Triple Integrals in Cylindrical and Spherical Coordinates

Evaluate the triple integral ${\displaystyle \underset{B}{\iiint }\left(zr\text{sin}\theta \right)}rdrd\theta dz$

where the cylindrical box $B$

is $B=\left\{\left(r,\theta ,z\right)\backslash |0\le r\le 2,0\le \theta \le \pi \text{/}2,0\le z\le 4\right\}.$

As stated in Fubini’s theorem, we can write the triple integral as the iterated integral

The evaluation of the iterated integral is straightforward. Each variable in the integral is independent of the others, so we can integrate each variable separately and multiply the results together. This makes the computation much easier:

Consider the region $E$

inside the right circular cylinder with equation $r=2\text{sin}\theta ,$

bounded below by the $r\theta$

-plane and bounded above by the sphere with radius $4$

centered at the origin ([link]). Set up a triple integral over this region with a function $f\left(r,\theta ,z\right)$

in cylindrical coordinates.

First, identify that the equation for the sphere is $r^2+z^2=16.$

We can see that the limits for $z$

are from $0$

to $z=\sqrt{16-r^2}.$

Then the limits for $r$

are from $0$

to $r=2\text{sin}\theta .$

Finally, the limits for $\theta$

are from $0$

to $\pi .$

Hence the region is

Therefore, the triple integral is

Let $E$

be the region bounded below by the cone $z=\sqrt{x^2+y^2}$

and above by the paraboloid $z=2-x^2-y^2.$

([link]). Set up a triple integral in cylindrical coordinates to find the volume of the region, using the following orders of integration:

1. $dzdrd\theta$
2. $drdzd\theta .$

   ---

   

1. The cone is of radius 1 where it meets the paraboloid. Since $z=2-x^2-y^2=2-r^2$

   and

   $z=\sqrt{x^2+y^2}=r$

   (assuming

   $r$

   is nonnegative), we have

   $2-r^2=r.$

   Solving, we have

   $r^2+r-2=\left(r+2\right)\left(r-1\right)=0.$

   Since

   $r\ge 0,$

   we have

   $r=1.$

   Therefore

   $z=1.$

   So the intersection of these two surfaces is a circle of radius

   $1$

   in the plane

   $z=1.$

   The cone is the lower bound for

   $z$

   and the paraboloid is the upper bound. The projection of the region onto the

   $xy$

   -plane is the circle of radius

   $1$

   centered at the origin.

   ---

   Thus, we can describe the region as

   ---

   $E=\left\{\left(r,\theta ,z\right)\backslash |0\le \theta \le 2\pi ,0\le r\le 1,r\le z\le 2-r^2\right\}.$

   ---

   Hence the integral for the volume is

   ---

   $V={\displaystyle \underset{\theta =0}{\overset{\theta =2\pi }{\int }}{\displaystyle \underset{r=0}{\overset{r=1}{\int }}{\displaystyle \underset{z=r}{\overset{z=2-r^2}{\int }}rdzdrd\theta }}}.$
2. We can also write the cone surface as $r=z$

   and the paraboloid as

   $r^2=2-z.$

   The lower bound for

   $r$

   is zero, but the upper bound is sometimes the cone and the other times it is the paraboloid. The plane

   $z=1$

   divides the region into two regions. Then the region can be described as

   ---

   $\begin{array}{cc}\hfill E& =\left\{\left(r,\theta ,z\right)\backslash |0\le \theta \le 2\pi ,0\le z\le 1,0\le r\le z\right\}\hfill \\ & \cup \left\{\left(r,\theta ,z\right)\backslash |0\le \theta \le 2\pi ,1\le z\le 2,0\le r\le \sqrt{2-z}\right\}.\hfill \end{array}$

   ---

   Now the integral for the volume becomes

   ---

   $V={\displaystyle \underset{\theta =0}{\overset{\theta =2\pi }{\int }}{\displaystyle \underset{z=0}{\overset{z=1}{\int }}{\displaystyle \underset{r=0}{\overset{r=z}{\int }}rdrdzd\theta +}}}{\displaystyle \underset{\theta =0}{\overset{\theta =2\pi }{\int }}{\displaystyle \underset{z=1}{\overset{z=2}{\int }}{\displaystyle \underset{r=0}{\overset{r=\sqrt{2-z}}{\int }}rdrdzd\theta }}}.$

