t))\cdot \mathbf{r}'(t)\, dt \\ &= \int_0^1 t^2+(1-t)^2\, dt \\ &= \left\{\frac{t^3}{3} - \frac{(1-t)^3}{3}\right\}\Bigg|_0^1 \\ &= \frac{1}{3}-\left(-\frac{1}{3}\right) = \frac{2}{3}. \end{align*} \]

Therefore,

\[ \int_C \mathbf{F}\cdot d\mathbf{r} = \int_{C_1} \mathbf{F}\cdot d\mathbf{r} + \int_{C_2} \mathbf{F}\cdot d\mathbf{r} + \int_{C_3} \mathbf{F}\cdot d\mathbf{r} = \frac{8}{3}-\frac{4}{3}+\frac{2}{3} = 2. \]

Problem 10

10. Let \(f(x,y) = \sqrt{|xy|}\). Write out details showing:

  1. (a) \(\dfrac{\partial f}{\partial x}(0,0)\) and \(\dfrac{\partial f}{\partial y}(0,0)\) exist.
  2. (b) \(f(x,y)\) is not differentiable at \((0,0)\).
  3. (c) Part (b) does not contradict part (a).

Solution. (a) \(\dfrac{\partial f}{\partial x}(0,0) = \lim_{h\to 0} \dfrac{f(0+h,0)-f(0,0)}{h} = \lim_{h\to 0} \dfrac{0}{h} = 0\). The calculation for \(\dfrac{\partial f}{\partial y}(0,0)\) is similar.

(b) From (a), the linear approximation to \(f(x,y)\) at \((0,0)\) is \(L(x,y)=0\). Therefore, \(f(x,y)-L(x,y) = f(x,y)\). In order for \(f(x,y)\) to be differentiable at \((0,0)\), the limit

\[ \lim_{(x,y)\to(0,0)} \frac{f(x,y)-L(x,y)}{\sqrt{x^2+y^2}} = \lim_{(x,y)\to(0,0)} \frac{f(x,y)}{\sqrt{x^2+y^2}} \]
should be 0. Taking the limit along the line \(y=x\), we have
\[ \lim_{(x,y)\to(0,0)} \frac{f(x,y)}{\sqrt{x^2+y^2}} = \lim_{x\to 0} \frac{\sqrt{|x^2|}}{\sqrt{2}\sqrt{x^2}} = \frac{1}{\sqrt{2}} \ne 0. \]
Therefore, \(f(x,y)\) is not differentiable at \((0,0)\).

(c) Part (b) does not contradict part (a) because the first order partial derivatives of \(f(x,y)\) are not continuous at \((0,0)\). In fact, the partial derivatives of \(f(x,y)\) are not defined at all points \((x,y)\) near \((0,0)\), so we cannot evaluate the required limit to test continuity. To see this, let's try to calculate \(\dfrac{\partial f}{\partial x}\) along the line \(x=0\), with \(y\ne 0\).

\[ \frac{\partial f}{\partial x}(0,y) = \lim_{h\to 0} \frac{f(0+h,y)-f(0,y)}{h} = \lim_{h\to 0} \frac{\sqrt{|hy|}-0}{h} = \sqrt{|y|}\lim_{h\to 0} \frac{\sqrt{|h|}}{h}, \]
which does not exist.

Problem 11

11. Evaluate \(\int\!\!\int_S \text{Curl}\,\mathbf{F}\cdot d\mathbf{S}\), for \(\mathbf{F} = (-y+z\sin(x),\, x,\, z^3)\) and \(S\) the surface defined by the equation \(x^2+\dfrac{y^2}{4}+z^2+z^4x^2=1\), with \(z\ge 0\).

Solution. We use Stokes' Theorem. The surface lies above the \(xy\)-plane, and intersects the \(xy\)-plane along the ellipse \(x^2+\dfrac{y^2}{4}=1\). By Stokes' Theorem, \(\int\!\!\int_S \text{Curl}\,\mathbf{F}\cdot d\mathbf{S} = \int_C \mathbf{F}\cdot d\mathbf{r}\), for \(C: (\cos(t), 2\sin(t), 0)\), with \(0\le t\le 2\pi\). Note that the orientation of \(C\) is consistent with what is required by Stokes' Theorem.

