# Math Help - Surface Area

1. ## Surface Area

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)

I set up my work like this:
double integral of (sin a) ((sin a)^2 - 2(cos a)^2) db * da
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...

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

I 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.

Help appreciated thanks!

2. ## jeanie's thread on Surface Area

jeanie wrote
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)
Okay, the "differential of surface area" for the unit sphere is $sin(\phi)d\theta\d\phi$ or, in your symbols, $sin(a)dbda$

I set up my work like this:
double integral of (sin a) ((sin a)^2 - 2(cos a)^2) db * da
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).

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...
No, your setup is correct so you must have integrated wrong- your error is in the part you did did not post.

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 $\phi$.