Multivariable calculus and vector analysis

Triple integral change of variables examples

Spherical coordinates example

For spherical coordinates, the change of variables function is

$$\begin{array}{llll}\hfill \left(x,y,z\right)& =\mathbf{T}\left(\rho ,\theta ,\varphi \right)& \hfill & \end{array}$$

where the components of $\mathbf{T}$ are given by

$$\begin{array}{llll}\hfill x& =\rho sin\varphi cos\theta & \hfill & \\ \hfill y& =\rho sin\varphi sin\theta & \hfill & \\ \hfill z& =\rho cos\varphi .& \hfill & \end{array}$$

We can compute that

$$\begin{array}{llll}\hfill \frac{\partial \left(x,y,z\right)}{\partial \left(\rho ,\theta ,\varphi \right)}& =\left|\begin{array}{ccc}\hfill \frac{\partial x}{\partial \rho }\hfill & \hfill \frac{\partial x}{\partial \theta }\hfill & \hfill \frac{\partial x}{\partial \varphi }\hfill \\ \hfill \frac{\partial y}{\partial \rho }\hfill & \hfill \frac{\partial y}{\partial \theta }\hfill & \hfill \frac{\partial y}{\partial \varphi }\hfill \\ \hfill \frac{\partial z}{\partial \rho }\hfill & \hfill \frac{\partial z}{\partial \theta }\hfill & \hfill \frac{\partial z}{\partial \varphi }\hfill \\ \hfill \hfill \end{array}\right|& \hfill & \\ \hfill & =\left|\begin{array}{ccc}\hfill sin\varphi cos\theta \hfill & \hfill -\rho sin\varphi cos\theta \hfill & \hfill \rho cos\varphi cos\theta \hfill \\ \hfill sin\varphi sin\theta \hfill & \hfill \rho sin\varphi cos\theta \hfill & \hfill \rho cos\varphi sin\theta \hfill \\ \hfill cos\varphi \hfill & \hfill 0\hfill & \hfill -\rho sin\varphi \hfill \end{array}\right|& \hfill & \\ \hfill & =-{\rho }^{2}sin\varphi .& \hfill & \end{array}$$

The change of variable factor is the absolute value of the determinant

$$\begin{array}{lll}\hfill \left|\frac{\partial \left(x,y,z\right)}{\partial \left(\rho ,\theta ,\varphi \right)}\right|={\rho }^{2}sin\varphi .& & \hfill \end{array}$$

This means that the map from spherical coordinates to rectangular coordinates changes volume by the factor ${\rho }^{2}sin\varphi$. For this reason, you need to do the above calculation only once. Now, you can just remember that the factor for spherical coordinates is ${\rho }^{2}sin\varphi$.

Now we’re ready for the example: find the mass of a star that is a ball of radius 3 centered at the origin if the density of the star is $g\left(x,y,z\right)=10-x^2-y^2-z^2$.

If we try to compute the integral directly in rectangular coordinates, it isn’t so easy:

$$\begin{array}{llll}\hfill & {\iiint }_{S}g\left(x,y,z\right)dxdydz& \hfill & \\ \hfill & ={\int \; <!--nolimits\; -->}_{-3}^3{\int \; <!--nolimits\; -->}_{-\sqrt{9-z^2}}^{\sqrt{9-z^2}}{\int \; <!--nolimits\; -->}_{-\sqrt{9-z^2-y^2}}^{\sqrt{9-z^2-y^2}}\left(10-x^2-y^2-z^2\right)dxdydz& \hfill & \end{array}$$

A mess.

We can find the mass of the star more easily in spherical coordinates.

The star density $g\left(x,y,z\right)=10-x^2-y^2-z^2$ becomes $g\left(\mathbf{T}\left(\rho ,\theta ,\varphi \right)\right)=10-{\rho }^2$.

