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

\begin{equation} -\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 \end{equation}

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 $2$.

  • 3
    $\begingroup$ You have to integrate from 0 to $2\pi$, not from zero to $\pi$. $\endgroup$ – stein Apr 18 at 20:11

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