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

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*}

• There is an independent $\int_0^\pi d\theta = \pi$, so the answer must be of the form $x\pi$, thus the extra lone -3/2 term in the final answer almost surely means it is wrong. – kennytm Feb 28 '15 at 15:41

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

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

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

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 $\frac{1}{3}$.

• We had $\frac{8}{3} - \frac{\sec^3 (\phi)}{3}$. I took the $1/3$ factor out, then the $8$ factor out. – Kamil Feb 28 '15 at 15:43
• Sure, but you messed up a lot of things there. I suggest you re-read your solution carefully; the 2nd step in particular. – user207710 Feb 28 '15 at 15:46
• Or, try to solve it again. – user207710 Feb 28 '15 at 15:48
• Can you check it now? I edited. – Kamil Feb 28 '15 at 15:57
• @Kamil $A(B-C) = AB - AC$, not $AB - C$. Please check your second step again. – kennytm Mar 1 '15 at 11:14