# Having trouble with a surface integral

1. Jun 1, 2009

### coderdave

I'm self learning and working my way through the book "Div Grad Curl and All That". On one of the pages (27) the author says
$$\int_{ }^{ } \int_{ }^{ } z^2 dS = \int_{ }^{ } \int_{ }^{ } \sqrt[ ]{ 1 - x^2 - y^2 } dx dy$$

"This is an ordinary double integral, and you should verify that its value is pi / 6.

The hint was to convert to polar coordinates and the furthest I got was

$$\int_{0}^{ \pi/2} \int_{0}^{1} \sqrt[ ]{ 1 - r^2cos^2( \theta) - r^2sin^2(\theta) } r dr d\theta$$

Any ideas getting past this point?

2. Jun 1, 2009

### nicksauce

What's cos(theta)^2 + sin(theta)^2?

3. Jun 1, 2009

=(

So obvious!