Review of Spherical Coordinates

In three-dimensional space ${\mathbb{R}^{3}}$ in the spherical coordinate system, we specify a point $P$ by its distance ${\rho}$ from the origin, the polar angle ${\theta}$ from the positive $x$-axis (same as in the cylindrical coordinate system), and the angle ${\varphi}$ from the positive $z$-axis and the line $OP$ (Figure 1). Note that ${\rho} \ {\geq} \ {0}$ and ${0} \ {\leq} \ {\varphi} \ {\leq} \ {\pi}$. (Refer to Cylindrical and Spherical Coordinates for a review.) Spherical coordinates are useful for triple integrals over regions that are symmetric with respect to the origin.

Figure 1. The spherical coordinate system locates points with two angles and a distance from the origin.

Recall the relationships that connect rectangular coordinates with spherical coordinates.

From spherical coordinates to rectangular coordinates:

${{x} = {\rho} \ {\sin} \ {\varphi} \ {\cos} \ {\theta}}, {{y} = {\rho} \ {\sin} \ {\varphi} \ {\sin} \ {\theta}}, \ {\text{and}} \ {{z} = {\rho} \ {\cos} \ {\varphi}}.$

From rectangular coordinates to spherical coordinates:

${\rho}^{2} = {x^2} + {y^2} + {z^2}, \ {\tan}{\theta} = {\frac{y}{x}}, {\varphi} = {\arccos}{\left ( {\frac{z}{\sqrt{{x^2} + {y^2} + {z^2}}}} \right )}.$

Other relationships that are important to know for conversions are

• ${r} = {\rho} \ {\sin} \ {\varphi}$
• ${\theta} = {\theta}$                                                     These equations are used to convert from spherical coordinates to cylindrical coordinates
• ${z} = {\rho} \ {\cos} \ {\varphi}$

Figure 2. Spherical coordinates are especially convenient for working with solids bounded by these types of surfaces. (The letter $c$ indicates a constant.)

Integration in Spherical Coordinates

We now establish a triple integral in the spherical coordinate system, as we did before in the cylindrical coordinate system. Let the function ${f}{({\rho},{\theta},{\varphi})}$ be continuous in a bounded spherical box, ${B} = {\left \{ {({\rho},{\theta},{\varphi})}{\mid}{a} \ {\leq} \ {\rho} \ {\leq} \ {{b},{\alpha}} \ {\leq} \ {\theta} \ {\leq} \ {{\beta},{\gamma}} \ {\leq} \ {\varphi} \ {\leq} \ {\psi} \right \}}$. We then divide each interval into $l$, $m$, and $n$ subdivisions such that ${{\Delta}{\rho} = {\frac{{b}-{a}}{l}}}, \ {{\Delta}{\theta} = {\frac{{\beta}-{\alpha}}{m}}}, \ {{\Delta}{\varphi} = {\frac{{\psi}-{\gamma}}{n}}}$.

Now we can illustrate the following theorem for triple integrals in spherical coordinates with ${\left ( {{\rho}^{*}_{ijk}}, {{\theta}^{*}_{ijk}}, {{\varphi}^{*}_{ijk}} \right )}$ being any sample point in the spherical subbox ${B}_{ijk}$. For the volume element of the subbox ${\Delta}{V}$ in spherical coordinates, we have ${\Delta}{V} = {({\Delta}{\rho})}{({\rho}{\Delta}{\varphi})}{({\rho} \ {\sin} \ {\varphi}{\Delta}{\theta})}$, as shown in the following figure.

Figure 3. The volume element of a box in spherical coordinates.

As with the other multiple integrals we have examined, all the properties work similarly for a triple integral in the spherical coordinate system, and so do the iterated integrals. Fubini’s theorem takes the following form.

As stated before, spherical coordinate systems work well for solids that are symmetric around a point, such as spheres and cones. Let us look at some examples before we consider triple integrals in spherical coordinates on general spherical regions.

