# Volume of solid bounded

I'm trying to find the volume of the solid bounded by the cone $z^2=x^2+y^2$ and the sphere $x^2+y^2+z^2=a^2$ (this is the solid outside of the cone and inside the sphere.)

I got the triple integral $$2\int_0^{2\pi }\int_{\pi /4}^{3\pi /4} \int_0^{a} r^2dr\sin \theta d\theta d\phi .$$ Is this correct (I'm taking $\phi$ as the angle in the x-y plane)?

• How the "outside" of an infinite cone is defined? – Jack D'Aurizio Feb 28 '17 at 21:46
• As in the points satisfying $z^2\geq x^2+y^2$ and the points satisfying $x^2+y^2+z^2\leq a^2$. – Anon Feb 28 '17 at 21:53
• If I am interpreting this correctly, this is the left and right sides of the sphere with radius $a$ after the cone has cut it out on the top and bottom? If so the integral looks almost correct. The $2$ on the outside doesn't appear to be needed. – WaveX Feb 28 '17 at 21:57
• @WaveX: the OP is probably accounting for points such that $z^2\geq x^2+y^2$ but $z$ is negative, too. – Jack D'Aurizio Feb 28 '17 at 22:13

If you are accounting for points with a negative $z$-coordinate too, your integral is correct.
However, it is probably easier to compute such integral by computing the areas of the $z$-sections. If $0\leq z\leq \frac{a}{\sqrt{2}}$, the section is a circle with area $\pi z^2$. If $\frac{a}{\sqrt{2}}\leq z\leq a$, the section is a circle with area $\pi(a^2-z^2)$. It follows that the wanted volume equals

$$2\pi \left(\int_{0}^{a/\sqrt{2}}z^2\,dz + \int_{a/\sqrt{2}}^a(a^2-z^2)\,dz\right)=\frac{2\pi a^3}{3}(2-\sqrt{2}).$$

• How come my answer differs from yours. I get $\frac{2}{3}a^3\pi \sqrt 2$? – Anon Mar 1 '17 at 17:42
• @Ben: how can I possibly reply to you, since I do not know the steps you took? – Jack D'Aurizio Mar 1 '17 at 17:58
• I just evaluated the integral that I posted above - which you said was correct - but it doesn't even resemble your final answer. – Anon Mar 1 '17 at 18:31
• @Ben: I understand, I am just curious about how you performed such evaluation. I might be of some help if you show your steps. – Jack D'Aurizio Mar 1 '17 at 18:42

Why to integrate if it is the volume that we have to find.

The volume is

$$V=V_{\text{sphere}}-V_{\text{cone}}.$$

The radius of the sphere is $a$. The volume is

$$V_{\text{sphere}}=\frac43\pi a^3.$$

For the cone, let's calculate the height first. The cone intersects the sphere at the $z$ for which

$$z^2+x^2+y^2=a^2\ \text{ and }\ x^2+y^2=z^2,$$

so, $$2z^2=a^2\ \text{ or } \ z=h_{\text{cone}}=\frac1{\sqrt{2}}a.$$

Then

$$x^2+y^2+\frac12a^2=a^2 \ \text{ that is, }\ r_{\text{cone}}=\frac1{\sqrt2}a$$

and

$$V_{\text{cone}}=\frac13r_{\text{cone}}^2\pi h_{\text{cone}}=\frac16a^3\pi\frac1{\sqrt2}.$$

Finally

$$V=\frac43\pi a^3-\frac16a^3\pi\frac1{\sqrt2}=a^3\pi\frac23\left(2-\frac1{4\sqrt2}\right).$$

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

Use Cylindrical Coordinates. The '$\color{#f00}{upper}$ volume' $\ds{V}$ is given by:

\begin{align} V & \,\,\,\stackrel{\mrm{def.}}{=}\,\,\, \iiint_{\mathbb{R}^{3}}\bracks{r^{2} + z^{2} > a^{2}} \bracks{z^{2} < r^{2}}\bracks{\color{#f00}{z > 0}}r\,\dd r\,\dd\theta\,\dd z \\[5mm] & = 2\pi\int_{0}^{\infty}\int_{0}^{\infty}\bracks{{\root{2} \over 2}\,a < r < a} \bracks{\root{a^{2} - r^{2}} < z < r}r\,\dd r\,\dd z \\[5mm] & = 2\pi\int_{\root{2}a/2}^{a}\pars{r - \root{a^{2} - r^{2}}}r\,\dd r = \bbx{\ds{{1\over 3}\,\pars{2 - \root{2}}\pi a^{3}}} \end{align}