# Homework Help: Evaluate the double integral

1. Nov 7, 2012

### whynot314

1. The problem statement, all variables and given/known data

$\int^{\pi}_{0}$ $\int^{1-sin\theta}_{0}$ r$^{2}$ cos$\theta$ drd$\theta$

I keep getting an answer of 0 but i am most certain that i am getting my trig messed up somewhere.
1/3 $\int^{\pi}_{0}$ r$^{3}$ cos$\theta$$d[itex]\theta$ from 0 to 1-sin\theta

then i get

1/3 $\int^{\pi}_{0}$ (1-sin\theta)^3 cos\theta d\theta

I then use substitution,u for 1-sin\theta then get 1/3 times -1/4(u)^4

substitute back the 1-sin\theta and evaluate from 0 to \pi and I keep getting zero please help thanks

2. Nov 7, 2012

### tiny-tim

hi whynot314!

looks ok …

the integrand is anti-symmetric about π/2, so it should be zero

3. Nov 7, 2012

### whynot314

thanks, I was just concerned bc these area ones are usually never turn out to be zero thanks.