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Evaluate the integral $$\iiint_{D} e^{(x^{2}+y^{2}+z^{2})^{3/2}} \, dV$$ where $D$ is the region above the cone $z=\sqrt{x^{2}+y^{2}}$ and below the hemisphere $z=\sqrt{1-x^{2}-y^{2}}$ using spherical coordinates.

A. $\frac{\pi}{3}(2-\sqrt{2})(e-1)$

B. $\frac{2\pi}{3}(2-\sqrt{2})(e-1)$

C. $\frac{\pi}{3}(2+\sqrt{2})(e-1)$

D. $\frac{\pi}{3}(\sqrt{2}-2)(e-1)$

E. None of the above

The answer key says the correct answer is A, but I am not able to get a difference between two numbers in the middle term. Have I made a mistake in the bounds for phi? Or is there a mistake in the answer key

$$\iiint_\limits{D} e^{\rho^3} \rho^2 \sin\phi $$

$$\phi = \frac{\pi}{4} $$ $$\rho^2 = 1 $$

$$\int_0^{2\pi} \int_{\large\frac{\pi}{4}}^{\large\frac{\pi}{2}} \int_0^1 e^{\rho^3} \rho^2 \sin\phi \ DV $$

$$ 2\pi \left( -\cos\phi \bigg\rvert_{\large\frac{\pi}{4}}^{\large\frac{\pi}{2}} \right) \left( \frac13 e^{\rho^3} \bigg\rvert_0^1 \right) \\ = -2\pi \left( \frac{1}{\sqrt{2}} \right) \left( \frac{e - 1}{3} \right) $$

Reference photo of problem and work:

problem

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  • $\begingroup$ "Below the hemisphere, above the cone"; your integral only satisfies the first condition. $\endgroup$ Commented Nov 7 at 17:51
  • $\begingroup$ Normally pictures of text and formulas are very much disliked here. A couple of us have made the effort to convert the text and formulas in your picture into the preferred format for this site, apparently because we consider this to be fun. Next time you really should do it yourself. You can click "edit" here to see how we did it, and there are also how-to pointers starting at math.stackexchange.com/help/notation $\endgroup$ Commented Nov 8 at 1:29
  • $\begingroup$ There are a some oddities in your handwritten solution such as omitting the $dV$ in one integral and writing it with a capital $D$ in the other; actually, even worse than the capital $D,$ you don't show the order of integration (e.g., which comes first, $d\rho$ or $d\phi$?). You are welcome to fix this by using the "Edit" link. $\endgroup$ Commented Nov 8 at 1:32
  • $\begingroup$ The other thing that you really should do when asking a question is to "talk" us through your work by explaining in words why you did each step the way you did, not just write the equations. We don't know what you think $\phi$ means, for example, so we can't really say for sure whether $\sin\phi$ is a mistake or whether the bounds $\pi/4,\pi/2$ are a mistake. (People are making good guesses in the answers, but they are still guessing when you should be telling.) Telling us what you were thinking is essential to learning how to think better. $\endgroup$ Commented Nov 8 at 1:36
  • $\begingroup$ By the way, lest all this advice on how to write a question seem like some hazing ritual peculiar to this site, it's actually good life skills. When you are asking for help, explain your though processes and write clearly, consistently, and legibly. This will help anybody help you. $\endgroup$ Commented Nov 8 at 1:42

3 Answers 3

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The domain of integration is wrong. It should be

$$ D = \left\{ (\rho, \phi, \theta) \ \bigg| \ 0 \le \theta \le 2\pi, \ 0 \le \rho \le 1, \ 0 \le \phi \le \frac{\pi}{4} \right\} $$

Just with Desmos:

enter image description here

and Geogebra

enter image description here

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To recap...

$$\int_0^{2 \pi} \int_0^{\pi /4} \int_0^1 r^2 e^{r^3} \sin{\theta} \ dr\ d\theta\ d\varphi $$

$$=\int_0^{2 \pi} d\varphi \int_0^{\pi /4} \sin{\theta} \ d\theta\ \int_0^1 r^2 e^{r^3} dr $$

$$=2\pi\ (-\cos(\pi/4)+\cos(0))\ \frac{e-1}{3} $$ $$= 2\pi\ \left(1-\frac{1}{\sqrt{2}}\right)\ \frac{e-1}{3}$$

Symbols are according to
https://en.wikipedia.org/wiki/Spherical_coordinate_system

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  • $\begingroup$ Perhaps Wikipedia takes colatitude in the explanation. $\endgroup$ Commented Nov 7 at 20:06
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When you integrate $\sin\theta\space d\theta$, you get $-\cos\theta$ which must then be evaluated as a difference between $\theta=0$ and $\theta=\pi/4$. Note that neither of these function values is zero ($\cos(0)=1,\cos(\pi/4)=1/\sqrt2$), so with proper evaluation you should get a difference in the $\theta$-integral factor.

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