\(\mathbf{r}'(t) = (-\sin(t), 2\cos(t), 0)\) and \(\mathbf{F}(\mathbf{r}(t)) = (-2\sin(t), \cos(t), 0)\). Therefore,

\[ \begin{align*} \int\!\!\int_S \text{Curl}\,\mathbf{F}\cdot d\mathbf{S} &= \int_C \mathbf{F}\cdot d\mathbf{r} \\ &= \int_0^{2\pi} \mathbf{F}(\mathbf{r}(t))\cdot \mathbf{r}'(t)\, dt \\ &= \int_0^{2\pi} 2\, dt \\ &= 4\pi. \end{align*} \]

Problem 12

12. Verify the Divergence Theorem for \(\mathbf{F} = (-x^2, y^2, -z^2)\) and \(S\) the rectangular box \([0,3]\times[-1,2]\times[1,2]\).

Solution. \(\text{Div}(\mathbf{F}) = -2x+2y-2z\), therefore if \(B\) is the solid contained in the given rectangular box,
\[ \begin{align*} \int\!\!\int\!\!\int_B \text{Div}\,\mathbf{F}\, dV &= \int_1^2\!\int_{-1}^2\!\int_0^3 -2x+2y-2z\, dx\,dy\,dz \\ &= \int_1^2\!\int_{-1}^2 (-x^2+2xy-2xz)\Big|_{x=0}^{x=3}\, dy\,dz \\ &= \int_1^2\!\int_{-1}^2 -9+6y-6z\, dy\,dz \\ &= \int_1^2 (-9y+3y^2-6yz)\Big|_{y=-1}^{y=2}\, dz \\ &= \int_1^2 -18-18z\, dz \\ &= (-18z-9z^2)\Big|_{z=1}^{z=2} \\ &= (-36-36)-(-18-9) \\ &= -45. \end{align*} \]

To calculate the surface integral, we must sum the integrals over each face of the given rectangular box.

Front Face, \(S_1\): \(S_1\) is given by \((3,y,z)\), with \(-1\le y\le 2\), \(1\le z\le 2\) and \(\mathbf{n}=\vi\). We will see the bounds on \(y\) and \(z\) are just used to calculate the area of the front face. The same will hold for the other five faces. So: \(\mathbf{F}\) on \(S_1\) is given by \((-9, y^2, -z^2)\) and thus \(\mathbf{F}\cdot\mathbf{n} = -9\). Therefore,

\[ \int\!\!\int_{S_1} \mathbf{F}\cdot d\mathbf{S} = \int\!\!\int_{S_1} -9\, dS = -9\cdot\text{area}(S_1) = -9\cdot 3 = -27. \]

Back Face, \(S_2\): \(S_2\) is given by \((0,y,z)\), with \(-1\le y\le 2\), \(1\le z\le 2\) and \(\mathbf{n}=-\vi\). \(\mathbf{F}\) on \(S_2\) is given by \((0,y^2,-z^2)\) and thus \(\mathbf{F}\cdot\mathbf{n} = 0\). Therefore, \(\int\!\!\int_{S_2} \mathbf{F}\cdot d\mathbf{S} = 0\).

Left Face, \(S_3\): \(S_3\) is given by \((x,-1,z)\), with \(0\le x\le 3\), \(1\le z\le 2\) and \(\mathbf{n}=-\vj\). \(\mathbf{F}\) on \(S_3\) is given by \((-x^2,1,-z^2)\) and thus \(\mathbf{F}\cdot\mathbf{n} = -1\). Therefore,

\[ \int\!\!\int_{S_3} \mathbf{F}\cdot d\mathbf{S} = -1\cdot\text{area}(S_3) = -1\cdot 3 = -3. \]

