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Given the set $$C=\{(x,y,z,) : x^2+y^2+z^2 \leq 1 \}$$ Using spherical coordinates, evaluate the function $$Q(C)=\iiint_C \sqrt{x^2+y^2+z^2} \, dx \, dy \, dz$$

So... I can see that the function easily converts to the triple integral of $\rho$.

My question is, what are the bounds of integration? My new textbook talk a lot about sets, rather than geometric shapes, as it is a statistics course. This looks almost identical to finding the volume of a sphere, except the inequality. Is this intuition correct?

The answer I came out with is $\pi$, but the textbook does not have a solution for this particular exercise.

Michael Hardy
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rocksNwaves
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    This should be $$ \int_0^1 r \left( 4\pi r^2 \right) , dr = \pi $$ since $4\pi r^2$ is the area of the sphere of radius $r. \qquad$ – Michael Hardy Jun 30 '17 at 16:26
  • @MichaelHardy So, I got $\pi$, but I didn't use radius... Not sure if I got lucky. The question statement in the book said to use spherical coordinates. Is what you used cylindrical? It's been a long time since I saw any of this stuff. – rocksNwaves Jun 30 '17 at 16:33
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    In my posted answer I used spherical coordinates. In my comment I did not use either spherical or cylindrical coordinates. $\qquad$ – Michael Hardy Jun 30 '17 at 16:54

3 Answers3

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If you have $dx\,dy\,dz = \rho^2 \sin\varphi \, d\rho\, d\theta\,d\varphi$ Then this is \begin{align} & \int_0^\pi \int_0^{2\pi} \int_0^1 \rho (\rho^2\sin\varphi \, d\rho \, d\theta \, d\varphi) \\[10pt] = {} & \int_0^\pi\left( \int_0^{2\pi} \left( \int_0^1 \rho^3 \sin\varphi \, d\rho \right) d\theta \right) d\varphi \\[10pt] = {} & \int_0^\pi\left( \sin\varphi \left( \int_0^{2\pi} \left( \int_0^1 \rho^3 \, d\rho \right) d\theta \right) \right) d\varphi \text{ since } \sin\varphi \text{ does not depend on } \rho \text{ or } \theta \\[10pt] = {} & \int_0^\pi \sin\varphi\,d\varphi \cdot \int_0^{2\pi} \int_0^1 \rho^3 \,d\rho\,d\theta \text{ since the latter integral does not depend on } \varphi \\[10pt] = {} & \int_0^\pi \sin\varphi\,d\varphi \cdot \int_0^1 \rho^3 \,d\rho \cdot \int_0^{2\pi} 1\,d\theta \text{ since the inner integral above does not depend on } \theta \\[10pt] = {} & 2\cdot \frac 1 4 \cdot 2\pi = \pi. \end{align}

Three times we used the fact that $\displaystyle \int \text{constant} \times f(\alpha)\,d\alpha = \text{constant} \times \int_a^b f(\alpha)\,d\alpha.$ “Constant” means something that does not change as $\alpha$ changes.

Michael Hardy
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Yes, you are correct. Since the integrand depends only on $r$, it is constant on each spherical layer and $$\iiint_C \sqrt{x^2+y^2+z^2}\, dx \, dy \, dz=\int_{r=0}^1 r\cdot (\mbox{surface area of the sphere of radius $r$})\, dr\\= \int_{r=0}^1 r\cdot (4\pi r^2)\, dr=4\pi\cdot\left[\frac{r^4}{4}\right]_0^1=\pi.$$

Robert Z
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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\iiint_{x^{2} + y^{2} + z^{2} < 1}\root{x^{2} + y^{2} + z^{2}}\, \dd x\,\dd y\,\dd z = \iiint_{r < 1}r\,\dd^{3}\vec{r} = \iiint_{r < 1}{\nabla\cdot\pars{r\vec{r} \over 4}}\,\dd^{3}\vec{r} \\[5mm] = &\ \iint_{r = 1}{r\ \vec{r} \over 4}\cdot\,\dd\vec{S}\qquad \pars{Gauss\ Divergence\ Theorem} \\[5mm] = &\ {1 \over 4}\iint_{r = 1}{\vec{r}\cdot\,\dd\vec{S} \over r^{2}} = {1 \over 4}\int_{\Omega_{\vec{r}}}\dd\Omega_{\vec{r}} = {1 \over 4}\,4\pi = \bbx{\pi}\qquad \pars{~\Omega_{\vec{r}}:\ Solid\ Angle~} \end{align}

Felix Marin
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