The concept of triple integration in spherical coordinates can be extended to integration over a general solid, using the projections onto the coordinate planes. Note that $dV$ and $dA$ mean the increments in volume and area, respectively. The variables $V$ and $A$ are used as the variables for integration to express the integrals.

The triple integral of a continuous function ${f}{({\rho},{\theta},{\varphi})}$ over a general solid region

${E} = {\left \{ {({\rho},{\theta},{\varphi})}{\mid}{({\rho},{\theta})} \ {\in} \ {{D},{{u_1} \ {({\rho},{\theta}}})} \ {\leq} \ {\varphi} \ {\leq} \ {{u_2} \ {({\rho},{\theta})}} \right \}}$

in ${\mathbb{R}}^{3}$, where $D$ is the projection of $E$ onto the ${\rho}{\theta}$-plane, is

$\underset{E}{\displaystyle\iiint}{f}{({\rho},{\theta},{\varphi})}{dV} = {\iint\limits_{D}}{\left [ \ {\int\limits^{{u_2}{({\rho},{\theta})}}_{{u_1}{({\rho},{\theta})}}} \ {f}{({\rho},{\theta},{\varphi})}{d}{\varphi} \right ]}{{d}{A}}.$

In particular, ${D} = {\left \{ {({\rho},{\theta})}{\mid}{g_1} \ {(\theta)} \ {\leq} \ {\rho} \ {\leq} \ {{g_2} \ {(\theta)},{\alpha}} \ {\leq} \ {\theta} \ {\leq} \ {\beta} \right \}}$, then we have

$\underset{E}{\displaystyle\iiint}{f}{({\rho},{\theta},{\varphi})}{dV} = {\displaystyle\int^{\beta}_{\alpha}} \ {\displaystyle\int^{{g_2}{({\theta})}}_{{g_1}{({\theta})}}} \ {\displaystyle\int^{{u_2}{({\rho},{\theta})}}_{{u_1}{({\rho},{\theta})}}} \ {f}{({\rho},{\theta},{\varphi})}{{\rho}^{2}}{\sin} \ {\varphi} \ {{d}{\varphi}} \ {{d}{\rho}} \ {{d}{\theta}}.$

Similar formulas occur for projections onto the other coordinate planes.

Example: setting up a triple integral in spherical coordinates

Set up an integral for the volume of the region bounded by the cone ${z} = {\sqrt{{3}{({x^2}+{y^2})}}}$ and the hemisphere ${z} = {\sqrt{{4}-{x^2}-{y^2}}}$ (see the figure below).

Figure 4. A region bounded below by a cone and above by a hemisphere.

Show Solution

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

For the cone: ${z} = {\sqrt{{3}{({x^2}+{y^2})}}}$ or ${\rho} \ {\cos} \ {\varphi} = {\sqrt{3}} \ {\rho} \ {\sin} \ {\varphi}$ or ${\tan} \ {\varphi} = {\frac{1}{\sqrt{3}}}$ or ${\varphi} = {\frac{\pi}{6}}$.

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 ${{V}{(E)}} = {\displaystyle\int^{{\theta} = {{2}{\pi}}}_{{\theta} = {0}}} \ \ {\displaystyle\int^{{\varphi} = {{\pi}/{6}}}_{{\phi} = {0}}} \ \ {\displaystyle\int^{{\rho} = {2}}_{{\rho} = {0}}} \ {{\rho}^{2}}{\sin} \ {\varphi} \ {{d}{\rho}} \ {{d}{\varphi}} \ {{d}{\theta}}$.

Example: interchanging order of integration in spherical coordinates

Let $E$ be the region bounded below by the cone ${z} = {\sqrt{{x^2}+{y^2}}}$ and above by the sphere ${z} = {x^2} + {y^2} + {z^2}$ (Figure 5). Set up a triple integral in spherical coordinates and find the volume of the region using the following orders of integration:

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

Figure 5. A region bounded below by a cone and above by a sphere.