Right Face, \(S_4\): \(S_4\) is given by \((x,2,z)\), with \(0\le x\le 3\), \(1\le z\le 2\) and \(\mathbf{n}=\vj\). \(\mathbf{F}\) on \(S_4\) is given by \((-x^2,4,-z^2)\) and thus \(\mathbf{F}\cdot\mathbf{n} = 4\). Therefore,

\[ \int\!\!\int_{S_4} \mathbf{F}\cdot d\mathbf{S} = 4\cdot\text{area}(S_4) = 4\cdot 3 = 12. \]

Top Face, \(S_5\): \(S_5\) is given by \((x,y,2)\), with \(0\le x\le 3\), \(-1\le y\le 2\) and \(\mathbf{n}=\vk\). \(\mathbf{F}\) on \(S_5\) is given by \((-x^2,y^2,-4)\) and thus \(\mathbf{F}\cdot\mathbf{n} = -4\). Therefore,

\[ \int\!\!\int_{S_5} \mathbf{F}\cdot d\mathbf{S} = -4\cdot\text{area}(S_5) = -4\cdot 9 = -36. \]

Bottom Face, \(S_6\): \(S_6\) is given by \((x,y,1)\), with \(0\le x\le 3\), \(-1\le y\le 2\) and \(\mathbf{n}=-\vk\). \(\mathbf{F}\) on \(S_6\) is given by \((-x^2,y^2,-1)\) and thus \(\mathbf{F}\cdot\mathbf{n} = 1\). Therefore,

\[ \int\!\!\int_{S_6} \mathbf{F}\cdot d\mathbf{S} = 1\cdot\text{area}(S_6) = 1\cdot 9 = 9. \]

Putting these all together we have \(\int\!\!\int_S \mathbf{F}\cdot d\mathbf{S} = -27+0-3+12-36+9 = -45\).

Problem 13

13. Let \(\mathbf{F} = (z^2, x^2, -y^2)\). Evaluate \(\int_C \mathbf{F}\cdot d\mathbf{r}\), where \(C\) is the path traversing counterclockwise the square with sides of length \(s\) centered at \((x_0,y_0,0)\). Then divide this number by the area of the square and take the limit as \(s\to 0\). Compare this with \((\text{Curl}\,\mathbf{F})(x_0,y_0,0)\cdot\mathbf{k}\).

Solution. In this problem we are using the limit definition to calculate \((\text{Curl}\,\mathbf{F})(x_0,y_0,0)\cdot\mathbf{k}\). For this, we must compute a line integral of \(\mathbf{F}\) over each side of the square \(C\).

Bottom side, \(C_1\): \(C_1\) is given by \(\mathbf{r}(t) = (x_0-\tfrac{s}{2}, y_0-\tfrac{s}{2}, 0)+t(s,0,0)\), with \(0\le t\le 1\). \(\mathbf{r}'(t)=(s,0,0)\), \(\mathbf{F}(\mathbf{r}(t)) = (0, (x_0-\tfrac{s}{2})^2, -(y_0-\tfrac{s}{2})^2)\).

\[ \int_{C_1} \mathbf{F}\cdot d\mathbf{r} = \int_0^1 0\, dt = 0. \]

Top side, \(C_2\): \(C_2\) is given by \(\mathbf{r}(t) = (x_0+\tfrac{s}{2}, y_0+\tfrac{s}{2}, 0)+t(-s,0,0)\), with \(0\le t\le 1\). \(\mathbf{r}'(t)=(-s,0,0)\), \(\mathbf{F}(\mathbf{r}(t)) = (0, (x_0+\tfrac{s}{2})^2, -(y_0+\tfrac{s}{2})^2)\).

\[ \int_{C_2} \mathbf{F}\cdot d\mathbf{r} = \int_0^1 0\, dt = 0. \]

Right side, \(C_3\): \(C_3\) is given by \(\mathbf{r}(t) = (x_0+\tfrac{s}{2}, y_0-\tfrac{s}{2}, 0)+t(0,s,0)\), with \(0\le t\le 1\). \(\mathbf{r}'(t)=(0,s,0)\), \(\mathbf{F}(\mathbf{r}(t)) = (0, (x_0+\tfrac{s}{2})^2, -(y_0-\tfrac{s}{2})^2)\).

