Quiz 2 Problem 3

Problem 3 (Mass w.r.t. a Density). A lopsided cone with height \(\frac{3}{2}R\) and base radius \(R\) (shown below) has mass density

$$\delta \left( x,y,z \right) = 1+\frac{2\left|y\right|}{R^2}.$$

Compute the total mass of this cone by evaluating a triple integral.

lopsided cone

Solution. The total mass is

$$\iiint_{\mathrm{cone}}\delta\left(x,y,z\right)dV = \int_0^{\frac{3}{2}R}\!\!\!\int_{-\frac{2}{3}z}^{\frac{2}{3}z}\!\!\int_{-\sqrt{\frac{4}{9}z^2-x^2}}^{\sqrt{\frac{4}{9}z^2-x^2}}\left(1+\frac{2\left|y\right|}{R^2}\right)\,dy\,dx\,dz.$$

(Brief explanation of how the triple integral is written out: For the outer-most integral we slice along the \(z\) direction, since this will yield a disk for each fixed value of \(z\), for which we can apply polar coordinates and make the most out of symmetry. Thus the \(\int_0^{\frac{3}{2}R}\cdots dz\). For each fixed value of \(z\), the disk is of radius \(\frac{2}{3}z\) as seen from the right figure, which is the trace of the cone in the \(\left(y,z\right)\)-plane, showing that \(y=\frac{2}{3}z\) for any fixed value of \(z\). The double integral inside is the double integral over a disk in the \(\left(x,y\right)\)-plane with radius \(\frac{2}{3}z\); if one prefers, it can be cast into polar coordinates as \(\int_0^{2\pi}\!\!\int_0^{\frac{2}{3}z}\delta\left(x,y,z\right)\,rdr\,d\theta\).)

By symmetry,

\begin{align*} \int_0^{\frac{3}{2}R}\!\!\!\int_{-\frac{2}{3}z}^{\frac{2}{3}z}\!\!\int_{-\sqrt{\frac{4}{9}z^2-x^2}}^{\sqrt{\frac{4}{9}z^2-x^2}}\left(1+\frac{2\left|y\right|}{R^2}\right)\,dy\,dx\,dz &= 2\int_0^{\frac{3}{2}R}\!\!\!\int_{-\frac{2}{3}z}^{\frac{2}{3}z}\!\!\int_{0}^{\sqrt{\frac{4}{9}z^2-x^2}}\left(1+\frac{2\left|y\right|}{R^2}\right)\,dy\,dx\,dz\\ &= 2\int_0^{\frac{3}{2}R}\!\!\!\int_{-\frac{2}{3}z}^{\frac{2}{3}z}\!\!\int_{0}^{\sqrt{\frac{4}{9}z^2-x^2}}\left(1+\frac{2y}{R^2}\right)\,dy\,dx\,dz.\\ &\quad \textrm{(getting rid of the absolute value around $y$)} \end{align*}

Direct computation gives

\begin{align*} &\int_0^{\frac{3}{2}R}\!\!\!\int_{-\frac{2}{3}z}^{\frac{2}{3}z}\!\!\int_{0}^{\sqrt{\frac{4}{9}z^2-x^2}}\left(1+\frac{2y}{R^2}\right)\,dy\,dx\,dz=\int_0^{\frac{3}{2}R}\!\!\!\int_{-\frac{2}{3}z}^{\frac{2}{3}z}\left(y+\frac{y^2}{R^2}\right)\Bigg|_{y=0}^{y=\sqrt{\frac{4}{9}z^2-x^2}}dx\,dz\\ &\quad=\int_0^{\frac{3}{2}R}\!\!\!\int_{-\frac{2}{3}z}^{\frac{2}{3}z}\left(\sqrt{\frac{4}{9}z^2-x^2}+\frac{1}{R^2}\left(\frac{4}{9}z^2-x^2\right)\right)dx\,dz\\ &\quad=\int_0^{\frac{3}{2}R}\!\!\!\int_{-\frac{2}{3}z}^{\frac{2}{3}z}\sqrt{\frac{4}{9}z^2-x^2}\,dy\,dx+\frac{1}{R^2}\int_0^{\frac{3}{2}R}\!\!\!\int_{-\frac{2}{3}z}^{\frac{2}{3}z}\left(\frac{4}{9}z^2-x^2\right) dx\,dz \end{align*}

