# Solid volume using double integral

I have to calculate the volume of the solid bounded by $$x^{2}+y^{2}+z^{2}=6 \qquad z=x^{2}+y^{2} \qquad z\geq0$$ using double integrals.

When I drew it, I could see that a part of the sphere is above the paraboloid. So, for me, it should be something like that: $$V=\iint_{B}\left[\sqrt{6-x^{2}-y^{2}}-\left(x^{2}+y^{2}\right)\right]dx\,dy$$ But my problem is to find these integrals. I think I'm supposed to not use spheric coordinates (since I'm calculating a double integral, not a triple one), but even the polar ones, I don't know how to find them.

• Alternately the solid is a sum of parabolid and sphereical segment, fo r which easy formulas exist. Commented Apr 10, 2018 at 3:25

## 4 Answers

Polar coordinates

$x = r\cos \theta\\ y = r\sin \theta\\ dx\ dy = r \ dr\ dz$

$\iint (\sqrt {6-r^2} - r^2) r\ dr\ d\theta$

Limits:

Find where the two curves intersect.

Substitute $x^2+ y^2 = z$ into the equation of the sphere.

$z + z^2 = 6\\ (z + 3)(z-2) = 0\\ z = 2\\ r^2 = 2$

$\int_0^{2\pi}\int_0^{\sqrt 2} r\sqrt {6-r^2} - r^3 \ dr\ d\theta$

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

Given the problem symmetry, it's convenient to use cylindrical coordinates $\ds{\pars{\rho, \phi, z}}$. Hereafter, $\ds{\bracks{\cdots}}$ is an Iverson Bracket which is a $\textsf{quite convenient 'tool' to handle constraints}$.

\begin{align} V & = \iiint_{\mathbb{R}^{3}}\bracks{x^{2} + y^{2} + z^{2} < 6} \bracks{z > x^{2} + y^{2}}\bracks{z \geq 0}\dd x\,\dd y\,\dd z \\[5mm] & = \int_{0}^{\infty}\int_{0}^{2\pi}\int_{0}^{\infty} \bracks{\rho^{2}+ z^{2} < 6} \bracks{z > \rho^{2}}\rho\,\dd\rho\,\dd\phi\,\dd z \end{align} Since the integrand is $\ds{\phi}$-independent $\ds{\pars{~\mbox{In addittion, I'll make the change}\ \rho^{2}\ \mapsto\ \rho~}}$: \begin{align} V & = \pi\int_{0}^{\infty}\int_{0}^{\infty} \bracks{z < \root{6}}\bracks{\rho < 6 - z^{2}}\bracks{\rho < z} \,\dd\rho\,\dd z \\[5mm] & = \pi\int_{0}^{\root{6}}\int_{0}^{\infty} \bracks{\rho < 6 - z^{2}}\bracks{\rho < z}\,\dd\rho\,\dd z \\[5mm] & = \pi\int_{0}^{\root{6}}\ \overbrace{\bracks{6 - z^{2} < z}} ^{\ds{\pars{z < - 3}\ \mbox{or}\ \pars{z > 2}}}\ \int_{0}^{6 - z^{2}}\,\dd\rho\,\dd z\ +\ \pi\int_{0}^{\root{6}}\ \overbrace{\bracks{6 - z^{2} > z}}^{\ds{-3 < z < 2}}\ \int_{0}^{z}\,\dd\rho\,\dd z \\[5mm] & = \pi\, \underbrace{\int_{2}^{\root{6}}\pars{6 - z^{2}}\dd z} _{\ds{4\root{6} - {28 \over 3}}}\ +\ \pi\ \underbrace{\int_{0}^{2}z\,\dd z}_{\ds{2}} = \bbx{4\root{6}\pi - {22 \over 3}\,\pi} \approx 7.7429 \end{align}

In the case that the solid in question is bounded between the graphs of two functions, say $f$ and $g$, calculating the volume with double integrals (integrating $f-g$) is the same as using triple integrals (integrating the function $1$ on an appropriate set).

Anyhow, to use polar coordinates, you need to to find the maximum possible value of the radius, since it is geometrically clear that $$0<\theta<2\pi$$ $$0<r<R$$ for some $R$. To find this $R$, substitute the second equation in the first. This should give you a circle in $x$ and $y$, which is to be expected (draw a picture if this is not clear). The radius of said circle is your $R$.

So your integral you write down is correct. And it is easy to asses that the radius of the circle you have to integrate over is $R=\sqrt{2}$ since :

$$x^2+y^2 = \sqrt{6-x^2-y^2},$$ $$r^2 = \sqrt{6-r^2}$$

Thus your integral is now directly solvable :

$$\iint_\mathcal{B} \sqrt{6-x^2-y^2} - x^2-y^2 \textrm{d}x\textrm{d}y,$$ $$\int_0^{2\pi}\int_0^R \left(\sqrt{6-r^2} - r^2\right)r\, \textrm{d}r\textrm{d}\theta,$$ $$2\pi\left(2\sqrt{6}-\frac{\sqrt{6-R^2}^3}{3}-\frac{R^4}{4}\right)$$

Which leads to the final result :

$$I=\pi\left(4\sqrt 6-\frac{22}3\right)$$

In the second step we switched to polar coordinates using:

$$\iint f(x,y)\,\textrm{d}x\textrm{d}y = \iint f(x(r,\theta),y(r,\theta)) r\,\textrm{d}r\textrm{d}\theta .$$