\[ \int_{C_3} \mathbf{F}\cdot d\mathbf{r} = \int_0^1 (x_0+\tfrac{s}{2})^2 s\, dt = x_0^2 s + x_0 s^2 + \frac{s^3}{4}. \]

Left side, \(C_4\): \(C_4\) is given by \(\mathbf{r}(t) = (x_0-\tfrac{s}{2}, y_0+\tfrac{s}{2}, 0)+t(0,-s,0)\), with \(0\le t\le 1\). \(\mathbf{r}'(t)=(0,-s,0)\), \(\mathbf{F}(\mathbf{r}(t)) = (0, (x_0-\tfrac{s}{2})^2, -(y_0+\tfrac{s}{2})^2)\).

\[ \int_{C_4} \mathbf{F}\cdot d\mathbf{r} = \int_0^1 -(x_0-\tfrac{s}{2})^2 s\, dt = -x_0^2 s+x_0 s^2-\frac{s^3}{4}. \]

We now have

\[ \int_C \mathbf{F}\cdot d\mathbf{r} = \int_{C_1}+\int_{C_2}+\int_{C_3}+\int_{C_4} = 0+0+\left(x_0^2 s+x_0 s^2+\frac{s^3}{4}\right)+\left(-x_0^2 s+x_0 s^2-\frac{s^3}{4}\right) = 2s^2x_0. \]
Therefore,
\[ \lim_{s\to 0} \frac{1}{\text{area}(S)}\int_C \mathbf{F}\cdot d\mathbf{r} = \lim_{s\to 0} \frac{1}{s^2}2s^2x_0 = 2x_0. \]

To Check:

\[ \text{Curl}\,\mathbf{F} = \begin{vmatrix} \vi & \vj & \vk \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ z^2 & x^2 & -y^2 \end{vmatrix} = (-2y, 2z, 2x). \]
Thus, \(\text{Curl}\,\mathbf{F}(x_0,y_0,z_0) = (-2y_0, 2z_0, 2x_0)\) and \(\text{Curl}\,\mathbf{F}(x_0,y_0,z_0)\cdot \mathbf{k} = 2x_0\).

Problem 14

14. Let \(C\) be the curve obtained by intersecting the cylinder \(x^2+y^2=1\) with the plane \(x+y+z=1\), and \(\mathbf{F} = -y^3\vi+x^3\vj-z^3\vk\). Set up the line integral \(\int_C \mathbf{F}\cdot d\mathbf{r}\) as a single integral over an interval of the form \([a,b]\). Now evaluate this line integral by using Stokes' Theorem.

Solution. The curve \(C\) lies on the plane \(z=1-x-y\) but also on the cylinder \(x^2+y^2=1\). So a parametrization and tangent of \(C\) are
\[ \mathbf{r}(t) = (\cos(t), \sin(t), 1-\cos(t)-\sin(t)) \quad \text{and} \quad \mathbf{r}'(t) = (-\sin(t), \cos(t), \sin(t)-\cos(t)). \]
\(\mathbf{F}(\mathbf{r}(t)) = (-\sin^3(t), \cos^3(t), -(1-\cos(t)-\sin(t))^3)\).
\[ \mathbf{F}(\mathbf{r}(t))\cdot \mathbf{r}'(t) = \sin^4(t)+\cos^4(t)-(1-\cos(t)-\sin(t))^3\cdot(\sin(t)-\cos(t)). \]
Integrating this last expression from 0 to \(2\pi\) is doable … but not much fun.

