Question

Evaluate the iterated integral. $$ \int_{0}^{2 \pi} \int_{0}^{\pi / 4} \int_{0}^{\cos \phi} \rho^{2} \sin \phi d \rho d \phi d \theta $$

Short answer

The value of the iterated integral is \(\pi/6\).

Step-by-step solution

Analyze the Integral Limits

The limits of integration are given by: \(0 \leq \theta \leq 2\pi\), for the angle in the xy-plane; \(0 \leq \phi \leq \pi/4\), for the angle from the z-axis; and \(0 \leq \rho \leq \cos\phi\), for the distance from the origin.

Integrate over \(\rho\)

The formula is \(\int \rho^2 \sin\phi\) d\rho. With the limits \(\rho\) from 0 to \(\cos\phi\), this results in: \((\cos\phi)^3 / 3\sin\phi\).

Integrate over \(\phi\)

Integrating \((\cos\phi)^3 / 3\sin\phi\) with respect to \(\phi\), from 0 to \(\pi/4\), we get: \(\int_0^{\pi/4} (\cos\phi)^3 / 3\sin\phi d\phi = 1/12\).

Integrate over \(\theta\)

The last step is to integrate the constant \(1/12\) with respect to \(\theta\), over the range \(0 \leq \theta \leq 2\pi\). The result is \(2\pi / 12 = \pi/6\).