Evaluate $\int_{0}^{\pi} \int_{0}^{\pi/3} \int_{\sec \phi}^{2} 5\rho^2 \sin(\phi) \mathrm d\rho \mathrm d\phi \mathrm d\theta$

Question

Evaluate the following \begin{align*} \int_{0}^{\pi} \int_{0}^{\pi/3} \int_{\sec \phi}^{2} 5\rho^2 \sin(\phi) \mathrm d\rho \mathrm d\phi \mathrm d\theta \end{align*}

Attempt at solution: We have \begin{align*} 5 \int_{0}^{\pi} d\theta \int_{0}^{\pi/3} \sin(\phi) \ d\phi \Big[\frac{8}{3} - \frac{\sec^3 \phi}{3} \Big] \\ = \frac{40}{3} \int_{0}^{\pi} d\theta \int_{0}^{\pi/3} \sin(\phi) \ d\phi - \int_{0}^{\pi/3} \sin(\phi) \sec^3 (\phi) \ d\phi \\ = \frac{40}{3} \int_{0}^{\pi} d\theta \big[-\cos(\pi/3) +1 \big] - \int_{0}^{\pi/3} \tan(\phi) \sec^2 (\phi) d\phi \end{align*}

The expression to the left of the minus sign becomes $\frac{40}{3}\int_{0}^{\pi} d\theta \frac{1}{2} = \frac{40 \pi}{6}$. For the expression on the right side we let $\tan (\phi) = u$. If we adapt the integration bounds, that integral then simplifies to $\int_{0}^{\sqrt3} u \ du$., which equals $3/2$.

So the final answer is $\frac{40 \pi}{6} - 3/2$. This is what I answered, and the online learning platform (where I encountered this integral) said it was wrong. So where did I go wrong?

Any help would be appreciated.

Edit: I'll try again from second step. \begin{align*} \frac{5}{3} \Big[ \int_{0}^{\pi} d\theta \int_{0}^{\pi/3} 8 \sin(\phi) d\phi - \int_{0}^{\pi/3} \sin(\phi) \sec^3 (\phi) \Big] \\ =\frac{40}{3} \int_{0}^{\pi} d\theta \int_{0}^{\pi/3} \sin(\phi) d\phi - \frac{5}{3} \int_{0}^{\pi/3} \tan(\phi) \sec^2 (\phi) d\phi \\ = \frac{40 \pi}{6} - (\frac{5}{3} \cdot \frac{3}{2}) = \frac{40 \pi - 15}{6} \end{align*}

Answer 1

Why the $\frac{40}{3}$? That's your mistake.

Edit: Did you separate the integrals in that step? In this case you forgot to multiply the RHS one by $\pi$ and $5$. Moreover you forgot the $\frac{1}{3}$.

Answer 2

$\displaystyle \int_{0}^{\pi} \int_{0}^{\pi/3} \int_{\sec \phi}^{2} 5\rho^2 \sin(\phi) \ d\rho \ d\phi \ d\theta=\frac{5}{3}\int_0^{\pi}\int_0^{\pi/3}(8-\sec^3\phi)\sin\phi\;d\phi d\theta$

$\displaystyle=\frac{5}{3}\int_0^{\pi}\int_0^{\pi/3}(8\sin\phi-\tan\phi\sec^2\phi)\;d\phi d\theta=\frac{5}{3}\pi\left[-8\cos\phi-\frac{1}{2}\tan^2\phi\right]_0^{\pi/3}$.

(Now use the values you have above to finish evaluating.)