Change of variables in triple integrals: spherical and cylindrical coordinates

Multiple integrals: Triple integrals

Change of variables in triple integrals: spherical and cylindrical coordinates

Suppose $x=x(u,v,w)$ , $y=y(u,v,w)$ and $z=z(u,v,w)$ is an invertible mapping from a bounded region $S$ in the $uvw$ -plane to a region $R$ in the $xyz$ plane and suppose the functions $x$ , $y$ and $z$ have continuous derivatives on $S$ . If the triple integral $\displaystyle \iint_R f(x,y,z)\,\dd(x,y,z)$ exists (i.e. $f(x,y,z)$ is integrable on $R$ ) and if $g(u,v,w)=f\bigl(x(u,v,w), y(u,v,w), z(u,v,w)\bigr)$ , then $g(u,v,w)$ is integrable on $S$ and $\iint_R f(x,y,z)\,\dd(x,y,z) =\iint_S g(u,v,w)\cdot \bigl|J(u,v,w)\bigr|\,\dd u\,\dd v\,\dd w$ with the Jacobian $J(u,v,w)$ defined as $\begin{aligned}J(u,v,w) &=\det\matrix{\dfrac{\partial x}{\partial u} &\dfrac{\partial x}{\partial v} &\dfrac{\partial x}{\partial w}\\ \dfrac{\partial y}{\partial u} & \dfrac{\partial y}{\partial v} & \dfrac{\partial y}{\partial w}\\ \dfrac{\partial z}{\partial u} & \dfrac{\partial z}{\partial v} & \dfrac{\partial z}{\partial w}}\\[0.25cm] &= \dfrac{\partial x}{\partial u}\cdot \det\matrix{\dfrac{\partial y}{\partial v} & \dfrac{\partial y}{\partial w}\\ \dfrac{\partial z}{\partial v} & \dfrac{\partial z}{\partial w}} - \dfrac{\partial y}{\partial u}\cdot \det\matrix{\dfrac{\partial x}{\partial v} & \dfrac{\partial x}{\partial w}\\ \dfrac{\partial z}{\partial v} & \dfrac{\partial z}{\partial w}}+\dfrac{\partial z}{\partial u}\cdot \det\matrix{\dfrac{\partial x}{\partial v} & \dfrac{\partial x}{\partial w}\\ \dfrac{\partial y}{\partial v} & \dfrac{\partial y}{\partial w}} \\[0.25cm] &\phantom{abcuva}\blue{\text{expansion of the determinant along the first column}} \end{aligned}$

Also if $u$ , $v$ and $w$ are known as functions of $x$ , $y$ , and $z$ you can determine the Jacobian $J(u,v,w)$ . After all, $J(u,v,w)\cdot \det\matrix{\dfrac{\partial u}{\partial x} &\dfrac{\partial u}{\partial y} &\dfrac{\partial u}{\partial z}\\ \dfrac{\partial v}{\partial x} & \dfrac{\partial v}{\partial y} & \dfrac{\partial v}{\partial z}\\ \dfrac{\partial w}{\partial x} & \dfrac{\partial w}{\partial y} & \dfrac{\partial w}{\partial z}}=1$ can be applied.

Suppose you want to calculate the volume $V$ of an ellipsoid $E=\{(x,y,x)\mid \frac{x^2}{a^2}+ \frac{y^2}{b^2}+ \frac{z^2}{c^2}\le 1]$ with $a>0, b>0, c>0$ . Under the transformation $x=a\,u, y=b\,v, z=c\,w$ the unit sphere $R=\{(u,v,w)\mid u^2+v^2+w^2\le 1\}$ is mapped onto the ellipsoid $E$ . The Jacobian $J(u,v,w)$ can easily be calculated: $\begin{aligned}J(u,v,w) &=\det\matrix{a & 0 & 0\\ 0 & b & 0 \\ 0 & 0 & c}\\[0.25cm] &= a\,b\,c\end{aligned}$ The volume $V$ of an ellipsoid can now be calculated as follows: $\begin{aligned} V &= \iiint_E \dd(x,y,z)\\[0.25cm] &= \iiint_{R} a\,b\,c\,\dd u\,\dd v\,\dd w\\[0.25cm] &= a\,b\,c\cdot \mathrm{volume of }(R)\\[0.25cm] &= \frac{4}{3}\pi\,a\,b\,c\end{aligned}$