To apply Stokes' Theorem we will integrate \(\nabla\times\mathbf{F}\) over \(S\), that portion of the given plane lying above the disk \(D: 0\le x^2+y^2\le 1\) in the \(xy\)-plane.

\[ \nabla\times\mathbf{F} = \begin{vmatrix} \vi & \vj & \vk \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ -y^3 & x^3 & -z^3 \end{vmatrix} = (3x^2+3y^2)\vk = (0,0,3x^2+3y^2). \]
\(S\) is given by \(G(u,v) = (u,v,1-u-v)\), with \(0\le u^2+v^2\le 1\). Thus,
\[ \mathbf{T}_u\times\mathbf{T}_v = \begin{vmatrix} \vi & \vj & \vk \\ 1 & 0 & -1 \\ 0 & 1 & -1 \end{vmatrix} = \vi+\vj+\vk = (1,1,1). \]
Moreover, \(\nabla\times\mathbf{F}\) on \(S\) is \((0,0,3u^2+3v^2)\). Thus, on \(S\),
\[ (\nabla\times\mathbf{F})\cdot \mathbf{T}_u\times\mathbf{T}_v = (0,0,3u^2+3v^2)\cdot(1,1,1) = 3u^2+3v^2. \]
Therefore:
\[ \begin{align*} \int\!\!\int_S \nabla\times\mathbf{F}\cdot d\mathbf{S} &= \int\!\!\int_D 3u^2+3v^2\, dA \\ &= 3\int_0^{2\pi}\!\int_0^1 r^2\cdot r\, dr\,d\theta \\ &= 6\pi\int_0^1 r^3\, dr \\ &= \frac{6\pi}{4} = \frac{3\pi}{2}. \end{align*} \]

Problem 15

15. Verify Stokes' Theorem for \(\mathbf{F} = (z^2, -y^2, 0)\) and \(C\) the square of side 1 oriented as shown, lying in the \(xz\)-plane and \(S\) the open box with sides \(S_1, S_2, S_3, S_4, S_5\). What happens, if instead, you take \(S\) to be the square enclosed by \(C\)?

Diagram of an open box (cube with one face missing) sitting above the xz-plane, with faces labeled S1 through S5, the curve C traced along the left face in the xz-plane with an arrow indicating orientation, and side lengths of 1 marked along each edge.
Solution. Both terms in Stokes' Theorem require computing several integrals. We start by computing \(\int_C \mathbf{F}\cdot d\mathbf{r}\), where \(C = C_1\cup C_2\cup C_3\cup C_4\) is the curve indicated in the diagram.

\(C_1: \mathbf{r}(t) = (1-t,0,0)\), \(0\le t\le 1\), \(\mathbf{r}'(t)=(-1,0,0)\), \(\mathbf{F}(\mathbf{r}(t)) = (0,0,0)\).

\[ \int_{C_1} \mathbf{F}\cdot d\mathbf{r} = \int_0^1 0\, dt = 0. \]

\(C_2: \mathbf{r}(t) = (0,0,t)\), \(0\le t\le 1\), \(\mathbf{r}'(t)=(0,0,1)\), \(\mathbf{F}(\mathbf{r}(t)) = (t^2,0,0)\).

\[ \int_{C_2} \mathbf{F}\cdot d\mathbf{r} = \int_0^1 0\, dt = 0. \]

\(C_3: \mathbf{r}(t) = (t,0,1)\), \(0\le t\le 1\), \(\mathbf{r}'(t)=(1,0,0)\), \(\mathbf{F}(\mathbf{r}(t)) = (1,0,0)\).

\[ \int_{C_3} \mathbf{F}\cdot d\mathbf{r} = \int_0^1 1\, dt = 1. \]

\(C_4: \mathbf{r}(t) = (1,0,1-t)\), \(0\le t\le 1\), \(\mathbf{r}'(t)=(0,0,-1)\), \(\mathbf{F}(\mathbf{r}(t)) = ((1-t)^2,0,0)\).

\[ \int_{C_4} \mathbf{F}\cdot d\mathbf{r} = \int_0^1 0\, dt = 0. \]

We now have, \(\int_C \mathbf{F}\cdot d\mathbf{r} = 0+0+1+0 = 1\).

To calculate the curl of \(\mathbf{F}\):

\[ \text{Curl}\,\mathbf{F} = \begin{vmatrix} \vi & \vj & \vk \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ z^2 & -y^2 & 0 \end{vmatrix} = (0,2z,0). \]
For the surface integral \(\int\!\!\int_S \text{Curl}\,\mathbf{F}\cdot d\mathbf{S}\), we sum the surface integrals of the curl of \(\mathbf{F}\) over the five faces indicated in the diagram.

