Stokes' theorem gives different results

[lriuui0x0]

Given surface $S$ in $\mathbb{R}^3$:

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

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:

$$
\begin{aligned}
&\phantom{{}={}} \int_{C_1} \mathbf{A} \cdot d\mathbf{C}_1 - \int_{C_2} \mathbf{A} \cdot d\mathbf{C}_2 \\
&= \int_0^{2\pi} \begin{pmatrix}6\sin\theta\\-2\cos\theta\\2\sin\theta\end{pmatrix} \cdot \frac{d}{d\theta} \begin{pmatrix}2\cos\theta\\2\sin\theta\\1\end{pmatrix} d\theta - \int_0^{2\pi} \begin{pmatrix}3\sin\theta\\-4\cos\theta\\16\sin\theta\end{pmatrix} \cdot \frac{d}{d\theta} \begin{pmatrix}\cos\theta\\\sin\theta\\4\end{pmatrix} d\theta \\
&= \int_0^{2\pi} \begin{pmatrix}6\sin\theta\\-2\cos\theta\\2\sin\theta\end{pmatrix} \cdot \begin{pmatrix}-2\sin\theta\\2\cos\theta\\0\end{pmatrix} d\theta - \int_0^{2\pi} \begin{pmatrix}3\sin\theta\\-4\cos\theta\\16\sin\theta\end{pmatrix} \cdot \begin{pmatrix}-\sin\theta\\\cos\theta\\0\end{pmatrix} d\theta \\
&= \int_0^{2\pi} -12\sin^2\theta -4\cos^2\theta +3\sin^2\theta +4\cos^2\theta d\theta \\
&= \int_0^{2\pi} -9\sin^2\theta d\theta \\
&= \int_0^{2\pi} \frac{9(\cos 2\theta - 1)}{2} d\theta \\
&= \frac{9}{2}\biggl[\frac{1}{2}\sin2\theta -\theta\biggr]_0^{2\pi} \\
&= -9\pi \\
\end{aligned}
$$

 

[ergospherical]

\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} \\\end{aligned}

Here you've made a mistake with the change of variables. You have, for the surface $\mathbf{r}(x,y) = (x,y,z(x,y))$,\begin{align*}
d\mathbf{S} = (\mathbf{r}_x \times \mathbf{r}_y) dx dy &= (2x, 2y, 1) \, dx dy \\
&= (2r\cos{\theta}, 2r\sin{\theta}, 1) \, rdrd\theta
\end{align*}in other words the Jacobian factor is $\dfrac{\partial(x,y)}{\partial(r,\theta)} = r$, which you missed out. You ought to be calculating$$\int_0^{2\pi} d\theta \int_1^2 dr \left[2r^6\cos\theta -20r^4\cos\theta +50r^2\cos\theta + 2r^3\cos^2\theta + r^3- 8r \right] \\
$$

 

[lriuui0x0]

Here you've made a mistake with the change of variables. You have, for the surface $\mathbf{r}(x,y) = (x,y,z(x,y))$,\begin{align*}
d\mathbf{S} = (\mathbf{r}_x \times \mathbf{r}_y) dx dy &= (2x, 2y, 1) \, dx dy \\
&= (2r\cos{\theta}, 2r\sin{\theta}, 1) \, rdrd\theta
\end{align*}in other words the Jacobian factor is $\dfrac{\partial(x,y)}{\partial(r,\theta)} = r$, which you missed out. You ought to be calculating$$\int_0^{2\pi} d\theta \int_1^2 dr \left[2r^6\cos\theta -20r^4\cos\theta +50r^2\cos\theta + 2r^3\cos^2\theta + r^3- 8r \right] \\
$$

Got you, thanks!

 

[ergospherical]

No worries
You could check it directly as well by putting $\mathbf{r}(r,\theta) = (r\cos{\theta}, r\sin{\theta}, 5-r^2)$ then\begin{align*}
d\mathbf{S} &= (\mathbf{r}_{r} \times \mathbf{r}_{\theta}) dr d\theta \\
&= (2r^2 \cos{\theta}, 2r^2 \sin{\theta}, r) dr d\theta = (2r \cos{\theta}, 2r \sin{\theta}, 1) r dr d\theta
\end{align*}