1. ## Stoke's Theorem

Q:let S be the surface x^2+y^2+z^2=1,z>=0.Use stokes theorem to evaluate
line integral of (2x-y)dx-ydy-zdz
where line integral is around circle x^2+y^2=1,z=0,oriented anti clockwise.

2. See attachment

3. Originally Posted by Calculus26
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every thing is clear to me except,how you calculate normal N

4. Originally Posted by Calculus26
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This answer of $\pi$ can't be right. If you calculate directly

$\int_c (2x-y)\,dx - y \,dy - z \,dz$ the vector field ${\bf F} = <2x-y, -y, -z>$ is conservative and hence the line integral is zero!

5. No the curlF is not 0 therefore the vector field is not conservative

If it were what would the potential function be

6. Originally Posted by danny arrigo
This answer of $\pi$ can't be right. If you calculate directly

$\int_c (2x-y)\,dx - y \,dy - z \,dz$ the vector field ${\bf F} = <2x-y, -y, -z>$ is conservative and hence the line integral is zero!
sorry F is not conservative?????

7. No F is not -- the condition is If F is conservative curl F = 0

If you calculate the line integral directly

z= 0 so F = (2x-y)i -yj

note this fails the test for conservative as well since d(2x-y)/dy does not equal d(-y)/dx

If you parameterize C by x =cos(t) y = sin(t)

you will also get pi

8. Originally Posted by Mathventure
sorry F is not conservative?????
Originally Posted by Calculus26
No F is not -- the condition is If F is conservative curl F = 0

If you calculate the line integral directly

z= 0 so F = (2x-y)i -yj

note this fails the test for conservative as well since d(2x-y)/dy does not equal d(-y)/dx

If you parameterize C by x =cos(t) y = sin(t)

you will also get pi
Your both right - how silly of me.