Show Solution

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

$\begin{aligned} {x^2} + {y^2} + \ & {z^2} = {z} \\ & {{\rho}^{2}} = {\rho} \ {\cos} \ {\varphi} \\ & {\rho} = {\cos} \ {\varphi}. \end{aligned}$

For the cone:

$\begin{aligned} {z} & = {\sqrt{{x^2}+{y^2}}} \\ {\rho} \ {\cos} \ {\varphi} & = {\sqrt{{{\rho}^{2}}{{\sin}^{2}}{\varphi}{{\cos}^{2}}{\phi}+{{\rho}^{2}}{{\sin}^{2}}{\varphi}{{\sin}^{2}}{\phi}}} \\ {\rho} \ {\cos} \ {\varphi} & = {\sqrt{{{\rho}^{2}}{{\sin}^{2}}{\varphi}{({{\cos}^{2}}{\phi}+{{\sin}^{2}}{\phi})}}} \\ {\rho} \ {\cos} \ {\varphi} & = {\rho} \ {\sin} \ {\varphi} \\ {\cos} \ {\varphi} & = {\sin} \ {\varphi} \\ {\varphi} & = {{\pi}/{4}}. \end{aligned}$

Hence the integral for the volume of the solid region $E$ becomes

${{V}{(E)}} = {\displaystyle\int^{{\theta} = {{2}{\pi}}}_{{\theta} = {0}}} \ {\displaystyle\int^{{\varphi} = {{\pi}/{4}}}_{{\varphi} = {0}}} \ {\displaystyle\int^{{\rho} = {{\cos}{\varphi}}}_{{\rho} = {0}}} \ {{{\rho}^{2}}{\sin} \ {\varphi} \ {{d}{\rho}} \ {{d}{\varphi}} \ {{d}{\theta}}}.$

 

2. Consider the ${\varphi}{\rho}$-plane. Note that the ranges for ${\varphi}$ and ${\rho}$ (from part a.) are

$\begin{aligned} {0} & \ {\leq} \ {\varphi} \ {\leq} \ {{\pi}/4} \\ {0} & \ {\leq} \ {\phi} \ {\leq} \ {{\cos}{\varphi}}. \end{aligned}$

The curve ${\rho} = {\cos}{\varphi}$ meets the line ${\varphi} = {{\pi}/{4}}$ at the point $({{\pi}/{4}},{{\sqrt{2}}/{2}})$. Thus, to change the order of integration, we need to use two pieces:

$\hspace{8cm}\begin{align}&0\leq\rho\leq\sqrt2/2 \\&0\leq\varphi\leq\pi/4\end{align}$ $\text{ and }$ $\begin{align}&\sqrt2/2\leq\rho\leq1 \\&0\leq\varphi\leq\cos^{-1}\rho\end{align}$

Hence the integral for the volume of the solid region $E$ becomes

${{V}{(E)}} = {\displaystyle\int^{{\theta} = {{2}{\pi}}}_{{\theta} = {0}}} \ {\displaystyle\int^{{\rho} = {{\sqrt{2}/{2}}}}_{{\rho} = {0}}} \ {\displaystyle\int^{{\varphi} = {{\pi}/{4}}}_{{\varphi} = {0}}} \ {{{\rho}^{2}}{\sin} \ {\varphi} \ {{d}{\varphi}} \ {{d}{\rho}} \ {{d}{\theta}}} \ + {\displaystyle\int^{{\theta} = {{2}{\pi}}}_{{\theta} = {0}}} \ {\displaystyle\int^{{\rho} = {1}}_{{\rho} = {{\sqrt{2}/{2}}}}} \ {\displaystyle\int^{{\varphi} = {{\cos}^{-1}}{\rho}}_{{\varphi} = {0}}} \ {{{\rho}^{2}}{\sin} \ {\varphi} \ {{d}{\varphi}} \ {{d}{\rho}} \ {{d}{\theta}}}.$

In each case, the integration results in ${{V}{(E)}} = {\frac{\pi}{8}}$.