In spherical coordinates, the integral over ball of radius 3 is the integral over the region

$$\begin{array}{lll}\hfill 0\le \rho \le 3,0\le \theta \le 2\pi ,0\le \varphi \le \pi & & \hfill \end{array}$$

The volume element is ${\rho }^{2}sin\varphi d\rho d\theta d\varphi$.

Therefore, the mass of the star is

$$\begin{array}{llll}\hfill & {\int \; <!--nolimits\; -->}_0^3{\int \; <!--nolimits\; -->}_0^{2\pi }{\int \; <!--nolimits\; -->}_0^{\pi }\left(1-{\rho }^2\right){\rho }^{2}sin\varphi d\varphi d\theta d\rho & \hfill & \\ \hfill & ={\int \; <!--nolimits\; -->}_0^3{\int \; <!--nolimits\; -->}_0^{2\pi }{\left.\left(10-{\rho }^2\right){\rho }^2\left(-cos\varphi \right)\right|}_{\varphi =0}^{\varphi =\pi }d\theta d\rho & \hfill & \\ \hfill & ={\int \; <!--nolimits\; -->}_0^3{\int \; <!--nolimits\; -->}_0^{2\pi }\left(10-{\rho }^2\right){\rho }^{2}2d\theta d\rho & \hfill & \\ \hfill & ={\int \; <!--nolimits\; -->}_0^{3}4\pi \left(10-{\rho }^2\right){\rho }^{2}d\rho =\frac{828\pi }{5}\approx 520& \hfill & \end{array}$$

Cylinderical coordinates example

For cylindrical coordinates, the change of variables function is

$$\begin{array}{llll}\hfill \left(x,y,z\right)& =\mathbf{T}\left(r,\theta ,w\right)& \hfill & \end{array}$$

where the components of $\mathbf{T}$ are given by

$$\begin{array}{llll}\hfill x& =rcos\theta & \hfill & \\ \hfill y& =rsin\theta & \hfill & \\ \hfill z& =w.& \hfill & \end{array}$$

We can compute that

$$\begin{array}{llll}\hfill \frac{\partial \left(x,y,z\right)}{\partial \left(\rho ,\theta ,\varphi \right)}& =\left|\begin{array}{ccc}\hfill \frac{\partial x}{\partial r}\hfill & \hfill \frac{\partial x}{\partial \theta }\hfill & \hfill \frac{\partial x}{\partial w}\hfill \\ \hfill \frac{\partial y}{\partial r}\hfill & \hfill \frac{\partial y}{\partial \theta }\hfill & \hfill \frac{\partial y}{\partial w}\hfill \\ \hfill \frac{\partial z}{\partial r}\hfill & \hfill \frac{\partial z}{\partial \theta }\hfill & \hfill \frac{\partial z}{\partial w}\hfill \\ \hfill \hfill \end{array}\right|& \hfill & \\ \hfill & =\left|\begin{array}{ccc}\hfill cos\theta \hfill & \hfill -rsin\theta \hfill & \hfill 0\hfill \\ \hfill sin\theta \hfill & \hfill rcos\theta \hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right|& \hfill & \\ \hfill & =r{cos}^2\theta +r{sin}^2\theta =r& \hfill & \end{array}$$

Volume changes by $\left|r\right|=r$ (just like area change for polar coordinates). Replace $dxdydz$ by $rdrd\theta dw$.

Example: Find the volume of cone of height 1 and radius one. Bounded by surface $z=\sqrt{x^2+y^2}$ and plane $z=1$.

Volume is

$$\begin{array}{lll}\hfill {\int \; <!--nolimits\; -->}_{-1}^1{\int \; <!--nolimits\; -->}_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}{\int \; <!--nolimits\; -->}_{\sqrt{x^2+y^2}}^{1}dzdydz& & \hfill \end{array}$$

Easier to compute in cylindrical coordinates.

