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\thetad\theta from 0 to 1-sin\theta<br /> <br /> then i get <br /> <br /> 1/3 \int^{\pi}_{0} (1-sin\theta)^3 cos\theta d\theta<br /> <br /> I then use substitution,u for 1-sin\theta then get 1/3 times -1/4(u)^4 <br /> <br /> 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.