Before we end this section, we present a couple of examples that can illustrate the conversion from rectangular coordinates to cylindrical coordinates and from rectangular coordinates to spherical coordinates.

Example: converting from rectangular coordinates to cylindrical coordinates

Convert the following integral into cylindrical coordinates:

${\displaystyle\int^{{y} = {1}}_{{y} = {-1}}}\displaystyle\int^{{x} = {\sqrt{{1} - {{y}^{2}}}}}_{{x} = {0}} \ {\displaystyle\int^{{z} = {\sqrt{{x^2}+{y^2}}}}_{{z} = {{x}^{2}}+{{y}^{2}}}} \ {{x}{y}{z}} \ {{d}{z}} \ {{d}{x}} \ {{d}{y}}.$

Show Solution

The ranges of the variables are

$\begin{aligned} {-1} & \ \ \ {\leq} \ \ \ {y}{\leq}{1} \\ {0} & \ \ \ {\leq} \ \ \ {x}{\leq}{\sqrt{{1} - {{y}^{2}}}} \\ {{x}^{2}} + {{y}^{2}} & \ \ \ {\leq} \ \ \ {z}{\leq}{\sqrt{{{x}^{2}}+{{y}^{2}}}}. \end{aligned}$.

The first two inequalities describe the right half of a circle of radius 1. Therefore, the ranges for ${\theta}$ and $r$ are

${-{\frac{\pi}{2}}} \ {\leq} \ {\theta} \ {\leq} \ {\frac{\pi}{2}} \ {\text{and}} \ {0} \ {\leq} \ {r} \ {\leq} \ {1}.$

The limits of $z$ are ${{r}^{2}}$, hence

${\displaystyle\int^{{y} = {1}}_{{y} = {-1}}}\displaystyle\int^{{x} = {\sqrt{{1} - {{y}^{2}}}}}_{{x} = {0}}\displaystyle\int^{{z} = {\sqrt{{{x}^{2}} + {{y}^{2}}}}}_{{z} = {{x}^{2}} + {{y}^{2}}} \ {{x}{y}{z}} \ {{d}{z}} \ {{d}{x}} \ {{d}{y}} = {\displaystyle\int^{{\theta} = {{\pi}/{2}}}_{{\theta} = {-{\pi}/{2}}}} \ {\displaystyle\int^{{r} = {1}}_{{r} = {0}}} \ {\displaystyle\int^{{z} = {r}}_{{z} = {{r}^{2}}}} \ {r}{({r}{\cos}{\theta})}{({r}{\sin}{\theta})}{z} \ {{d}{z}} \ {{d}{r}} \ {{d}{\theta}}$.

Example: converting from rectangular coordinates to spherical coordinates

Convert the following integral into spherical coordinates:

${\displaystyle\int^{{y} = {3}}_{{y} = {0}}}\displaystyle\int^{{x} = {\sqrt{{9} - {{y}^{2}}}}}_{{x} = {0}}\displaystyle\int^{{z} = {\sqrt{{18} - {{x}^{2}} - {{y}^{2}}}}}_{{z} = {\sqrt{{x}^{2} + {{y}^{2}}}}} \ {({{x}^{2}} + {{y}^{2}} + {{z}^{2}})} \ {{d}{z}} \ {{d}{x}} \ {{d}{y}}.$

Show Solution

The ranges of the variables are

$\begin{aligned} {0} \ \ \ & {\leq} \ \ \ {y} \ {\leq} \ {3} \\ {0} \ \ \ & {\leq} \ \ \ {x} \ {\leq} \ {\sqrt{{9} - {{y}^{2}}}} \\ {\sqrt{{{x}^{2}} + {{y}^{2}}}} \ \ \ & {\leq} \ \ \ {z} \ {\leq} \ {\sqrt{{18} - {{x}^{2}} - {{y}^{2}}}}. \end{aligned}$

The first two ranges of variables describe a quarter disk in the first quadrant of the $xy$-plane. Hence the range for ${\theta}$ is ${0} \ {\leq} \ {\theta} \ {\leq} \ {\frac{\pi}{2}}$.