Let E be the region bounded below by the $r\theta$

-plane, above by the sphere $x^2+y^2+z^2=4,$

and on the sides by the cylinder $x^2+y^2=1$

([link]). Set up a triple integral in cylindrical coordinates to find the volume of the region using the following orders of integration, and in each case find the volume and check that the answers are the same:

1. $dzdrd\theta$
2. $drdzd\theta .$

   ---

   

1. Note that the equation for the sphere is

   ---

   $x^2+y^2+z^2=4\text{or}{r}^2+z^2=4$

   ---

   and the equation for the cylinder is

   ---

   $x^2+y^2=1\text{or}{r}^2=1.$

   ---

   Thus, we have for the region

   $E$

   ---

   $E=\left\{\left(r,\theta ,z\right)\backslash |0\le z\le \sqrt{4-r^2},0\le r\le 1,0\le \theta \le 2\pi \right\}$

   ---

   Hence the integral for the volume is

   ---

   $\begin{array}{cc}\hfill V\left(E\right)& ={\displaystyle \underset{\theta =0}{\overset{\theta =2\pi }{\int }}{\displaystyle \underset{r=0}{\overset{r=1}{\int }}{\displaystyle \underset{z=0}{\overset{z=\sqrt{4-r^2}}{\int }}rdzdrd\theta }}}\hfill \\ & ={\displaystyle \underset{\theta =0}{\overset{\theta =2\pi }{\int }}{\displaystyle \underset{r=0}{\overset{r=1}{\int }}\left[{rz\backslash |}_{z=0}^{z=\sqrt{4-r^2}}\right]}}drd\theta ={\displaystyle \underset{\theta =0}{\overset{\theta =2\pi }{\int }}{\displaystyle \underset{r=0}{\overset{r=1}{\int }}\left(r\sqrt{4-r^2}\right)}}drd\theta \hfill \\ & ={\displaystyle \underset{0}{\overset{2\pi }{\int }}\left(\frac{8}{3}-\sqrt{3}\right)d\theta =2\pi \left(\frac{8}{3}-\sqrt{3}\right)}\text{cubic units}\text{.}\hfill \end{array}$
2. Since the sphere is $x^2+y^2+z^2=4,$

   which is

   $r^2+z^2=4,$

   and the cylinder is

   $x^2+y^2=1,$

   which is

   $r^2=1,$

   we have

   $1+z^2=4,$

   that is,

   $z^2=3.$

   Thus we have two regions, since the sphere and the cylinder intersect at

   $\left(1,\sqrt{3}\right)$

   in the

   $rz$

   -plane

   ---

   $E_1=\left\{\left(r,\theta ,z\right)\backslash |0\le r\le \sqrt{4-r^2},\sqrt{3}\le z\le 2,0\le \theta \le 2\pi \right\}$

   ---

   and

   ---

   $E_2=\left\{\left(r,\theta ,z\right)\backslash |0\le r\le 1,0\le z\le \sqrt{3},0\le \theta \le 2\pi \right\}.$

   ---

   Hence the integral for the volume is

   ---

   $\begin{array}{cc}\hfill V\left(E\right)& ={\displaystyle \underset{\theta =0}{\overset{\theta =2\pi }{\int }}{\displaystyle \underset{z=\sqrt{3}}{\overset{z=2}{\int }}{\displaystyle \underset{r=0}{\overset{r=\sqrt{4-r^2}}{\int }}rdrdzd\theta }}}+{\displaystyle \underset{\theta =0}{\overset{\theta =2\pi }{\int }}{\displaystyle \underset{z=0}{\overset{z=\sqrt{3}}{\int }}{\displaystyle \underset{r=0}{\overset{r=1}{\int }}rdrdzd\theta }}}\hfill \\ & =\sqrt{3}\pi +\left(\frac{16}{3}-3\sqrt{3}\right)\pi =2\pi \left(\frac{8}{3}-\sqrt{3}\right)\text{cubic units}.\hfill \end{array}$

Evaluate the iterated triple integral ${\displaystyle \underset{\theta =0}{\overset{\theta =2\pi }{\int }}{\displaystyle \underset{\phi =0}{\overset{\phi =\pi \text{/}2}{\int }}{\displaystyle \underset{p=0}{\overset{\rho =1}{\int }}{\rho }^2\text{sin}\phi d\rho d\phi d\theta }}}.$

As before, in this case the variables in the iterated integral are actually independent of each other and hence we can integrate each piece and multiply:

Set up an integral for the volume of the region bounded by the cone $z=\sqrt{3\left(x^2+y^2\right)}$

and the hemisphere $z=\sqrt{4-x^2-y^2}$

(see the figure below).