Cylindrical coordinates are actually polar coordinates $x=r\cos\varphi$ and $y=r\sin\varphi$ with an extra dimension $z$ . As with polar coordinates, $\bigl|J(r,\varphi,z\bigr|=r$ and therefore $\iiint_R f(x,y,z)\dd(x,y,z)=\iiint_Sf(r\cos\varphi, r\sin\varphi,z)\,r\,\dd r\,\dd\varphi\,\dd z$ apply

You can also calculate the Jacobian $J(r,\varphi,z)$ directly based on the coordinate transformation and the definition of the Jacobian: $\begin{aligned}J(r,\varphi,z) &=\det\matrix{\dfrac{\partial x(r,\varphi,z)}{\partial r} &\dfrac{\partial x(r,\varphi,z)}{\partial \varphi} &\dfrac{\partial x(r,\varphi,z)}{\partial z}\\ \dfrac{\partial y(r,\varphi,z)}{\partial r} & \dfrac{\partial y(r,\varphi,z)}{\partial \varphi} & \dfrac{\partial y(r,\varphi,z)}{\partial z}\\ \dfrac{\partial z(r,\varphi,z)}{\partial r} & \dfrac{\partial z(r,\varphi,z)}{\partial \varphi} & \dfrac{\partial z(r,\varphi,z)}{\partial z}}\\[0.25cm] &= \det\matrix{\cos\varphi & \sin\varphi &0 \\ -r\sin\varphi & r\cos\varphi & 0\\ 0 & 0 & 1}\\[0.25cm] &= r\,(\cos^2\varphi + \sin^2\varphi)\\[0.25cm] &= r\end{aligned}$

Using cylindrical coordinates, calculate the triple integral $\iiint_R (x^2+y^2+ z^2)\,\dd(x,y,z)$ where $R=\{(x,y,z)\mid 0\le x^2+y^2\le 4,\;\; 0\le z\le 3\}$

$\iiint_B (x^2+y^2+ z^2)\,\dd(x,y,z)={}$ $30$

First note that the region of integration $R$ can be described in cylindrical coordinates with the region of integration $S=\{(r,\varphi,z)\mid 0\le r\le 2,\;\; 0\le \varphi\le 2\pi,\;\; 0\le z\le 3\}$ . The requested triple integral can now be calculated as an iterated integral in the following way: $\begin{aligned} \iiint_R (x^2+y^2+ z^2)\,\dd(x,y,z) &= \iiint_S (r^2+ z^2)\,r\,\dd r\,\dd\varphi\,\dd z \\[0.25cm] &= \int_{z=0}^{z=3}\left(\int_{\varphi=0}^{\varphi=2\pi}\left(\int_{r=0}^{r=2}(r^3+z^2\,r)\,\dd r\right)\dd \varphi\right)\dd z\\[0.25cm] &= \int_{z=0}^{z=3}\left(\int_{\varphi=0}^{\varphi=2\pi} \biggl[\frac{1}{4}r^4+\frac{1}{2}z^2\,r^2\biggr]_{r=0}^{r=2}\;\dd\varphi\right)\dd z\\[0.25cm] &= \int_{z=0}^{z=3}\left(\int_{\varphi=0}^{\varphi=2\pi}\Bigl(4 +2z^2\Bigr)\,\dd\varphi\right)\dd z \\[0.25cm] &= \int_{z=0}^{z=3}\biggl[\Bigl(4 +2z^2\Bigr)\varphi\biggr]_{\varphi=0}^{\varphi=2\pi}\;\dd z\\[0.25cm] &= 2\pi \int_{z=0}^{z=3}\Bigl(4 +2z^2\Bigr)\dd z\\[0.25cm] &= \biggl[4z+2\times \frac{1}{3}z^3\biggr]_{z=0}^{z=3}\\[0.25cm] &= \biggl[4z+{{2}\over{3}}z^3\biggr]_{z=0}^{z=3}\\[0.25cm] &=4\times 3+{{2}\over{3}}\times 3^3\\[0.25cm] &= 30\end{aligned}$

New example

Spherical coordinates $(r,\varphi,\theta)$ are defined as in the figure below, with the following formulas $\begin{aligned}x(r,\varphi,\theta)&=r\cos\varphi\sin\theta\\ y(r,\varphi,\theta)&=r\sin\varphi\sin\theta\\ z(r,\varphi,\theta) &= r\cos\theta\end{aligned}$