Cone is

$$\begin{array}{lll}\hfill 0\le \theta \le 2\pi ,0\le r\le 1& & \hfill \\ \hfill r\le z\le 1& & \hfill \end{array}$$

Volume of cone is

$$\begin{array}{llll}\hfill & {\int \; <!--nolimits\; -->}_0^1{\int \; <!--nolimits\; -->}_0^{2\pi }{\int \; <!--nolimits\; -->}_r^{1}rdzd\theta dr& \hfill & \\ \hfill & ={\int \; <!--nolimits\; -->}_0^1{\int \; <!--nolimits\; -->}_0^{2\pi }\left(1-r\right)rd\theta dr& \hfill & \\ \hfill & ={\int \; <!--nolimits\; -->}_0^{1}2\pi \left(1-r\right)rdr=\frac{\pi }{3}& \hfill & \end{array}$$

Ice cream cone revisited

Armed with the knowledge of how to change to spherical coordinates, we can now revisit the ice cream cone example.

Earlier we tried to find the volume of an ice cream cone

and discovered volume was

$$\begin{array}{lll}\hfill {\iiint }_{W}dV={\int \; <!--nolimits\; -->}_{-1∕\sqrt{2}}^{1∕\sqrt{2}}{\int \; <!--nolimits\; -->}_{-\sqrt{1∕2-x^2}}^{\sqrt{1∕2-x^2}}{\int \; <!--nolimits\; -->}_{\sqrt{x^2+y^2}}^{\sqrt{1-x^2-y^2}}dzdydx& & \hfill \end{array}$$

We need to describe the bounds in terms of spherical coordinates. Since the cone is symmetrical around the $z$ axis, $\theta$ is easy. $0\le \theta \le 2\pi$ in the cone and the ranges of the other variables don’t depend on $\theta$. Also, we see that $0\le r\le 1$, since a given line from the origin extends until it hits the sphere $z=\sqrt{1-x^2-y^2}$, which is a sphere of radius 1. Lastly, $\varphi$ is determined by requiring that we are in the cone $z\ge \sqrt{x^2+y^2}$. (To see this is exactly a condition on $\varphi$, look at the reading on spherical coordinates.) On the cone, $z=\sqrt{x^2+y^2}$, $\varphi =\pi ∕4$. Consquently, the condition on $\varphi$ is $0\le \varphi \le \pi ∕4$.

In summary, the ice cream cone is described by

$$\begin{array}{lll}\hfill 0\le \theta \le 2\pi ,0\le r\le 1,0\le \varphi \le \pi ∕4& & \hfill \end{array}$$

Changing to spherical coordinates, we calculate that the volume of the ice cream cone is

$$\begin{array}{llll}\hfill {\int \; <!--nolimits\; -->}_0^1{\int \; <!--nolimits\; -->}_0^{2\pi }{\int \; <!--nolimits\; -->}_0^{\pi ∕4}{\rho }^{2}sin\varphi d\varphi d\theta d\rho & ={\int \; <!--nolimits\; -->}_0^1{\int \; <!--nolimits\; -->}_0^{2\pi }\left[cos\left(0\right)-cos\left(\pi ∕4\right)\right]{\rho }^{2}d\theta d\rho & \hfill & \\ \hfill & ={\int \; <!--nolimits\; -->}_0^1{\int \; <!--nolimits\; -->}_0^{2\pi }\frac{2-\sqrt{2}}{2}{\rho }^{2}d\theta d\rho & \hfill & \\ \hfill & ={\int \; <!--nolimits\; -->}_0^1\pi \left(2-\sqrt{2}\right){\rho }^{2}d\rho & \hfill & \\ \hfill & =\pi \frac{2-\sqrt{2}}{3}& \hfill & \end{array}$$

Contents by topic

1. Geometry in higher dimensions

2. Differential Calculus

3. Integral calculus

3a. Double integrals

3b. Triple Integrals

3c. Changing variables

• Change of variables in double integrals
• Double integral change of variables examples
• Change of variables in triple integrals
• Triple integral change of variables examples

4. Vector calculus

5. The fundamental theorems of vector calculus

6. Review material