# $-\frac{\rho_0}{4\pi\epsilon_0}\int\limits_{0}^{2\pi}\int\limits_{a}^{2a}\int\limits_{0}^{\frac{\pi}{4}}(\cos{\varphi}\sin{\theta}+\sin{\varphi}…$

So I have the following integral that pops up when trying to solve a physics problem:

$$$$-\frac{\rho_0}{4\pi\epsilon_0}\int\limits_{0}^{2\pi}\int\limits_{a}^{2a}\int\limits_{0}^{\frac{\pi}{4}}(\cos{\varphi}\sin{\theta}+\sin{\varphi}\sin{\theta}+\cos{\theta})\sin{\theta} \ d\theta \ dr \ d \varphi.\tag1$$$$

Now, instead of actually computing this integral term by term, one can do some simplifications. In my book they say that the first two terms in the integrand, both will have a $$\cos{\varphi}$$ and $$\sin{\varphi}$$ respectively that will be integrated between $$0$$ and $$\pi$$ so both of them will me $$0$$.

I don't find this obvious. Can someone explain? If integrate $$\cos{\varphi}$$ and $$\sin{\varphi}$$ between $$0$$ and $$\pi$$ I will not get zero in both cases. For the sin I get $$2$$.

• You have to integrate from 0 to $2\pi$, not from zero to $\pi$. – stein Apr 18 at 20:11