Fall 2025 Math 147: Quiz 12 Solutions

Problem 1

1. Use a line integral to find the area contained in the ellipse \(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1\). (5 points)

Solution. This problem relies upon Green's theorem. We use the vector field \(\mathbf{F}(x,y) = \left(-\dfrac{y}{2}, \dfrac{x}{2}\right)\), as this has \(\text{Curl}\,\mathbf{F} = 1\). Then, applying Green's theorem, one has

\int_C \mathbf{F}\, d\mathbf{r} = \int\!\!\int_S \text{Curl}\,\mathbf{F}\cdot dA = \int\!\!\int_S dA = \text{Area}(S).

Thus, we calculate this line integral. We parameterize the ellipse in the standard way, that is \(\mathbf{r}(t) = (a\cos t, b\sin t)\), where \(t\) ranges from 0 to \(2\pi\). Note that this gives us \(\mathbf{r}'(t) = (-a\sin t, b\cos t)\). Then, \(\mathbf{F}(\mathbf{r}(t)) = \left(\dfrac{-b\sin t}{2}, \dfrac{a\cos t}{2}\right)\). We now have

\int_C \mathbf{F}\cdot d\mathbf{s} = \int_0^{2\pi} \left(\frac{-b\sin t}{2}, \frac{a\cos t}{2}\right)\cdot(-a\sin t, b\cos t)\, dt = \int_0^{2\pi} \frac{ab\sin^2 t}{2}+\frac{ab\cos^2 t}{2}\, dt = \int_0^{2\pi} \frac{ab}{2}\, dt = \pi ab.

Problem 2

2. Set \(\mathbf{F} = z^2\vi+x^2\vj-y^2\vk\). (5 points)

(i) Calculate \(\nabla\times\mathbf{F}\).
(ii) Let \(C\) be the square path with sides equal to \(a\) centered at the point \((x_0,0,z_0)\) lying in the \(xz\)-plane, oriented so that each side is parallel to the \(x\) or \(z\) axis. Calculate \(\displaystyle\int_C \mathbf{F}\cdot d\mathbf{r}\).
(iii) Divide your answer in (ii) by the area of the square and take the limit as \(a\) goes to zero.
(iv) Use your answer in (i) to corroborate your answer in (iii).

Solution. (i)

\nabla\times\mathbf{F} = \begin{vmatrix} \vi & \vj & \vk \\ d/dx & d/dy & d/dz \\ z^2 & x^2 & -y^2 \end{vmatrix} = (-2y, 2z, 2x).

(ii) We break down the path \(C\) into 4 separate lines for ease of calculation. Let \(C_1(t) = (x_0+\tfrac{a}{2}, 0, z_0+at)\), \(C_2(t) = (x_0-at, 0, z_0+\tfrac{a}{2})\), \(C_3(t) = (x_0-\tfrac{a}{2}, 0, z_0-at)\), and \(C_4(t) = (x_0+at, 0, z_0-\tfrac{a}{2})\), all individually with domain \(-1/2\le t\le 1/2\). Note this parameterization keeps the surface positively oriented. Note that for all of these lines, \(\mathbf{r}'(t) = (0,0,\pm a)\) or \((\pm a, 0, 0)\). Now, we can integrate over the path. One has

\begin{align*} \int_C \mathbf{F}\cdot d\mathbf{r} &= \sum_{i=1}^4 \int_{C_i} \mathbf{F}\cdot d\mathbf{r} \\ &= \int_{-1/2}^{1/2} \left((z_0+at)^2, \left(x_0+\tfrac{a}{2}\right)^2, 0\right)\cdot(0,0,a)\, dt + \int_{-1/2}^{1/2} \left(\left(z_0+\tfrac{a}{2}\right)^2, (x_0-at)^2, 0\right)\cdot(-a,0,0)\, dt \\ &\quad + \int_{-1/2}^{1/2} \left((z_0-at)^2, \left(x_0-\tfrac{a}{2}\right)^2, 0\right)\cdot(0,0,-a)\, dt + \int_{-1/2}^{1/2} \left(\left(z_0-\tfrac{a}{2}\right)^2, (x_0+at)^2, 0\right)\cdot(a,0,0)\, dt \\ &= \int_{-1/2}^{1/2} -a\left(z_0+\tfrac{a}{2}\right)^2 dt + \int_{-1/2}^{1/2} a\left(z_0-\tfrac{a}{2}\right)^2 dt \\ &= -a\left(z_0^2+z_0a+\frac{a^2}{4}\right) + a\left(z_0^2-z_0a+\frac{a^2}{4}\right) \\ &= -2z_0a^2. \end{align*}

(iii) Dividing by the area gives \(-2z_0\), and taking the limit keeps it as \(-2z_0\).

(iv) The process of surrounding a point with a path, taking the line integral of said path, dividing by the area, and then using limits to send the area to zero, is precisely the definition of the curl of a vector field, along a vector normal to the surface enclosed. As the normal vector to our square is \((0,1,0)\), it is expected that our answer in part (iii) is the \(y\)-component of the curl calculated in part (i), once the point \((x_0,0,z_0)\) is substituted in.