Let $R$ consist of the points lying inside of the sphere $$ x^2 + y^2 + z^2 = 3^2 $$ and inside the cone $$ z = \cot(\alpha)\sqrt{x^2 +y^2} $$ where $\alpha$ is $\arccos(\frac15)$ Find the volume of $R$.

So I used cylindrical coordinates and set the two equations for $z$ equal to each other, and got $z = \pm \sqrt{9-r^2}$ as the bounds for $z$. Then for $r$ I replaced the expression for $z^2$ which I got from the the cone equation with the $z^2$ in the sphere equation. I got $r = \pm \frac{3}{\sqrt{1+\cot^2(\alpha)}}$ as the bounds. I think the bounds for r is where I made some mistake but I don't know what exactly.

I'm aware you can solve this using spherical coordinates as well, but I would like to solve the problem using cylindrical coordinates, and particularly learn where i went wrong in this problem.

The answer should be some rational number times $\pi$.

  • $\begingroup$ Note that $\sqrt{1+\cot^2(\alpha)} = \csc(\alpha)$ and $1/\csc(\alpha)$ is quite simple. Note that in polar and cylindrical coordinates you should have $0\le r<\infty$, so only the positive value of $r$ you wrote down is the appropriate limit of integration. Check your $z$ bounds, too. $\endgroup$ – Ted Shifrin Mar 6 '18 at 21:25


  • The intersection between the cone and the sphere is for $z=\cot \alpha \sqrt{9-z^2}\ge0\\\implies z=\frac{3\cot \alpha}{\sqrt{1+\cot^2 \alpha}}$
  • For symmetry we can set up the integral in two parts for $z\ge0$, notably

$$V=\int_0^{2\pi} \int_0^{\frac{3\cot \alpha}{\sqrt{1+\cot^2 \alpha}}}\int_0^{r_1(z)}r\,dr\,dz\,d\theta+\int_0^{2\pi} \int_{\frac{3\cot \alpha}{\sqrt{1+\cot^2 \alpha}}}^3\int_0^{r_2(z)}r\,dr\,dz\,d\theta$$

  • $\begingroup$ How did you find the bounds for $z$? I don't see how you can find an expression for $z$ that doesen't include $r$. We have $$r^2 + cot^2(\alpha)r^2 = 9$$ $\endgroup$ – Pame Mar 6 '18 at 21:59
  • $\begingroup$ @Pame we have $z = \cot(\alpha)\sqrt{x^2 +y^2}$ and $x^2 + y^2=9 - z^2$ the we can solve for $z$. $\endgroup$ – gimusi Mar 6 '18 at 22:04
  • $\begingroup$ Ok i see how you got to $z = cot \alpha \sqrt{9-z^2}$ but then which substitution did you use for $z$ to get $$z=\frac{\pm 3\cot \alpha}{\sqrt{1+\cot^2 \alpha}}$$ ? $\endgroup$ – Pame Mar 6 '18 at 22:22
  • $\begingroup$ I've squared both sides $z^2 = (cot \alpha \sqrt{9-z^2})^2$ $\endgroup$ – gimusi Mar 6 '18 at 22:24
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    $\begingroup$ I think i figured it out. $$ \phi = arctan(\frac{r}{z}) = arctan(\frac{3sin(\alpha)}{3cos(\alpha)} = arctan(tan(\alpha)) = \alpha.$$ $\endgroup$ – Pame Mar 8 '18 at 21:15

In spherical coordinates we should have simply

$$V=\int_0^{2\pi} \int_0^{\alpha}\int_0^{3}r^2\sin \varphi\,dr\,d\varphi\,d\theta$$

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  • $\begingroup$ I don't think you need the $2$ in front, since the cone has only nonnegative values for $z$. $\endgroup$ – Pame Mar 8 '18 at 21:27
  • $\begingroup$ @Pame ah ok if we are considering only the upper part of course we cal eliminate the factor 2. This is also true for the previous integral! $\endgroup$ – gimusi Mar 8 '18 at 21:31

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