# Thread: surface integrals for sphere

1. ## surface integrals for sphere

Hello,

I am having trouble integrating this one ...especially because there is (sinθsinφsin2φ)/2.. when one tries to integrate the following ..
∫∫2cosφsinθsinθdφdθ
inner integral:
φ:-pi/2 to to pi/2

theta :0 to pi/2
How does one integrate this... ??

thanks

2. ## Re: surface integrals for sphere

That "$\displaystyle sin(\theta)sin(\theta)$" looks peculiar. Is it really [tex]sin^2(\theta)$or did you mean to write "[tex]sin(\theta)cos(\theta)$"?

Either way, it's a pretty straightforward integral.
$\displaystyle \int_0^{\pi/2}\int_{-\pi/2}^{\pi/2} 2cos(\phi)sin^2(\theta)d\theta d\theta d\phi= 2\int_0^{\pi/2}cos(\phi)d\phi \int_{-\pi/2}^{\pi/2} sin^2(\theta) d\theta$
Of course, the integral of "[tex]cos(\phi)[/itex]" is "$\displaystyle sin(\phi)$" and to integrate "$\displaystyle sin^2(\theta)$" use the identity $\displaystyle sin^2(\theta)= \frac{1}{2}(1- cos(2\theta))$.

If you intended $\displaystyle sin(\theta)cos(\theta)$ rather than $\displaystyle sin^2(\theta)$,
$\displaystyle \int_0^{\pi/2}\int_{-\pi/2}^{\pi/2} 2cos(\phi)sin(\theta)cos(\theta)d\theta d\theta d\phi= 2\int_0^{\pi/2}cos(\phi)d\phi \int_{-\pi/2}^{\pi/2} sin(\theta)cos(\theta) d\theta$
Of course, the integral of "$\displaystyle cos(\phi)$" is "$\displaystyle sin(\phi)$" and to integrate "$\displaystyle sin(\theta)cos(\theta)$" use the substitution $\displaystyle u= sin(\theta)$.