How to evaluate $\int _0^{2\pi }\int _0^{\frac{\pi }{4}}\int _0^{\frac{1}{\cos\phi }}\:\:\:\rho ^2\sin\left(\phi \right)d\rho \:d\phi \:d\theta $? can you help me with this integral, can you also show the solution method, because I don't even know how to start. Thank you
$$\int _0^{2\pi }\int _0^{\frac{\pi }{4}}\int _0^{\frac{1}{\cos\phi }}\:\:\:\rho ^2\sin\left(\phi \right)d\rho \:d\phi \:d\theta $$
 A: The integral is to be properly grouped according to the $3$ variables $\rho,\theta $ and $\phi$, each variable with its differential and respective limits for integration. 
So we have that
$$\begin{align}
I &= \int _0^{2\pi }\int _0^{\frac{\pi }{4}}\int _0^{\frac{1}{\cos\phi }}\rho ^2\sin\left(\phi \right)d\rho \:d\phi \:d\theta \\
&=\left[\int _0^{2\pi }\left\{\int _0^{\frac{\pi }{4}}\left(\int _0^{\frac{1}{\cos\phi }}\rho ^2d\rho\right)  \sin\left(\phi \right) \:d\phi \right\} \:d\theta \right] \\
&=\left[\int _0^{2\pi }\left\{\int _0^{\frac{\pi }{4}}\left(\frac{1}{3\cos^3 (\phi)}\right)  \sin\left(\phi \right) \:d\phi \right\} \:d\theta \right] \\
&=\frac{1}{3}\left[\int _0^{2\pi }\left\{\int _0^{\frac{\pi }{4}}\sec^2 (\phi)  \tan (\phi) \:d\phi \right\} \:d\theta \right] \\
&=\frac{1}{3}\left[\int _0^{2\pi }\left\{\int _0^1 \tan (\phi)   \:d(\tan (\phi)) \right\} \:d\theta \right] \\
&=\frac{1}{3}\left[\int _0^{2\pi }  \frac{1}{2} \cdot  \:d\theta \right] \\
&=\frac{1}{6}\left[\int _0^{2\pi } \:d\theta \right] \\
&=\frac{\pi}{3}
\end{align}$$
Hope this helps.
A: To get started, the inner integral is
$$\int_0^{1/\cos(\phi)} \rho^2 \sin(\phi) d\rho=\frac13 \frac{\sin(\phi)}{\cos(\phi)^3} = \frac13 \frac{\tan(\phi)}{\cos(\phi)^2}$$
Now substitute $u=\tan(\phi)$ in the next integral to get
$$\int_0^{\pi/4} \frac13\frac{\tan(\phi)}{\cos(\phi)^2} d\phi = \frac13\int_0^1 u du = \frac16.$$
Finally the last integration variable doesn't even appear, so the integral is altogether $2\pi\cdot \frac16=\frac{\pi}{3}$.
A: Hint:
The triple integral can be calculated as a sequence of simple integrals as:
$$
\int _0^{2\pi } \left[
\int _0^{\frac{\pi }{4}} \sin\left(\phi \right)\left(\int _0^{\frac{1}{\cos\phi }}\rho ^2d\rho \right)\:d\phi
\right]\:d\theta
$$