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In this case the Jacobian of the coordinate transformation is $\bigl|J(r,\varphi,z\bigr|=r^2\cos\theta$ and therefore $\iiint_R f(x,y,z)\dd(x,y,z)=\iiint_S f(r\cos\varphi\cos\theta, r\sin\varphi\cos\theta,r\sin\theta)\,r^2\sin\theta\,\dd r\,\dd\varphi\,\dd\theta$

You can directly calculate the Jacobian $J(r,\varphi,z)$ based on the coordinate transformation and the definition of the Jacobian: $\begin{aligned}J(r,\varphi,\theta) &=\det\matrix{\dfrac{\partial x}{\partial r} &\dfrac{\partial x}{\partial \varphi} &\dfrac{\partial x}{\partial \theta}\\ \dfrac{\partial y}{\partial r} & \dfrac{\partial y}{\partial \varphi} & \dfrac{\partial y}{\partial \theta}\\ \dfrac{\partial z}{\partial r} & \dfrac{\partial z}{\partial \varphi} & \dfrac{\partial z}{\partial \theta}}\\[0.25cm] &= \det\matrix{\cos\varphi \sin\theta& -r\sin\varphi\sin\theta & r\cos\varphi\cos\theta \\ \sin\varphi\sin\theta & r\cos\varphi\sin\theta & r\sin\varphi\cos\theta\\ \cos\theta & 0 & -r\sin\theta}\\[0.25cm] &= \cos\theta\cdot\det\matrix{-r\sin\varphi\sin\theta & r\cos\varphi\cos\theta \\ r\cos\varphi\sin\theta & r\sin\varphi\cos\theta} -r\sin\theta\cdot\det\matrix{\cos\varphi \sin\theta& -r\sin\varphi\sin\theta \\ \sin\varphi\sin\theta & r\cos\varphi\sin\theta } \\[0.25cm] &= -r\cos^2\theta\sin\theta\,(\sin^2\varphi+\cos^2\varphi) - r\sin^3\theta\,(\cos^2\varphi+\sin^2\varphi)\\[0.25cm] &= -r^2(\cos^2\theta+\sin^2 \theta)\sin\theta\\[0.25cm] &= -r^2\sin\theta\end{aligned}$ So: $\bigl|J(r,\varphi,\theta)\bigr|= r^2\sin\theta$

Using spherical coordinates, calculate the triple integral $\iiint_R \frac{1}{x^2+y^2+z^2}\,\dd(x,y,z)$ where $R=\{(x,y,z)\mid 0\le x^2+y^2+z^2\le 1\}$

$\iiint_R \frac{1}{x^2+y^2+z^2}\,\dd(x,y,z)={}$ $4\pi$

First note that the region of integration $R$ can be described in spherical coordinates with the region of integration $S=\{(r,\varphi,\theta)\mid 0\le r\le 1,\;\; 0\le \varphi\le 2\pi,\;\; 0\le \theta \le \pi\}$ The requested triple integral can now be calculated as iterated integral in the following way:
$\begin{aligned} \iiint_R \frac{1}{x^2+y^2+z^2}\,\dd(x,y,z) &=\int_{\theta=0}^{\theta=\pi}\left(\int_{\varphi=0}^{\varphi=2\pi}\left(\int_{r=0}^{r=1}\frac{1}{r^2}\cdot r^2\sin\theta\,\dd r\right)\dd\varphi\right)\dd \theta\\[0.25cm] &=\int_{\theta=0}^{\theta=\pi}\left(\int_{\varphi=0}^{\varphi=2\pi}\left(\int_{r=0}^{r=1}\sin\theta\,\dd r\right)\dd\varphi\right)\dd \theta\\[0.25cm] &=\int_{\theta=0}^{\theta=\pi}\left(\int_{\varphi=0}^{\varphi=2\pi}\biggl[r\sin\theta\biggr]_{r=0}^{r=1}\;\dd\varphi\right)\dd \theta\\[0.25cm] &=\int_{\theta=0}^{\theta=\pi}\left(\int_{\varphi=0}^{\varphi=2\pi}\sin\theta\,\dd\varphi\right)\dd \theta\\[0.25cm] &=\int_{\theta=0}^{\theta=\pi}2\pi\sin\theta\,\dd \theta\\[0.25cm] &=\biggl[-2\pi\cos\theta\biggr]_{\theta=0}^{\theta=\pi}\\[0.25cm]&= 4\pi\end{aligned}$

New example