# Stoke's Theorem

• Apr 11th 2008, 06:20 AM
prescott2006
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.
• Apr 11th 2008, 10:30 PM
TheEmptySet
Quote:

Originally Posted by prescott2006
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

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

$\vec F(x,y,z)=$

then

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

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

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

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

$-\int_{0}^{2\pi}\frac{3}{4}+\frac{1}{4}\cos(4t)dt =-\frac{3}{4}t-\frac{1}{16}\sin(4t)|_{0}^{2\pi}=-\frac{3}{2}\pi$
• Apr 11th 2008, 11:49 PM
prescott2006
$

\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
$

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)?
• Apr 12th 2008, 12:37 AM
TheEmptySet
Quote:

Originally Posted by prescott2006
$

\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
$

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

Attachment 5813

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