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 ${0} \ {\leq} \ {\rho} \ {\leq} \ {\sqrt{18}}$, which is ${0} \ {\leq} \ {\rho} \ {\leq} \ {3{\sqrt{2}}}$.

For the ranges of ${\varphi}$, we need to find where the cone and the sphere intersect, so solve the equation

$\begin{aligned} {{{r}^{2}} + {{z}^{2}}} \ \ \ & = \ \ \ {18} \\ {({\sqrt{{{x}^{2}} + {{y}^{2}}}})^2} + {{z}^{2}} \ \ \ & = \ \ \ {18} \\ {{z}^{2}} + {{z}^{2}} \ \ \ & = \ \ \ {18} \\ {{2}{{z}^{2}}} \ \ \ & = \ \ \ {18} \\ {{z}^{2}} \ \ \ & = \ \ \ {9} \\ {z} \ \ \ & = \ \ \ {3}. \end{aligned}$

This gives

$\hspace{10cm}\begin{align} 3\sqrt2\cos\varphi&=3 \\ \cos\varphi&=\frac1{\sqrt2} \\ \varphi&=\frac{\pi}4. \end{align}$

Putting this together, we obtain

$\displaystyle\int_{y=0}^{y=3}\displaystyle\int_{x=0}^{x=\sqrt{9-y^2}}\displaystyle\int_{z=\sqrt{x^2+y^2}}^{z=\sqrt{18-x^2-y^2}}(x^2+y^2+z^2)dz \ dx \ dy=\displaystyle\int_{\varphi=0}^{\varphi=\pi/4}\displaystyle\int_{\theta=0}^{\theta=\pi/2}\displaystyle\int_{\rho=0}^{\rho=3\sqrt2}\rho^4\sin\varphi \ d\rho \ d\theta \ d\varphi.$

try it

Use rectangular, cylindrical, and spherical coordinates to set up triple integrals for finding the volume of the region inside the sphere $x^2+y^2+z^2=4$ but outside the cylinder $x^2+y^2=1$.

Figure 6.

Show Solution

Rectangular: $\displaystyle\int_{x=-2}^{x=2}\displaystyle\int_{y=-\sqrt{4-x^2}}^{y=\sqrt{4-x^2}}\displaystyle\int_{z=-\sqrt{4-x^2-y^2}}^{z=\sqrt{4-x^2-y^2}}dz \ dy \ dx - \displaystyle\int_{x=-1}^{x=1}\displaystyle\int_{y=-\sqrt{1-x^2}}^{y=\sqrt{1-x^2}}\displaystyle\int_{z=-\sqrt{4-x^2-y^2}}^{z=\sqrt{4-x^2-y^2}}dz \ dy \ dx.$

Cylindrical: $\displaystyle\int_{\theta=0}^{\theta=2\pi}\displaystyle\int_{r=1}^{r=2}\displaystyle\int_{z=-\sqrt{4-r^2}}^{z=\sqrt{4-r^2}}r \ dz \ dr \ d\theta.$

Spherical: $\displaystyle\int_{\varphi=\pi/6}^{\varphi=5\pi/6}\displaystyle\int_{\theta=0}^{\theta=2\pi}\displaystyle\int_{\rho=\csc\varphi}^{\rho=2} \rho^2 \sin{\varphi} d\rho \ d\theta \ d\varphi$

Watch the following video to see the worked solution to the above Try It

You can view the transcript for “CP 5.32” here (opens in new window).

Now that we are familiar with the spherical coordinate system, let’s find the volume of some known geometric figures, such as spheres and ellipsoids.

Example: chapter opener: finding the volume of L’hemisphÈric

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 $x^2+y^2+z^2=r^2$.

