How do I evaluate a surface integral with parametric equations?

[fk378]

Homework Statement

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)

Homework Equations

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

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?

 

[Defennder]

When I solved for the normal vector, I crossed r_u X r_v and got 5u^4.

I didn't get 5u^4.

 

[Dick]

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

 

[Defennder]

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

 

[HallsofIvy]

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".