Solving Surface Integral: Find I Over Half Sphere A

[flaxstrax]

Homework Statement

Find integral I = ∫∫xz^2 dydz + (x^2y − z^3) dzdx + (2xy + y^2z) dxdy (Integrate over A)
if A is half a sphere(radius is a). Sphere is given with equation z=(a-x^2-y^2)^1/2 and z=0.

Homework Equations

The excercise is in 2 parts , find it with just integrating and b) applying gauss's law.

The Attempt at a Solution

I just can't understand how i get it ... Every way i can think of , gives me wrong answer, I have to find scalar. If i substitute x from the sphere equation , then in integration bounds it still remains ?
I know this aint much to go on but help me. Just tell me what i can substitute so i can find this ingtegral or is it even possible ? I don't need whole excercise, i can integrate myself.
sorry for bad english.

 

[HallsofIvy]

I would do this as three separate integrals: Doing the "dydz" integral y will go from -a to a and, for each y, z will go from 0 to $\sqrt{a^2- y^2}$. And, of course, for each y and z, $x= \sqrt{a^2- y^2- z^2}$. The first integral is
$$\int\int xz^2 dydz= \int_{y=-a}^a\int_{z= 0}^\sqrt{a^2- y^2} z^2\sqrt{a^2- y^2- z^2}dzdy$$
and similarly for the other two integrals.

 

[flaxstrax]

Okay i have 2 solutions. Gauss divergence is correct but the other one is false. In Gauss divergence i took φ bounds incorrectly , its actually 0 to pi/2 , that gives me 2/5 * pi * a^5 . Thats correct answer. But i fail somewhere in direct integrating ...
Can someone tell me where i fail?

http://www.upload.ee/image/3301562/20130510_221421.jpg

http://www.upload.ee/image/3301570/20130510_221443.jpg

 

[flaxstrax]

Did it with spherical coordinates now . Tthis is so impossible :(
http://www.upload.ee/image/3302135/20130511_041747.jpg
http://www.upload.ee/image/3302136/20130511_041718.jpg