How Do You Apply Stokes' Theorem to Evaluate a Line Integral?

[filter54321]

Homework Statement

Use Stokes's Theorem to evaluate $$\int F · dr$$
In this case, C (the curve) is oriented counterclockwise as viewed from above.

Homework Equations

F(x,y,z) = xyzi + yj + zk, x² + y² ≤ a²
S: the first-octant portion of z = x² over x² + y² = a²

The Attempt at a Solution

"Use Stoke's" is code for "stick the dot product of curl F and the normal vector into the integral". If this problem behaves nicely, this should become a double integral like every other Stoke's problem in the book and will need a new area of integration as well.

curl F would be <0, xy-yz, -xz>
G(x,y) = z-x²
the normal is <-G_x,-G_y,1>
which is <-2x, 0, 1>
F · N = 0+0-xz = -xz because of convenient canceling

So now -xz has to be integrated over some area. The "first octant" is simple enough, and x² + y² ≤ a² is a cylinder of infinite height and radius a centered around the Z-axis (a fixed circle for every Z). I'm picturing a a parabola-cum-trough looking thing that got stamped by a circular cookie cutter.

If that's the case, the x's get restricted which means the z's get restricted to a constant. How do you set up the integral to get rid of all variables (the radius a is a constant and obviously stays undefined).

Thanks!

 

[mighty2000]

Thank you for your post. You are on the right track with your approach to using Stokes's Theorem. To evaluate the integral, you will indeed need to find the curl of F, which you have correctly calculated to be <0, xy-yz, -xz>. However, the normal vector that you have calculated is incorrect. The normal vector should be <2x, 0, 1>, as can be seen by taking the partial derivatives of G(x,y)=z-x^2. You can then take the dot product of F and the normal vector to get -2xz.

To set up the integral, you are correct in thinking that you will need to integrate over the area of the first octant portion of the paraboloid z=x^2 over the circle x^2+y^2=a^2. To do this, you can use polar coordinates. Since the circle is centered at the origin and has a radius of a, you can let x=rcosθ and y=rsinθ, where r is the radius and θ is the angle. The limits of integration for r will be from 0 to a, and the limits for θ will be from 0 to π/2.

Your final integral should look something like this:

∫∫-2xz dA = ∫0^π/2∫0^a -2rcosθsinθ(r^2cosθ) drdθ

I hope this helps! Let me know if you have any further questions or if you need clarification on anything. Good luck with your problem.