jeanie wrote

Okay, the "differential of surface area" for the unit sphere is or, in your symbols,Hi, just having some issues with this problem...

Integral over the unit sphere of (x^2 + y^2 - 2z^2) dA

I changed all the x, y, z into spherical coordinates (using a and b for the angles) with r = 1

Calculated the determinant to be (sin a)

Which is the same as (sin a)sin^2(a)+ cos^2(a)- 3 cos^2(a)= 1- 3cos^2(a)[/tex]. The way I looked at it was x^2+ y^2- 2z^2= x^2+ y^2+ z^2- 3z^2= 1- 3 cos^2(a).I set up my work like this:

double integral of (sin a) ((sin a)^2 - 2(cos a)^2) db * da

No, your setup is correct so you must have integrated wrong- your error is in the part you did did not post.taking b from 2pi to 0, and a from pi to 0

Skipping the work I did (I presume I did something wrong in the setup...)

My answer came out to be

2pi * ((cos pi)^3 - cos pi - (cos 0)^3 + cos 0)

But this gives me zero. I don't understand...

What? That's non-sense. You are asked to find an integral over the unit sphere. The answer must be a number, not a function of a (or .The answer given is

(2pi/3) * ((1 + a^2)^(3/2) - 1)

Please go back, read the problem again and postI suppose I don't understand something because I'm not sure where this a is supposed to come from//be since its obviously not the same as what I did.exactlywhat it said.

Help appreciated thanks!