- #1

lriuui0x0

- 101

- 25

$$

z = 5-x^2-y^2, 1<z<4

$$

For a vector field ##\mathbf{A} = (3y, -xz, yz^2)##. I'm trying to calculate the surface flux of the curl of the vector field ##\int \nabla \times \mathbf{A} \cdot d\mathbf{S}##. By Stokes's theorem, this should be equal the the line integral of the vector field on the boundary of the surface. But I got different result. I'm sure I made some mistakes somewhere, but I couldn't spot.

The curl ##\nabla \times \mathbf{A}## is:

$$

\begin{aligned}

&\phantom{{}={}} \nabla \times \mathbf{A} \\

&= \begin{pmatrix}\frac{\partial yz^2}{\partial y} - \frac{\partial -xz}{\partial z} \\ \frac{\partial 3y}{\partial z} - \frac{\partial yz^2}{\partial x} \\ \frac{\partial -xz}{\partial x} - \frac{\partial 3y}{\partial y} \\ \end{pmatrix} \\

&= \begin{pmatrix}z^2 + x \\ 0 \\ -z-3\end{pmatrix} \\

&= \begin{pmatrix}r^4-10r^2+25 + r\cos\theta \\ 0 \\ r^2-8\end{pmatrix}

\end{aligned}

$$

The surface normal is:

$$

\begin{aligned}

&\phantom{{}={}} \frac{\partial}{\partial x}\begin{pmatrix}x\\y\\5-x^2-y^2\end{pmatrix} \times \frac{\partial}{\partial y}\begin{pmatrix}x\\y\\5-x^2-y^2\end{pmatrix} \\

&= \begin{pmatrix}1\\0\\-2x\end{pmatrix} \times \begin{pmatrix}0\\1\\-2y\end{pmatrix} \\

&= \begin{pmatrix}2x\\2y\\1\end{pmatrix} \\

&= \begin{pmatrix}2r\cos\theta\\2r\sin\theta\\1\end{pmatrix} \\

\end{aligned}

$$

Calculating surface integral:

$$

\begin{aligned}

&\phantom{{}={}} \int_S (\nabla \times \mathbf{A}) \cdot d\mathbf{S} \\

&= \int_0^{2\pi}d\theta \int_1^2 dr \begin{pmatrix}r^4-10r^2+25 + r\cos\theta \\ 0 \\ r^2-8\end{pmatrix} \cdot \begin{pmatrix}2r\cos\theta\\2r\sin\theta\\1\end{pmatrix} \\

&= \int_0^{2\pi}\int_1^2 2r^5\cos\theta -20r^3\cos\theta +50r\cos\theta + 2r^2\cos^2\theta + r^2- 8dr d\theta \\

&= \int_0^{2\pi}\biggl[\frac{1}{3}\cos\theta r^6 -5\cos\theta r^4 +25\cos\theta r^2 +\frac{2}{3}\cos^2\theta r^3 + \frac{1}{3}r^3 -8r\biggr]_1^2 d\theta \\

&= \int_0^{2\pi} \frac{14}{3}\cos^2\theta + 21\cos\theta - \frac{17}{3} d\theta \\

&= \int_0^{2\pi} \frac{7}{3}(1+\cos2\theta) + 21\cos\theta - \frac{17}{3} d\theta \\

&= \biggl[\frac{7}{6}\sin2\theta + 21\sin\theta -\frac{10}{3}\theta\biggr]_0^{2\pi} \\

&= -\frac{20}{3}\pi \\

\end{aligned}

$$

Yet if we calculate through the line integral:

$$

\begin{aligned}

&\phantom{{}={}} \int_{C_1} \mathbf{A} \cdot d\mathbf{C}_1 - \int_{C_2} \mathbf{A} \cdot d\mathbf{C}_2 \\

&= \int_0^{2\pi} \begin{pmatrix}6\sin\theta\\-2\cos\theta\\2\sin\theta\end{pmatrix} \cdot \frac{d}{d\theta} \begin{pmatrix}2\cos\theta\\2\sin\theta\\1\end{pmatrix} d\theta - \int_0^{2\pi} \begin{pmatrix}3\sin\theta\\-4\cos\theta\\16\sin\theta\end{pmatrix} \cdot \frac{d}{d\theta} \begin{pmatrix}\cos\theta\\\sin\theta\\4\end{pmatrix} d\theta \\

&= \int_0^{2\pi} \begin{pmatrix}6\sin\theta\\-2\cos\theta\\2\sin\theta\end{pmatrix} \cdot \begin{pmatrix}-2\sin\theta\\2\cos\theta\\0\end{pmatrix} d\theta - \int_0^{2\pi} \begin{pmatrix}3\sin\theta\\-4\cos\theta\\16\sin\theta\end{pmatrix} \cdot \begin{pmatrix}-\sin\theta\\\cos\theta\\0\end{pmatrix} d\theta \\

&= \int_0^{2\pi} -12\sin^2\theta -4\cos^2\theta +3\sin^2\theta +4\cos^2\theta d\theta \\

&= \int_0^{2\pi} -9\sin^2\theta d\theta \\

&= \int_0^{2\pi} \frac{9(\cos 2\theta - 1)}{2} d\theta \\

&= \frac{9}{2}\biggl[\frac{1}{2}\sin2\theta -\theta\biggr]_0^{2\pi} \\

&= -9\pi \\

\end{aligned}

$$