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Math Help - Stoke's Theorem

  1. #1
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    Stoke's Theorem

    F=y^3i-x^3j+z^3k;C the trace of the cylinder x^2+y^2=1 in the plane x+y+z=1 and i arrive at -3integrate 0 to pi/2 integrate 0 to 1 r^4 dr dθ and obtain the answer -3pi/8 but the answer given is -3pi/2.This imply that it integrate from 0 to 2pi.But why this happen since the region is generated from 0 to pi/2 only i think.Thanks.
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  2. #2
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    Quote Originally Posted by prescott2006 View Post
    F=y^3i-x^3j+z^3k;C the trace of the cylinder x^2+y^2=1 in the plane x+y+z=1 and i arrive at -3integrate 0 to pi/2 integrate 0 to 1 r^4 dr dθ and obtain the answer -3pi/8 but the answer given is -3pi/2.This imply that it integrate from 0 to 2pi.But why this happen since the region is generated from 0 to pi/2 only i think.Thanks.
    using stokes theorem we need to find the parametric rep of the space curve

    we know that
    x=\cos(t),y=\sin(t)

    using the equation of the plane we get

    x+y+z=1 \iff z=1-x-y \iff z=1-\cos(t) -\sin(t)

    so finally

    \vec r (t)=<\cos(t),\sin(t),1-\cos(t)-\sin(t)>

    then

    d\vec r (t) =<-\sin(t),\cos(t),\sin(t)-\cos(t)>dt

    \vec F(x,y,z)=<y^3,-x^3,z^3>

    then

    \vec F(r(t)=<\sin^{3}(t),-\cos^{3}(t),(1-\cos^3(t)-\sin^3(t))^3>

    \int_{0}^{2\pi}\vec F \cdot d\vec r =\int_{0}^{2\pi}-\sin^4(t)-\cos^4(t)+(\sin(t)-\cos(t))(1-\cos(t)-\sin(t))^3dt

    -\int_{0}^{2\pi}(\sin^4(t)+\cos^4(t))dt+\underbrace  {\int_{0}^{2\pi}(\sin(t)-\cos(t))(1-\cos(t)-\sin(t))^3dt}_{=0(WHY?)}

    -\int_{0}^{2\pi}\frac{1}{4}(1-\cos(2t))^2+\frac{1}{4}(1+\cos(2t))^2dt

    -\int_{0}^{2\pi}\frac{3}{4}+\frac{1}{4}\cos(4t)dt =-\frac{3}{4}t-\frac{1}{16}\sin(4t)|_{0}^{2\pi}=-\frac{3}{2}\pi
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    <br /> <br />
\int_{0}^{2\pi}\vec F \cdot d\vec r =\int_{0}^{2\pi}-\sin^4(t)-\cos^4(t)+(\sin(t)-\cos(t))(1-\cos(t)-\sin(t))^3dt<br />
    why you integrate from 0 to 2 pi and not from 0 to pi/2?since C trace the plane x+y+z=1 from 0 to pi/2 only.

    By the way,can you show me the way to evaluate it in (double integral curlF.n ds)?
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    Quote Originally Posted by prescott2006 View Post
    <br /> <br />
\int_{0}^{2\pi}\vec F \cdot d\vec r =\int_{0}^{2\pi}-\sin^4(t)-\cos^4(t)+(\sin(t)-\cos(t))(1-\cos(t)-\sin(t))^3dt<br />
    why you integrate from 0 to 2 pi and not from 0 to pi/2?since C trace the plane x+y+z=1 from 0 to pi/2 only.

    By the way,can you show me the way to evaluate it in (double integral curlF.n ds)?
    This should answer you question. Did you sketch the surface.

    The integral should go from 0 to 2 Pi

    Here is a graph of the surface and its projection into the xy plane

    Stoke's Theorem-capture.jpg

    your integral should be

    -3\int_{0}^{2\pi}\int_{0}^{1}r^3drd\theta
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