Front Face, \(S_1\): \(S_1\) is given by \((1,y,z)\), with \(0\le y\le 1\), \(0\le z\le 1\) and \(\mathbf{n}=\vi\). \(\text{Curl}\,\mathbf{F}\) on \(S_1\) is \((0,2z,0)\), so \(\text{Curl}\,\mathbf{F}\cdot\mathbf{n} = 0\). Therefore, \(\int\!\!\int_{S_1} \text{Curl}\,\mathbf{F}\cdot d\mathbf{S} = 0\).

Back Face, \(S_2\): \(S_2\) is given by \((0,y,z)\), with \(0\le y\le 1\), \(0\le z\le 1\) and \(\mathbf{n}=-\vi\). \(\text{Curl}\,\mathbf{F}\) on \(S_2\) is \((0,2z,0)\), so \(\text{Curl}\,\mathbf{F}\cdot\mathbf{n} = 0\). Therefore, \(\int\!\!\int_{S_2} \text{Curl}\,\mathbf{F}\cdot d\mathbf{S} = 0\).

Top Face, \(S_3\): \(S_3\) is given by \((x,y,1)\), with \(0\le x\le 1\), \(0\le y\le 1\) and \(\mathbf{n}=\vk\). \(\text{Curl}\,\mathbf{F}\) on \(S_3\) is \((0,2,0)\), so \(\text{Curl}\,\mathbf{F}\cdot\mathbf{n} = 0\). Therefore, \(\int\!\!\int_{S_3} \text{Curl}\,\mathbf{F}\cdot d\mathbf{S} = 0\).

Bottom Face, \(S_4\): \(S_4\) is given by \((x,y,0)\), with \(0\le x\le 1\), \(0\le y\le 1\) and \(\mathbf{n}=-\vk\). \(\text{Curl}\,\mathbf{F}\) on \(S_4\) is \((0,0,0)\), so \(\text{Curl}\,\mathbf{F}\cdot\mathbf{n} = 0\). Therefore, \(\int\!\!\int_{S_4} \text{Curl}\,\mathbf{F}\cdot d\mathbf{S} = 0\).

Right Face, \(S_5\): \(S_5\) is given by \((x,1,z)\), with \(0\le x\le 1\), \(0\le z\le 1\) and \(\mathbf{n}=\vj\). \(\text{Curl}\,\mathbf{F}\) on \(S_5\) is \((0,2z,0)\), so \(\text{Curl}\,\mathbf{F}\cdot\mathbf{n} = 2z\). Therefore,

\[ \int\!\!\int_{S_5} \text{Curl}\,\mathbf{F}\cdot d\mathbf{S} = \int_0^1\!\int_0^1 2z\, dz\,dx = \int_0^1 2z\, dz = 1. \]

Putting these all together we have \(\int\!\!\int_S \text{Curl}\,\mathbf{F}\cdot d\mathbf{S} = 0+0+0+0+1 = 1\), as expected, thereby confirming Stokes' Theorem.

Finally, an important consequence of Stokes' Theorem is that the surface integrals of \(\text{Curl}\,\mathbf{F}\) over two surfaces sharing a common oriented boundary are the same. Let \(S_0\) be the left face of the square in the diagram, so that \(S\) and \(S_0\) share the same oriented boundary.

\(S_0\) is given by \((x,0,z)\), with \(0\le x\le 1\), \(0\le z\le 1\) and \(\mathbf{n}=\vj\). Note that \(\vj\) is the correct normal when considering \(S_0\) as an open, oriented surface with boundary \(C\). If we were considering \(S_0\) as the 6th side of the cube, we would take \(-\vj\) as the unit normal. \(\text{Curl}\,\mathbf{F}\) on \(S_0\) is \((0,2z,0)\), so \(\text{Curl}\,\mathbf{F}\cdot\mathbf{n} = 2z\). Therefore,

\[ \int\!\!\int_{S_0} \text{Curl}\,\mathbf{F}\cdot d\mathbf{S} = \int_0^1\!\int_0^1 2z\, dz\,dx = \int_0^1 2z\, dz = 1 = \int\!\!\int_S \text{Curl}\,\mathbf{F}\cdot d\mathbf{S}. \]