Using the conversion formulas from rectangular coordinates to spherical coordinates, we have:

For the cone: $z=\sqrt{3\left(x^2+y^2\right)}$

or $\rho \text{cos}\phi =\sqrt{3}\rho \text{sin}\phi$

or $\text{tan}\phi =\frac{1}{\sqrt{3}}$

or $\phi =\frac{\pi }{6}.$

For the sphere: $z=\sqrt{4-x^2-y^2}$

or $z^2+x^2+y^2=4$

or ${\rho }^2=4$

or $\rho =2.$

Thus, the triple integral for the volume is $V\left(E\right)={\displaystyle \underset{\theta =0}{\overset{\theta =2\pi }{\int }}{\displaystyle \underset{\varphi =0}{\overset{\phi =\pi \text{/}6}{\int }}{\displaystyle \underset{\rho =0}{\overset{\rho =2}{\int }}{\rho }^2\text{sin}\phi d\rho d\phi d\theta }}}.$

Let $E$

be the region bounded below by the cone $z=\sqrt{x^2+y^2}$

and above by the sphere $z=x^2+y^2+z^2$

([link]). Set up a triple integral in spherical coordinates and find the volume of the region using the following orders of integration:

1. $d\rho d\varphi d\theta ,$
2. $d\phi d\rho d\theta .$

   ---

   

1. Use the conversion formulas to write the equations of the sphere and cone in spherical coordinates.

   ---

   For the sphere:

   ---

   $\begin{array}{ccc}\hfill x^2+y^2+z^2& =\hfill & z\hfill \\ \hfill {\rho }^2& =\hfill & \rho \text{cos}\phi \hfill \\ \hfill \rho & =\hfill & \text{cos}\phi .\hfill \end{array}$

   ---

   For the cone:

   ---

   $\begin{array}{ccc}\hfill z& =\hfill & \sqrt{x^2+y^2}\hfill \\ \hfill \rho \text{cos}\phi & =\hfill & \sqrt{{\rho }^2{\text{sin}}^2\phi {\text{cos}}^2\varphi +{\rho }^2{\text{sin}}^2\phi {\text{sin}}^2\varphi }\hfill \\ \hfill \rho \text{cos}\phi & =\hfill & \sqrt{{\rho }^2{\text{sin}}^2\phi \left({\text{cos}}^2\varphi +{\text{sin}}^2\varphi \right)}\hfill \\ \hfill \rho \text{cos}\phi & =\hfill & \rho \text{sin}\phi \hfill \\ \hfill \text{cos}\phi & =\hfill & \text{sin}\phi \hfill \\ \hfill \phi & =\hfill & \pi \text{/}4.\hfill \end{array}$

   ---

   Hence the integral for the volume of the solid region

   $E$

   becomes

   ---

   $V(E)={\displaystyle \underset{\theta =0}{\overset{\theta =2\pi }{\int }}{\displaystyle \underset{\phi =0}{\overset{\phi =\pi \text{/}4}{\int }}{\displaystyle \underset{\rho =0}{\overset{\rho =\text{cos}\phi }{\int }}{\rho }^2\text{sin}\phi d\rho d\phi d\theta }}}.$
2. Consider the $\phi \rho$

   -plane. Note that the ranges for

   $\phi$

   and

   $\rho$

   (from part a.) are

   ---

   $\begin{array}{c}0\le \phi \le \pi \text{/}4\hfill \\ 0\le \rho \le \text{cos}\phi .\hfill \end{array}$

   ---

   The curve

   $\rho =\text{cos}\phi$

   meets the line

   $\phi =\pi \text{/}4$

   at the point

   $\left(\pi \text{/}4,\sqrt{2}\text{/}2\right).$

   Thus, to change the order of integration, we need to use two pieces:

