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Math Help - verification of Stoke's Theorem

  1. #1
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    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

    -----------------------------------------------------------------------------------------------------------------
    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
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  2. #2
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    Here is some starting work:

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

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

    Also,

    \int\int_S \mbox{curl } \mathbf{F}\cdot\mathbf{n}\,dS=\int\int_R\left|<br />
\begin{array}{ccc}<br />
\mathbf{i} & \mathbf{j} & \mathbf{k}\\<br />
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z}\\<br />
x-yz & xz & 0<br />
\end{array}\right|<br />
\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.
    Last edited by Scott H; December 7th 2009 at 06:03 PM.
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  3. #3
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    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
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  4. #4
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    Quote Originally Posted by Scott H View Post
    Here is some starting work:

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

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

    Also,

    \int\int_S \mbox{curl } \mathbf{F}\cdot\mathbf{n}\,dS=\int\int_R\left|<br />
\begin{array}{ccc}<br />
\mathbf{i} & \mathbf{j} & \mathbf{k}\\<br />
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z}\\<br />
x-yz & xz & 0<br />
\end{array}\right|<br />
\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.
    Last edited by mohammadmurtaza; December 6th 2009 at 11:42 PM. Reason: grammar
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  5. #5
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    40\pi is correct.
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  6. #6
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    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
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  7. #7
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    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.
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