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Consider the cone $z^2=x^2+y^2$ between $z=0$ and $z=1$. Find the volume of the region above this cone and inside the sphere of radius $\sqrt2$ centered at the origin that encloses the cone.


The straightforward approach to this problem would have been a triple integral in spherical coordinates, but for practice I tried using the cone as a surface and using a surface integral to find the volume between the cone and the sphere. I first parameterized the surface using cylindrical coordinates $x=r\cos\theta$, $y=r\sin\theta$, $z=r$ (since $z^2=x^2+y^2$), and then found the intersection points using the equations $z=r$ and $r^2+z^2=2$: $$2z^2=2\Rightarrow z=1\Rightarrow r=1$$ Now that the bounds of integration have been found I used the function $z=\sqrt{2-r^2}$ as the function to integrate and set up my surface integral as: $$\iint_S \sqrt{2-r^2}\Vert\vec{r_r}\times\vec{r_\theta}\Vert\,\text{d}S$$ I then computed the cross product, for $\vec{r}(r,\theta)=\left<r\cos\theta,r\sin\theta,r\right>$: $$\begin{vmatrix} \hat{\imath}&\hat{\jmath}&\hat{k}\\ \cos\theta&\sin\theta&1\\ -r\sin\theta&r\cos\theta&0 \end{vmatrix}=\hat{\imath}(-r\cos\theta)-\hat{\jmath}(-r\sin\theta)+\hat{k}(r\cos^2\theta+r\sin^2\theta)$$ So the surface integral becomes $$\int_0^{2\pi}\int_0^1\sqrt{\left(2-r^2\right)\left(r^2\cos^2\theta+r^2\sin^2\theta+r^2\right)}\,\text dr\,\text d\theta$$ $$=\int_0^{2\pi}\int_0^1\sqrt{\left(2-r^2\right)r^2\left(\cos^2\theta+\sin^2\theta+1\right)}\,\text dr\,\text d\theta$$ $$=\int_0^{2\pi}\int_0^1\sqrt{2r^2(2-r^2)}\,\text dr\text d\theta$$ $$=2\sqrt2\pi\int_0^1r\sqrt{2-r^2}\,\text dr$$ Using a $u$-substitution of $u=2-r^2$: $$\sqrt2\pi\int_1^2 \sqrt{u}\,\text du=\sqrt2\pi\left.\left(\frac{2u\sqrt u}{3}\right)\right|_1^2=\sqrt2\pi\left(\frac{4\sqrt2-2}{3}\right)$$ $$=\frac{2\pi}{3}\left(4-\sqrt2\right)$$ However, the answer given in the answer key says that the volume is $\frac{4\pi}{3}(\sqrt2-1)$, so I would like to know why my answer is incorrect, whether that be a computational mistake or faulty reasoning as to why a surface integral would work here to find the volume.

EDIT: Here is the posted solution: Solution

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  • $\begingroup$ Surface integral to find a volume??? $\endgroup$ – William Elliot Dec 6 '18 at 6:52
  • $\begingroup$ @WilliamElliot it was obviously not the most practical way to evaluate it but if my understanding is correct it should work, since the volume is contained between the cone (used as the surface) and the sphere (above the surface) $\endgroup$ – csch2 Dec 6 '18 at 7:00
  • $\begingroup$ @WilliamElliot We use line integrals to find areas all the time. It's how most of us learn about integrals in the first place. And if you use triple integrals to find a volume, then once you've done the innermost integral, what you're left with is a surface integral to find a volume. So it's not strange at all. $\endgroup$ – Arthur Dec 6 '18 at 7:05
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We can also calculate by cap area and its solid angle.

Denoting radius by $ R={\sqrt 2}$

Cap Area $A= 2 \pi R. R (1-1/\sqrt 2 ) $

Solid Angle of Cap = $\dfrac{A}{ R^2} = 2 \pi (1-1/\sqrt 2 )$

Remaining Solid Angle $V_1 = 4 \pi -\dfrac{A}{ R^2} = 2 \pi (1+1/\sqrt 2 )$

Remaining Volume $V= \dfrac{R^3}{3}. V_1 = \dfrac{4 \pi }{3}(\sqrt 2 +1)$

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  • $\begingroup$ Thank you for your answer. I thought of using this method as well, and it confirms that the answer given in the answer key is correct. However I am not so much looking for the correct answer as I am why the method I used is incorrect $\endgroup$ – csch2 Dec 6 '18 at 7:02
  • $\begingroup$ My answer still does not tally with the key ! $\endgroup$ – Narasimham Dec 6 '18 at 7:07
  • $\begingroup$ I did not notice the different sign! That is strange; maybe I should post the given solution for comparison $\endgroup$ – csch2 Dec 6 '18 at 7:08
  • $\begingroup$ Maybe. Btw, you should also have indicated your preferred method at the start. $\endgroup$ – Narasimham Dec 6 '18 at 7:11
  • $\begingroup$ I have edited the title to include the preferred method $\endgroup$ – csch2 Dec 6 '18 at 7:12

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