# Thread: verification of Stoke's Theorem

1. ## verification of Stoke's Theorem

Hello everybody,
I am stuck on my homework. This question is giving me a headache, I've literally been staring at it for over 30 minutes. Perhaps someone here may be able to answer it (I hope), any help will be greatly appreciated:

Legend:
S= integral
SS= double integral
sub= boundary/limit a
#^n= number raised to exponent n
*= multiply

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Let S be the portion of the paraboloid: z = 9 - x^2 - y^2 which lies above the plane: z= 5, with normal vector N pointing upward (i.e. in the direction of increasing z), and let F be the vector field given by: F(x,y,z)= (x-yz)i + xzj. Verify Stoke's Theorem by evaluation each of the following:

a) The line integral: S sub c (F dr), where C is the boundary of s, oriented counter-clockwise relative to N.

b) The flux integral: SS sub s (curl F) * N dS

2. Here is some starting work:

$z=9-x^2-y^2=5\quad\mbox{when}\quad x^2+y^2=2^2$

\begin{aligned}
\mathbf{r}(t)&=2\cos t\mathbf{i}+2\sin t\mathbf{j}+5\mathbf{k}\\
\mathbf{r'}(t)&=-2\sin t\mathbf{i}+2\cos t\mathbf{j}\\
\mathbf{F}(\mathbf{r}(t))&=(2\cos t-10\sin t)\mathbf{i}+10\cos t\mathbf{j}\\
\oint_C\mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r'}(t )\,dt&=\int_0^{2\pi}(-4\sin t\cos t+20\sin^2 t+20\cos^2 t)\,dt\\
&=\int_0^{2\pi}(20-4\sin t\cos t)\,dt.
\end{aligned}

Also,

$\int\int_S \mbox{curl } \mathbf{F}\cdot\mathbf{n}\,dS=\int\int_R\left|
\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k}\\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z}\\
x-yz & xz & 0
\end{array}\right|
\cdot\left(-\frac{\partial}{\partial x}\mathbf{i}-\frac{\partial}{\partial y}\mathbf{j}+\mathbf{k}\right)\,dA$

$=\int\int_R (-x\mathbf{i}-y\mathbf{j}+2z\mathbf{k})\cdot(2x\mathbf{i}+2y\mat hbf{j}+\mathbf{k})\,dA$
$=\int\int_R (-2x^2-2y^2+2(9-x^2-y^2))\,dA$
$=\int_0^{2\pi}\int_0^2 (18-4r^2)\,r\,dr\,d\theta.$

We obtain the same answer for each.

Edit: Put in some equals signs, made $k$ a vector.

3. ## thanks scott

thank you scott. I really appreciate it. not only do i have a starting point but I now also can type in integrals

4. Originally Posted by Scott H
Here is some starting work:

$z=9-x^2-y^2=5\quad\mbox{when}\quad x^2+y^2=2^2$

\begin{aligned}
\mathbf{r}(t)&=2\cos t\mathbf{i}+2\sin t\mathbf{j}+5\mathbf{k}\\
\mathbf{r'}(t)&=-2\sin t\mathbf{i}+2\cos t\mathbf{j}\\
\mathbf{F}(\mathbf{r}(t))&=(2\cos t-10\sin t)\mathbf{i}+10\cos t\mathbf{j}\\
\oint_C\mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r'}(t )\,dt&=\int_0^{2\pi}(-4\sin t\cos t+20\sin^2 t+20\cos^2 t)\,dt\\
&=\int_0^{2\pi}(20-4\sin t\cos t)\,dt.
\end{aligned}

Also,

$\int\int_S \mbox{curl } \mathbf{F}\cdot\mathbf{n}\,dS=\int\int_R\left|
\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k}\\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z}\\
x-yz & xz & 0
\end{array}\right|
\cdot\left(-\frac{\partial}{\partial x}\mathbf{i}-\frac{\partial}{\partial y}\mathbf{j}+k\right)\,dA$

$=\int\int_R (-x\mathbf{i}-y\mathbf{j}+2z\mathbf{k})\cdot(2x\mathbf{i}+2y\mat hbf{j}+k)\,dA$
$\int\int_R (-2x^2-2y^2+2(9-x^2-y^2))\,dA$
$\int_0^{2\pi}\int_0^2 (18-4r^2)\,r\,dr\,d\theta.$

We obtain the same answer for each.
Hello again scott

Perhaps you (or anybody) may be able to tell me if I have the correct answer. I got 40 pi for both of the equations. Are these the correct answers for the problem? There is no answer in the back of the book since its an even numbered problem.

5. $40\pi$ is correct.

6. ## problem completed?

so 40pi is the answer to both questions? You also said to "Put in some equals signs, made k a vector." Is there more to this problem? Because I don't know how to do that I asked around and so far, nobody in my class has been able to solve this problem

7. Yes, the answer is $40 \pi$ in both cases.

Stokes' theorem states that

$\oint_C \mathbf{F}(\mathbf{r})\cdot\mathbf{dr}=\int\int_S \mbox{curl }\mathbf{F}\cdot\mathbf{n}\,dS.$

In our case,

$\oint_C \mathbf{F}(\mathbf{r})\cdot\mathbf{dr}=40\pi = \int\int_S \mbox{curl }\mathbf{F}\cdot\mathbf{n}\,dS.$

I put the Edit message in to show you that I've corrected some of my typos.