Double Integral with Trigonometric Functions: Troubleshooting and Evaluation

[whynot314]

Homework Statement

$\int^{\pi}_{0}$ $\int^{1-sin\theta}_{0}$ 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$\theta$[itex]d$\theta$ 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

 

[tiny-tim]

hi whynot314!

looks ok …

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

 

[whynot314]

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