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Thread: [SOLVED] Stokes' Theorem

  1. #1
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    [SOLVED] Stokes' Theorem

    Evaluate $\displaystyle \oint_{C}(y^3+cosx)dx+(siny+z^2)dy+xdz$, where C is the closed curve parameterized and oriented by the path $\displaystyle \vec{x}(t)=(cost,sint,sin2t)$.

    Hint: Show that the path $\displaystyle \vec{x}(t)$ lies on the surface $\displaystyle z=2xy$.
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  2. #2
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    Quote Originally Posted by davesface View Post
    Evaluate $\displaystyle \oint_{C}(y^3+cosx)dx+(siny+z^2)dy+xdz$, where C is the closed curve parameterized and oriented by the path $\displaystyle \vec{x}(t)=(cost,sint,sin2t)$.
    "Parameterized", yes, but this doesn't give an orientation. Are we to assume that t is going from 0 to $\displaystyle 2\pi$? That would give an orientation.

    [/quote]Hint: Show that the path $\displaystyle \vec{x}(t)$ lies on the surface $\displaystyle z=2xy$.[/QUOTE]
    Have you done the "hint"? Do you know any identity for "sin(2t)"?

    Stokes' theorem, since you title this thread "Stokes' Theorem", in its simplest form says that $\displaystyle \vec{f}(x,y,z)$ is a vector valued function then $\displaystyle \oint_C \vec{f}\cdot d\vec{\sigma}= \int\int \nabla\times d\vec{S}$.

    Here, $\displaystyle \vec{f}(x,y,z)= (y^3+ cos(x))\vec{i}+ (sin(y)+ z^2)\vec{j}+ x\vec{k}$ and $\displaystyle d\vec{\sigma}$ is the differential tangent vector, $\displaystyle (-sin(t)\vec{i}+ cos(t)\vec{j}+ 2cos(2t)\vec{k})dt$.

    Since you know that the path lies entirely in the plane z= 2xy, you can write that as the position vector $\displaystyle \vec{r}(x,y)= x\vec{i}+ y\vec{j}+ 2xy\vec{k}$. The two derivative vectors, $\displaystyle \vec{r}_x(x,y)= \vec{i}+ 2y\vec{k}$ and $\displaystyle \vec{r}_y(x,y)= \vec{j}+ 2x\vec{i}$ are in the tangent plane to the surface (which here is the plane itself) and their cross product, $\displaystyle \left|\begin{array}{ccc}\vec{i} & \vec{j} & \vec{k} \\ 1 & 0 & 2y \\ 0 & 1 & 2x\end{array}\right|= -2y\vec{i}- 2x\vec{j}+ \vec{k}$ is normal to the plane and its length gives the "differential of surface area". Assuming that the path integral was to be from 0 to $\displaystyle 2\pi$ (rather than from 0 to $\displaystyle -2\pi$), the normal is to be oriented by positive z so $\displaystyle d\vec{S}= (-2y\vec{i}- 2\vec{j}+ \vec{k})dy dx$.

    You still need to determine the bounds on x and y.
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    I don't know what to say about the orientation. I've posted all the information given about the problem. I'm guessing it's the negative orientation, though, since the given answer is $\displaystyle \frac{-3\pi}{4}$.

    I have no clue how they got that, as my work has boiled down to $\displaystyle \int\int_{S} (\nabla \times F) \cdot \vec{N} dxdy= \int \int_{D}(8x^2+2x-3y^2)dxdy$ by taking the cross product and dot product where indicated and then substituting in 2xy for the z value that showed up from the dot product.

    What am I missing here?
    Last edited by davesface; May 4th 2010 at 04:09 PM. Reason: Sign error
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  4. #4
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    Wow, you know what? It's way easier than I made it seem. All you have to do is convert it to polar and reverse the sign on the integral. r goes from 0 to 1 and theta goes from 0 to 2pi. Thanks for the help.
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