Figure 7. (credit: modification of work by Javier Yaya Tur, Wikimedia Commons)

Show Solution

We calculate the volume of the ball in the first octant, where $x\geq0$, $y\geq0$, and $z\geq0$, 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

$\large{0\leq\varphi\leq\frac{\pi}2, \ 0\leq\rho\leq{r}, \ 0\leq\theta\leq\frac{\pi}2}$

Therefore,

$\hspace{6cm}\begin{align} V&=\underset{D}{\displaystyle\iiint}dx \ dy \ dz = 8\displaystyle\int_{\theta=0}^{\theta=\pi/2}\displaystyle\int_{\rho=0}^{\rho=\pi}\displaystyle\int_{\varphi=0}^{\varphi=\pi/2}\rho^2\sin\theta \ d\varphi \ d\rho \ d\theta \\ &=8\displaystyle\int_{\varphi=0}^{\varphi=\pi/2}d\varphi\displaystyle\int_{\rho=0}^{\rho=\pi}\rho^2 \ d\rho\displaystyle\int_{\theta=0}^{\theta=\pi/2}\sin\theta \ d\theta \\ &=8\left(\frac{\pi}2\right)\left(\frac{r^3}3\right)(1) \\ &=\frac43\pi{r}^3 \end{align}$

This exactly matches with what we knew. So for a sphere with a radius of approximately $50$ ft, the volume is $\frac43\pi{r}^3\approx523,000\text{ ft.}^3$.

For the next example we find the volume of an ellipsoid.

Example: finding the volume of an ellipsoid

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

Show Solution

We again use symmetry and evaluate the volume of the ellipsoid using spherical coordinates. As before, we use the first octant $x\geq0$, $y\geq0$, and $z\geq0$ and then multiply the result by 8.

In this case the ranges of the variables are

$\large{0\leq\varphi\leq\frac{\pi}2, \ 0\leq\rho\leq\frac{\pi}2, \ 0\leq\rho\leq1, \ 0\leq\theta\leq\frac{\pi}2}$

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

$\large{x=a\rho\cos\varphi\sin\theta, \ y=b\rho\sin\varphi\sin\theta,\text{ and }z=c\rho\cos\theta}.$

Then the volume of the ellipsoid becomes

$\hspace{6cm}\begin{align} V&=\underset{D}{\displaystyle\iiint}dx \ dy \ dz \\ &=8\displaystyle\int_{\theta=0}^{\theta=\pi/2}\displaystyle\int_{\rho=0}^{\rho=1}\displaystyle\int_{\varphi=0}^{\varphi=\pi/2}abc\rho^2\sin\theta \ d\varphi \ d\rho \ d\theta \\ &=8abc\displaystyle\int_{\varphi=0}^{\varphi=\pi/2}d\varphi\displaystyle\int_{\rho=0}^{\rho=1}\rho^2 \ d\rho\displaystyle\int_{\theta=0}^{\theta=\pi/2}\sin\theta \ d\theta \\ &=8abc\left(\frac{\pi}2\right)\left(\frac13\right)(1) \\ &=\frac43\pi{abc} \end{align}$

Example: finding the volume of the space inside an ellipsoid and outside a sphere

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 $x^2+y^2+z^2=50^2$.

Show Solution

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 subtract.

First we find the volume of the ellipsoid using $a=75$ ft, $b=80$ ft, and $c=90$ ft in the result from Example “Finding the Volume of an Ellipsoid”. Hence the volume of the ellipsoid is

$\large{V_{\text{ellipsoid}}=\frac43\pi(75)(80)(90)\approx2,262,000\text{ ft.}^3}$

From Example “Chapter Opener: Finding the Volume of l’Hemisphèric”, the volume of the sphere is

$\large{V_{\text{sphere}}\approx523,600\text{ ft.}^3.}$

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

$\large{V_{\text{Hemisferic}}=V_{\text{ellipsoid}}-V_{\text{sphere}}=1,738,400\text{ ft.}^3.}$

Activity: hot air balloons

Hot air ballooning is a relaxing, peaceful pastime that many people enjoy. Many balloonist gatherings take place around the world, such as the Albuquerque International Balloon Fiesta. The Albuquerque event is the largest hot air balloon festival in the world, with over $500$ balloons participating each year.

