# Stokes Theorem;determine double integral!

1. Nov 25, 2012

### Aizek

1. The problem statement, all variables and given/known data
Let S be the surface defined by y=10 -x^2 -z^2 with y≥1, oriented with rightward-pointing normal. Let F=(2xyz+5z)i+ e^x Cos(yz) j +x^2 y k
Determine ∫∫s ∇×F dS (Hint: you will need an indirect approach)

2. Relevant equations
Stokes Theorem ∫∫s ∇×F dS

3. The attempt at a solution
If im correct, there is a rightward-pointing normal given by N=(2xi,1j,2zk) ( but Im not quite shure why, does it matter that the normal is rightward-pointing?) . And We have a radius of 3.

I find the ∇×F by cross derivation an get:
∇×F =(x^2 +e^x ysin(yz))i, 5j, (e^x Cos(yz)-2xz) k

I then multiply them together with the Stokes theorem

∫∫s ∇×F dS =∫∫(x^2 +e^x ysin(yz))i, 5j, (e^x Cos(yz)-2xz ) k $\ast$ (2xi,1j,2zk) dxdy

=∫∫(2x^3 +2xy e^x Sin(yz))i, 5j,(2z e^x Cos (yz) -4xz^2)k dxdy

This is the point where im not shure what to do. According to an example in my textbook that is simulair,they pick out the "easiest" one of i,j,k. At this example it will be j, that is 5. This is also what they do at the solution manual (that is kind of lacking of information). And the answer only says :∫∫5dxdy =45π. They dont explain why we can pick out one to find the surface integral, has it to do with symmetry of some kind? what is the boarders (why it gets 45π)?

2. Nov 25, 2012

### Staff: Mentor

I would expect that the integrals over i and k cancel due to symmetry, but did not check it. As an example, $\iint -4xz^2 dx dz = \int [-2x^2z^2]_{x=-f(z)}^{x=+f(z)} dz = \int 0 dz = 0$ (dz, not dy!)

Concerning the integral for the j component:
The boundary of your integral is given by y=1, which corresponds to $x^2+z^2=9$. This is a circle of radius 3, with an area of $9\pi$. 5 does not depend on x or z, so the integral is just 5 times the area, of $45 \pi$.