Stokes's Theorem showing 2 surface integrals are equal

[sandylam966]

Homework Statement

Let F = <z,x,y>. The plane D1: z = 2x +2y-1 and the paraboloid D2: z = x^2 + y^2 intersect in a closed curve. Stoke's Theorem implies that the surface integrals of the of either surface is equal since they share a boundary (provided that the orientations match).

Homework Equations

show each of the following integral and show them that they are equal.
∫∫[itex]_{D1}[/itex](∇[itex]\wedge[/itex]F).N dS
∫∫[itex]_{D2}[/itex](∇[itex]\wedge[/itex]F).N dS

The Attempt at a Solution

I first found that the boundary is given by ]r[/U]=(cosθ +1, sinθ +1, 2(sinθ + cosθ) +3), 0<θ<2pi. Then ∫F.dr on this boundary = -3pi. Then I will try to show the 2 surface integrals equal to this.

But I have trouble parameterising the surfaces.
for D1, r = (u cosθ +1, u sinθ +1, 2r(sinθ + cosθ) +3), 0<r<1, 0<θ<2pi
for D2, r = (x, y, x^2+y^2), ,0<x^2+y^2<2x+2y-1

I tried to solve both but got really complicated integrals which I could not solve. Could someone please tell me if I have parameterise them correctly?

 

[LCKurtz]

Homework Statement

Let F = <z,x,y>. The plane D1: z = 2x +2y-1 and the paraboloid D2: z = x^2 + y^2 intersect in a closed curve. Stoke's Theorem implies that the surface integrals of the of either surface is equal since they share a boundary (provided that the orientations match).

Homework Equations

show each of the following integral and show them that they are equal.
∫∫[itex]_{D1}[/itex](∇[itex]\wedge[/itex]F).N dS
∫∫[itex]_{D2}[/itex](∇[itex]\wedge[/itex]F).N dS

I presume the ##\wedge## symbol is supposed to be ##\times##, right?

The Attempt at a Solution

I first found that the boundary is given by ]r[/U]=(cosθ +1, sinθ +1, 2(sinθ + cosθ) +3), 0<θ<2pi. Then ∫F.dr on this boundary = -3pi. Then I will try to show the 2 surface integrals equal to this.

But I have trouble parameterising the surfaces.
for D1, r = (u cosθ +1, u sinθ +1, 2r(sinθ + cosθ) +3), 0<r<1, 0<θ<2pi

Don't use r for two different things. And I suppose the u is supposed to be r? So$$
D_1:~~\vec R = \langle 1 + r\cos\theta, 1 + r\sin\theta, 2r(\sin\theta+\cos\theta)+3\rangle$$

for D2, r = (x, y, x^2+y^2), ,0<x^2+y^2<2x+2y-1

I tried to solve both but got really complicated integrals which I could not solve. Could someone please tell me if I have parameterise them correctly?

You have done a good job choosing your parameters for ##x## and ##y## (for which you didn't show your steps). I would use the same ##r,\theta## parameters for ##D_2##. You might find it easier to calculate the curl in terms of the x and y variables, then make the substitution. Or better yet, use$$
\iint_S \nabla \times \vec F\cdot d\vec S =\pm \iint_{(r,\theta)}\vec F\cdot \vec R_r\times \vec R_\theta~drd\theta$$

 

[sandylam966]

Don't use r for two different things. And I suppose the u is supposed to be r? So$$
D_1:~~\vec R = \langle 1 + r\cos\theta, 1 + r\sin\theta, 2r(\sin\theta+\cos\theta)+3\rangle$$

Yes that's a mistake, u is the radius

You have done a good job choosing your parameters for ##x## and ##y## (for which you didn't show your steps). I would use the same ##r,\theta## parameters for ##D_2##. You might find it easier to calculate the curl in terms of the x and y variables, then make the substitution

So for D2 is (r cosθ, r sinθ, r^2), 0<θ<2pi, 0<r^2<2r(cosθ + sinθ) -1 correct?

 

[LCKurtz]

Yes that's a mistake, u is the radius



So for D2 is (r cosθ, r sinθ, r^2), 0<θ<2pi, 0<r^2<2r(cosθ + sinθ) -1 correct?

That is not the same parameterization of ##x## and ##y## as you used for ##D_1##.

[Edit, added]. For your problem ##\nabla \times \vec F = \langle 1,1,1\rangle##. There is a typo in my last formula in post #2 which is too late for me to edit. All you have to do now is calculate$$
\pm\iint_{(r,\theta)} \langle 1,1,1\rangle \cdot \vec R_r\times \vec R_\theta~drd\theta$$with the proper sign to verify your (correct) answer of ##-3\pi##.