   ---

   $\begin{array}{ccccccc}\begin{array}{c}0\le \rho \le \sqrt{2}\text{/}2\hfill \\ 0\le \phi \le \pi \text{/}4\hfill \end{array}\hfill & & & \text{and}\hfill & & & \begin{array}{ccc}\hfill \sqrt{2}\text{/}2& \le \hfill & \rho \le 1\hfill \\ \hfill 0& \le \hfill & \phi \le {\text{cos}}^{\mathrm{-1}}\rho .\hfill \end{array}\hfill \end{array}$

   ---

   Hence the integral for the volume of the solid region

   $E$

   becomes

   ---

   $V(E)={\displaystyle \underset{\theta =0}{\overset{\theta =2\pi }{\int }}{\displaystyle \underset{\rho =0}{\overset{\rho =\sqrt{2}\text{/}2}{\int }}{\displaystyle \underset{\phi =0}{\overset{\phi =\pi \text{/}4}{\int }}{\rho }^2\text{sin}\phi d\phi d\rho d\theta }}}+{\displaystyle \underset{\theta =0}{\overset{\theta =2\pi }{\int }}{\displaystyle \underset{\rho =\sqrt{2}\text{/}2}{\overset{\rho =1}{\int }}{\displaystyle \underset{\phi =0}{\overset{\phi ={\text{cos}}^{\mathrm{-1}}\rho }{\int }}{\rho }^2\text{sin}\phi d\phi d\rho d\theta }}}.$

   ---

   In each case, the integration results in

   $V\left(E\right)=\frac{\pi }{8}.$

Convert the following integral into cylindrical coordinates:

The ranges of the variables are

The first two inequalities describe the right half of a circle of radius $1.$

Therefore, the ranges for $\theta$

and $r$

are

The limits of $z$

are $r^2\le z\le r,$

hence

Convert the following integral into spherical coordinates:

The ranges of the variables are

The first two ranges of variables describe a quarter disk in the first quadrant of the $xy$

-plane. Hence the range for $\theta$

is $0\le \theta \le \frac{\pi }{2}.$

The lower bound $z=\sqrt{x^2+y^2}$

is the upper half of a cone and the upper bound $z=\sqrt{18-x^2-y^2}$

is the upper half of a sphere. Therefore, we have $0\le \rho \le \sqrt{18},$

which is $0\le \rho \le 3\sqrt{2}.$

For the ranges of $\phi ,$

we need to find where the cone and the sphere intersect, so solve the equation

This gives

Putting this together, we obtain

Find the volume of the spherical planetarium in l’Hemisphèric in Valencia, Spain, which is five stories tall and has a radius of approximately $50$

ft, using the equation $x^2+y^2+z^2=r^2.$

We calculate the volume of the ball in the first octant, where $x\ge 0,y\ge 0,$

and $z\ge 0,$

using spherical coordinates, and then multiply the result by $8$

for symmetry. Since we consider the region $D$

as the first octant in the integral, the ranges of the variables are

Therefore,

This exactly matches with what we knew. So for a sphere with a radius of approximately $50$

ft, the volume is $\frac{4}{3}\pi {(50)}^3\approx \mathrm{523,600}{\text{ft}}^3.$

Find the volume of the ellipsoid $\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1.$

We again use symmetry and evaluate the volume of the ellipsoid using spherical coordinates. As before, we use the first octant $x\ge 0,y\ge 0,$

and $z\ge 0$

and then multiply the result by $8.$

In this case the ranges of the variables are

Also, we need to change the rectangular to spherical coordinates in this way:

Then the volume of the ellipsoid becomes

Find the volume of the space inside the ellipsoid $\frac{x^2}{{75}^2}+\frac{y^2}{{80}^2}+\frac{z^2}{{90}^2}=1$

and outside the sphere $x^2+y^2+z^2={50}^2.$

This problem is directly related to the l’Hemisphèric structure. The volume of space inside the ellipsoid and outside the sphere might be useful to find the expense of heating or cooling that space. We can use the preceding two examples for the volume of the sphere and ellipsoid and then substract.

First we find the volume of the ellipsoid using $a=75\text{ft,}$

$b=80\text{ft,}$

and $c=90\text{ft}$

in the result from [link]. Hence the volume of the ellipsoid is

From [link], the volume of the sphere is

Therefore, the volume of the space inside the ellipsoid $\frac{x^2}{{75}^2}+\frac{y^2}{{80}^2}+\frac{z^2}{{90}^2}=1$

and outside the sphere $x^2+y^2+z^2={50}^2$

is approximately