# Stokes' theorem gives different results

• lriuui0x0
In summary, we discussed the calculation of the surface flux of a curl of a vector field over a given surface in the form of a function in three-dimensional space. We used Stokes's theorem to relate the surface integral to a line integral along the boundary of the surface, but obtained different results. Upon further inspection, we found that a mistake was made in the change of variables, which resulted in an incorrect Jacobian factor. Correcting the mistake led to the correct calculation of the surface integral, validating Stokes's theorem.
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}

lriuui0x0 said:
\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$$\int_0^{2\pi} d\theta \int_1^2 dr \left[2r^6\cos\theta -20r^4\cos\theta +50r^2\cos\theta + 2r^3\cos^2\theta + r^3- 8r \right] \\$$

Orodruin
ergospherical said:
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!

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*}

Orodruin

## 1. What is Stokes' theorem?

Stokes' theorem is a fundamental theorem in vector calculus that relates the surface integral of a vector field over a closed surface to the line integral of the same vector field around the boundary of the surface.

## 2. How does Stokes' theorem give different results?

Stokes' theorem can give different results when applied to different types of surfaces or when using different orientations for the surface and the boundary. It is important to carefully consider the orientation and type of surface being used in order to obtain accurate results.

## 3. Can you provide an example of when Stokes' theorem gives different results?

One example is when using a surface with a hole in it. If the surface is oriented in one direction, the line integral around the boundary will include the hole, whereas if the surface is oriented in the opposite direction, the hole will be excluded from the line integral. This can result in different values for the surface integral.

## 4. How can I ensure that I am using Stokes' theorem correctly?

To ensure that you are using Stokes' theorem correctly, it is important to carefully consider the orientation and type of surface being used, as well as the direction of the line integral around the boundary. Double-checking your calculations and using multiple methods to verify your results can also help to ensure accuracy.

## 5. Are there any limitations to Stokes' theorem?

Stokes' theorem is limited to vector fields in three-dimensional space and closed surfaces with smooth boundaries. It also assumes that the vector field is continuously differentiable within the region enclosed by the surface. These limitations should be taken into consideration when applying Stokes' theorem to a problem.

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