Question

Evaluate the iterated integral. $$ \int_{0}^{\pi / 2} \int_{0}^{\pi} \int_{0}^{2} e^{-\rho^{3}} \rho^{2} d \rho d \theta d \phi $$

Short answer

\(\pi^2 / 6 (1 - e^{-8})\)

Step-by-step solution

Integrate with respect to \(\rho\)

The integral becomes \(\int_{0}^{\pi/2} \int_{0}^{\pi} [-1/3 e^{-\rho^3}]_{0}^{2} d\theta d\phi\), which can be simplified to \(\int_{0}^{\pi/2} \int_{0}^{\pi} (-1/3 e^{-8} + 1/3) d\theta d\phi\).

Integrate with respect to \(\theta\)

The integral then becomes \(\int_{0}^{\pi/2} [(-\theta/3 e^{-8} + \theta/3)]_{0}^{\pi} d\phi\), further simplifying to \(\int_{0}^{\pi/2} (-\pi / 3 e^{-8} + \pi / 3) d\phi\).

Integrate with respect to \(\phi\)

Finally the integral becomes \([(-\phi \pi / 3 e^{-8} + \phi \pi / 3 )]_{0}^{\pi/2}\), that simplifies to \(-\pi^2 / 6 e^{-8} + \pi^2 / 6\).

Simplify the result

The result is written as \(-\pi^2 / 6 e^{-8} + \pi^2 / 6\). Due to the rule of addition we can rewrite the solution as \(\pi^2 / 6 (1 - e^{-8})\).