Figure 8. Balloons lift off at the 2001 Albuquerque International Balloon Fiesta. (credit: David Herrera, Flickr)

As the name implies, hot air balloons use hot air to generate lift. (Hot air is less dense than cooler air, so the balloon floats as long as the hot air stays hot.) The heat is generated by a propane burner suspended below the opening of the basket. Once the balloon takes off, the pilot controls the altitude of the balloon, either by using the burner to heat the air and ascend or by using a vent near the top of the balloon to release heated air and descend. The pilot has very little control over where the balloon goes, however—balloons are at the mercy of the winds. The uncertainty over where we will end up is one of the reasons balloonists are attracted to the sport.

In this project we use triple integrals to learn more about hot air balloons. We model the balloon in two pieces. The top of the balloon is modeled by a half sphere of radius $28$ feet. The bottom of the balloon is modeled by a frustum of a cone (think of an ice cream cone with the pointy end cut off). The radius of the large end of the frustum is $28$ feet and the radius of the small end of the frustum is $6$ feet. A graph of our balloon model and a cross-sectional diagram showing the dimensions are shown in the following figure.

Figure 9. (a) Use a half sphere to model the top part of the balloon and a frustum of a cone to model the bottom part of the balloon. (b) A cross section of the balloon showing its dimensions.

We first want to find the volume of the balloon. If we look at the top part and the bottom part of the balloon separately, we see that they are geometric solids with known volume formulas. However, it is still worthwhile to set up and evaluate the integrals we would need to find the volume. If we calculate the volume using integration, we can use the known volume formulas to check our answers. This will help ensure that we have the integrals set up correctly for the later, more complicated stages of the project.

1. Find the volume of the balloon in two ways.
   1. Use triple integrals to calculate the volume. Consider each part of the balloon separately. (Consider using spherical coordinates for the top part and cylindrical coordinates for the bottom part.)
   2. Verify the answer using the formulas for the volume of a sphere, $V=\frac43\pi{r}^3$, and for the volume of a cone, $V=\frac13\pi{r}^2h$.

   In reality, calculating the temperature at a point inside the balloon is a tremendously complicated endeavor. In fact, an entire branch of physics (thermodynamics) is devoted to studying heat and temperature. For the purposes of this project, however, we are going to make some simplifying assumptions about how temperature varies from point to point within the balloon. Assume that just prior to liftoff, the temperature (in degrees Fahrenheit) of the air inside the balloon varies according to the function
   

   $\large{T_0(r,\theta,z)=\frac{z-r}{10}+210}$

2. What is the average temperature of the air in the balloon just prior to liftoff? (Again, look at each part of the balloon separately, and do not forget to convert the function into spherical coordinates when looking at the top part of the balloon.)
   Now the pilot activates the burner for $10$ seconds. This action affects the temperature in a $12$-foot-wide column $20$ feet high, directly above the burner. A cross section of the balloon depicting this column in shown in the following figure.
   

   

   Figure 10. Activating the burner heats the air in a $20$-foot-high, $12$-foot-wide column directly above the burner.

   Assume that after the pilot activates the burner for $10$ seconds, the temperature of the air in the column described above increases according to the formula
   

   $\large{H(r,\theta,z)=-2z-48.}$

   
   Then the temperature of the air in the column is given by
   

   $\large{T_1(r,\theta,z)=\frac{z-r}{10}+210+(-2z-48)}$,

   
   while the temperature in the remainder of the balloon is still given by
   

   $\large{T_0(r,\theta,z)=\frac{z-r}{10}+210}$.

3. Find the average temperature of the air in the balloon after the pilot has activated the burner for $10$ seconds.