Let us compute these two integrals one by one. First,

\begin{align*} &\int_0^{\frac{3}{2}R}\!\!\!\int_{-\frac{2}{3}z}^{\frac{2}{3}z}\sqrt{\frac{4}{9}z^2-x^2}\,dx\,dz=2\int_0^{\frac{3}{2}R}\!\!\!\int_{0}^{\frac{2}{3}z}\sqrt{\frac{4}{9}z^2-x^2}\,dx\,dz\\ &\stackrel{x=\frac{2}{3}z\sin\theta}{=\!=\!=\!=\!=\!=\!=\!=}2\int_0^{\frac{3}{2}R}\!\!\!\int_{0}^{\frac{\pi}{2}}\frac{2}{3}z\cos\theta\cdot\frac{2}{3}z\cos\theta\,d\theta\,dz\\ &=\frac{8}{9}\int_0^{\frac{3}{2}R}z^2\,dz\int_{0}^{\frac{\pi}{2}}\cos^2\theta\,d\theta=\frac{8}{9}\cdot \frac{1}{3}\left(\frac{3}{2}R\right)^3\int_0^\frac{\pi}{2}\frac{1+\cos 2\theta}{2}\,d\theta\\ &=\frac{\pi}{4}R^3. \end{align*}

Second,

\begin{align*} \frac{1}{R^2}\int_0^{\frac{3}{2}R}\!\!\!\int_{-\frac{2}{3}z}^{\frac{2}{3}z}\left(\frac{4}{9}z^2-x^2\right)\,dx\,dz=\frac{1}{2}R^2. \end{align*}

Thus

$$\int_0^{\frac{3}{2}R}\!\!\!\int_{-\frac{2}{3}z}^{\frac{2}{3}z}\!\!\int_{0}^{\sqrt{\frac{4}{9}z^2-x^2}}\left(1+\frac{2y}{R^2}\right)\,dy\,dx\,dz=\frac{\pi}{4}R^3+\frac{1}{2}R^2$$

and hence

$$\int_0^{\frac{3}{2}R}\!\!\!\int_{-\frac{2}{3}z}^{\frac{2}{3}z}\!\!\int_{-\sqrt{\frac{4}{9}z^2-x^2}}^{\sqrt{\frac{4}{9}z^2-x^2}}\left(1+\frac{2\left|y\right|}{R^2}\right)\,dy\,dx\,dz=2\int_0^{\frac{3}{2}R}\!\!\!\int_{-\frac{2}{3}z}^{\frac{2}{3}z}\!\!\int_{0}^{\sqrt{\frac{4}{9}z^2-x^2}}\left(1+\frac{2y}{R^2}\right)\,dy\,dx\,dz=\frac{\pi}{2}R^3+R^2.$$

Alternative Solution using Cylindrical Coordinates.

\begin{align*} &\iiint_{\mathrm{cone}}\delta\left(x,y,z\right)dV = \int_0^{\frac{3}{2}R}\!\!\!\int_{0}^{2\pi}\!\!\int_{0}^{\frac{2}{3}z}\left(1+\frac{2r\left|\sin\theta\right|}{R^2}\right)\,rdr\,d\theta\,dz\\ &=2\int_0^{\frac{3}{2}R}\!\!\!\int_{0}^{\pi}\!\!\int_{0}^{\frac{2}{3}z}\left(1+\frac{2r\left|\sin\theta\right|}{R^2}\right)\,rdr\,d\theta\,dz\quad\textrm{(use symmetry in $\theta$)}\\ &=2\int_0^{\frac{3}{2}R}\!\!\!\int_{0}^{\pi}\!\!\int_{0}^{\frac{2}{3}z}\left(1+\frac{2r\sin\theta}{R^2}\right)\,rdr\,d\theta\,dz\quad\textrm{($\sin\theta\geq 0$ for $\theta\in\left[0,\pi\right]$)}\\ &=2\int_0^{\frac{3}{2}R}\!\!\!\int_{0}^{\pi}\!\!\int_{0}^{\frac{2}{3}z}\,rdr\,d\theta\,dz+\frac{4}{R^2}\int_0^{\frac{3}{2}R}\!\!\!\int_{0}^{\pi}\!\!\int_{0}^{\frac{2}{3}z}r\sin\theta\cdot rdr\,d\theta\,dz\\ &=2\cdot \frac{\pi}{4}R^3+\frac{4}{R^2}\int_0^{\frac{3}{2}R}\!\!\!\left(\int_{0}^{\pi}\sin\theta\,d\theta\right)\left(\int_{0}^{\frac{2}{3}z}r^2dr\right)\,dz=\frac{\pi}{2}R^3+\frac{4}{R^2}\int_0^{\frac{3}{2}R}2\left(\int_{0}^{\frac{2}{3}z}r^2dr\right)\,dz\\ &=\frac{\pi}{2}R^3+\frac{4}{R^2}\int_0^{\frac{3}{2}R}2\cdot\frac{1}{3}\left(\frac{2}{3}z\right)^3\,dz=\frac{\pi}{2}R^3+\frac{4}{R^2}\cdot\frac{16}{81}\int_0^{\frac{3}{2}R}z^3\,dz\\ &=\frac{\pi}{2}R^3+\frac{4}{R^2}\cdot\frac{16}{81}\cdot\frac{1}{4}\left(\frac{3}{2}R\right)^4\\ &=\frac{\pi}{2}R^3+R^2. \end{align*}

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