Problem 16

16. Calculate, without using Stokes' Theorem, \(\int\!\!\int_{S_1} \nabla\times\mathbf{F}\cdot d\mathbf{S}\), for \(\mathbf{F} = (3y^2+2y)\vi+3z^2\vj+3x^2\vk\) and \(S_1\) the inverted cone \(z=1-\sqrt{x^2+y^2}\), with vertex \((0,0,1)\), and \(z\ge 0\). Then calculate directly \(\int\!\!\int_{S_2} \nabla\times\mathbf{F}\cdot d\mathbf{S}\), for \(S_2\) the unit disk in the \(xy\)-plane. The answers you get should be the same. This shows the consequence of Stokes' Theorem, that surface integrals of the curl of a vector field over surfaces sharing the same boundary are independent of the surface.

Solution.
\[ \nabla\times\mathbf{F} = \begin{vmatrix} \vi & \vj & \vk \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ 3y^2+2y & 3z^2 & 3x^2 \end{vmatrix} = (-6z,-6x,-6y-2). \]
If we integrate directly over the cone, we use the parametrization \(G(u,v) = (v\cos(u), v\sin(u), 1-v)\), with \(0\le u\le 2\pi\), \(0\le v\le 1\). It follows that
\[ \nabla\times\mathbf{F}(G(u,v)) = (-6(1-v), -6v\cos(u), -6v\sin(u)-2). \]
Moreover, \(\mathbf{T}_u\times\mathbf{T}_v = (v\cos(u), v\sin(u), -v)\). Note that this vector has a negative \(z\)-component, and thus is inside of the inverted cone. If we flatten the cone, this would point downward, and contradict the right hand thumb rule. Thus, we need to use the vector \(-(\mathbf{T}_u\times\mathbf{T}_v)\) when integrating \(\nabla\times\mathbf{F}\). (An important point: We could also parametrize the inverted cone using \(H(u,v) = (u,v,1-\sqrt{u^2+v^2})\), with \(0\le u^2+v^2\le 1\), and in this case \(\mathbf{T}_u\times\mathbf{T}_v\) gives the correct normal vector.)

We now have

\[ \{\nabla\times\mathbf{F}(G(u,v))\}\cdot -(\mathbf{T}_u\times\mathbf{T}_v) = 6v(1-v)\cos(u)+6v^2\sin(u)\cos(u)-6v^2\sin(u)-2v. \]
When we calculate \(\int\!\!\int_S (\nabla\times\mathbf{F})\cdot d\mathbf{S}\), first integrating with respect to \(u\), the trig terms integrated from 0 to \(2\pi\) all become 0. We are left with,
\[ \int_0^1\!\int_0^{2\pi} -2v\, du\,dv = 2\pi\int_0^1 -2v\, dv = -4\pi\cdot \frac{v^2}{2}\Big|_0^1 = -2\pi. \]

To integrate over the disk we have \(G(u,v) = (u,v,0)\) with \(0\le u^2+v^2\le 1\) and \(\mathbf{T}_u\times\mathbf{T}_v = (0,0,1)\). \(\nabla\times\mathbf{F}(G(u,v)) = (0,-6u,-6v-2)\), so \(\nabla\times\mathbf{F}(G(u,v))\cdot \mathbf{T}_u\times\mathbf{T}_v = -6v-2\). Thus,

\[ \begin{align*} \int\!\!\int_{S_2} (\nabla\times\mathbf{F})\cdot d\mathbf{S} &= \int\!\!\int_{S_2} -6v-2\, dA \\ &= \int_0^{2\pi}\!\int_0^1 (-6r\sin(\theta)-2)\, r\,dr\,d\theta \\ &= \int_0^{2\pi}\!\int_0^1 -6r^2\sin(\theta)-2r\, dr\,d\theta \\ &= \int_0^{2\pi} -2\sin(\theta)-1\, d\theta \\ &= -2\pi, \end{align*} \]
which is what we want.