Calculating doubleintM (∇×F) ·dS

[Borat321]

Homework Statement

Let M be the capped cylindrical surface which is the union of two surfaces, a cylinder given by x^2 + y^2 = 9, 0 ≤ z ≤ 1, and a hemispherical cap defined by x^2 + y^2 + (z−1)^2 = 9, z ≥ 1. For the vector field F = (zx + z^2y + 2 y, z^3yx+ 8 x, z^4x^2), compute doubleintM (∇×F) ·dS in any way you like.

doubleintM (∇×F) ·dS = ?

Homework Equations

I thought the line integral would make the most sense in solving this - I wanted to take the line integral with respect to the bottom circle. parameterized, it is (3costheta, 3sintheta, 1).

Line int = F(r) (rprime)

The Attempt at a Solution

I have no idea what's going on here - I tried to take the line integral by saying r=(3sint,3cost,0) Since z=0 for circle the only parts that matter in the F is the 2y and the 8x, but I don't think that's right.

Integral w/ limits 0 to 2pi

(2(3sintheta),8(3costheta),0) dot product (-3sintheta, 3costheta, 0)

Can anyone help me with this one?

 

[HallsofIvy]

Homework Statement

Let M be the capped cylindrical surface which is the union of two surfaces, a cylinder given by x^2 + y^2 = 9, 0 ≤ z ≤ 1, and a hemispherical cap defined by x^2 + y^2 + (z−1)^2 = 9, z ≥ 1. For the vector field F = (zx + z^2y + 2 y, z^3yx+ 8 x, z^4x^2), compute doubleintM (∇×F) ·dS in any way you like.

doubleintM (∇×F) ·dS = ?

Homework Equations

I thought the line integral would make the most sense in solving this - I wanted to take the line integral with respect to the bottom circle. parameterized, it is (3costheta, 3sintheta, 1).

Not quite. That is the bottom of the hemispherical cap. The bottom of M is the circle in the xy-plane: (3 cos\theta , 3 sin\theta ,0)

Line int = F(r) (rprime)

The Attempt at a Solution

I have no idea what's going on here - I tried to take the line integral by saying r=(3sint,3cost,0) Since z=0 for circle the only parts that matter in the F is the 2y and the 8x, but I don't think that's right.

Oh! Okay, now you moved down to the correct circle!

Integral w/ limits 0 to 2pi

(2(3sintheta),8(3costheta),0) dot product (-3sintheta, 3costheta, 0)

Can anyone help me with this one?

F= (zx+ z^2y+ 2y, z^3yx+8x, z^4x) which, on z= 0 becomes
F= (2y, 8x, 0). Good. That's what you have. But s= (3 cos(\theta), 3 sin(\theta), 0) so ds= (-3 sin(\theta), 3 cos(\theta), 0)d\theta. Your integral should be:
\int_{\theta= 0}^{2\pi}(6sin(\theta ), 24cos(\theta ),0)\cdot (-3sin(\theta ),3cos(\theta ),0)d\theta.
Never forget the "d" in an integral!