# Surface integral evaluation

1. Jul 28, 2008

### fk378

1. The problem statement, all variables and given/known data
Evaluate the double integral of yz dS. S is the surface with parametric equations x=(u^2), y=usinv, z=ucosv, 0<u<1, 0<v<(pi/2)

(all the "less than" signs signify "less than or equal to" here)

2. Relevant equations

double integral of dot product of (F) and normal vector over the domain

3. The attempt at a solution
When I solved for the normal vector, I crossed r_u X r_v and got 5u^4.

Then I solved the double integral of (u^4)sinvcosv(5u^4) dudv. u is from 0-->1 and v is from 0-->pi/2

My final answer came out to be pi/12, but it's wrong. Can anyone help?

2. Jul 28, 2008

### Defennder

I didn't get 5u^4.

3. Jul 29, 2008

### Dick

Can I point out that you don't have an F vector? yz is a scalar.

4. Jul 29, 2008

### Defennder

I think he miswrote it. It was probably a scalar surface integral.

5. Jul 29, 2008

### HallsofIvy

Staff Emeritus
Recalculate $|\vec{r}_u\times \vec{r}_v|$ it is not 5u4. I think you have a "u4" at one point where you should have a "u2".