Stoke's Theorem Question with plane and sphere

[mlb2358]

Homework Statement

Let F =< yz + x, xz + 2x, xy + 3x >. Evaluate ∫F·dr where C is the intersection of the plane
2x + y − 3z = 0 and the sphere x²+ y² + z² = 4 oriented positively when viewed from above.

Homework Equations

The Attempt at a Solution

The main question I have about this problem regards the normal vector of the surface. Because the surface is just an isolated portion of the plane, I thought I could use the normal vector of the plane to calculate the surface integral. However, when I try this I don't get the right answer. Can someone explain why this shouldn't work and what the right way would be to find the normal vector?
Thanks!

 

[HallsofIvy]

You have misunderstood the problem. You don't have a "normal vector to the surface" because you don't have a surface! You are asked to integrate around the intersection of the given plane and surface- and that is a curve, not a surface.

The simplest thing to do, I think, is to solve the equation of the plane for y: y= 3z- 2x. Putting that into the equation of the sphere we have [itex]x^2+ (3z- 2x)^3+ z^2= x^2+ 9z^2- 12xz+ 4x^2+ z^2= 5x^2- 12xz+ 10z^2= 4[/itex] which is the equation of a circle in the xz-plane. That can be written in parametric equations and then you can integrate around the circle.

 

[mlb2358]

I thought that by Stoke's Theorem ∫F·dr = ∫∫(∇xF)dS, which means that if I take the curl of F, I can use the surface instead of the boundary.

 

[HallsofIvy]

Yes! I just noticed that the title of your thread says "Stokes' theorem"! Sorry about that. You want to, then, integrate over the surface of the sphere. The differental of surface area on a sphere is [itex]\rho^2 sin(\phi)d\theta d\phi[/itex] and the "unit normal" will be [itex](1/\sqrt{x^2+ y^2+ z^2})(\vec{i}+ \vec{j}+ \vec{k})[/itex]. The "hard part" of this problem, I think, will be finding the limits of integration. In spherical coordinates, [itex]x= \rho cos(\theta) sin(\phi)[/itex], [itex]y= \rho sin(\theta) sin(\phi)[/itex], and [itex]z= \rho cos(\phi)[/itex]. On this sphere, of course, [itex]\rho= 2[/itex]. putting those into the equation of the plane give [itex]4cos(\theta)sin(\phi)+ 2sin(\theta)sin(\phi)- 6cos(\phi)= 0[/itex] so that [itex]6cos(\phi)= (4cos(\theta)+ 2sin(\theta))sin(\phi)[/itex] or [itex]tan(\phi)= (2/3)cos(\theta)+ (1/3)sin(\theta)[/itex]. Your integral should take [itex]\phi[/itex] from 0 to that and [itex]\theta[/itex] from 0 to [itex]2\pi[/itex].