In three-dimensional space [latex]<\mathbb
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:
From rectangular coordinates to spherical coordinates:
Other relationships that are important to know for conversions are
The following figure shows a few solid regions that are convenient to express in spherical coordinates.
Figure 2. Spherical coordinates are especially convenient for working with solids bounded by these types of surfaces. (The letter [latex]c[/latex] indicates a constant.)
Now we can illustrate the following theorem for triple integrals in spherical coordinates with [latex]<\left ( <^_>, ^_>, ^_> \right )>[/latex] being any sample point in the spherical subbox [latex]_[/latex]. For the volume element of the subbox [latex][/latex] in spherical coordinates, we have [latex] = <()><()> <(\ \ )>[/latex], as shown in the following figure.
a blown up version of the box that shows it has sides Delta rho, rho Delta phi, and rho sin phi Delta theta, with overall volume Delta V = rho squared sin phi Delta rho Delta phi Delta theta." width="756" height="361" />
Figure 3. The volume element of a box in spherical coordinates.
The triple integral in spherical coordinates is the limit of a triple Riemann sum,
provided the limit exists.
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.
This iterated integral may be replaced by other iterated integrals by integrating with respect to the three variables in other orders.
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.
Evaluate the iterated triple integral
Show SolutionAs 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:
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 [latex]dV[/latex] and [latex]dA[/latex] mean the increments in volume and area, respectively. The variables [latex]V[/latex] and [latex]A[/latex] are used as the variables for integration to express the integrals.
The triple integral of a continuous function [latex]<(<\rho>,,)>[/latex] over a general solid region
in [latex]<\mathbb
Similar formulas occur for projections onto the other coordinate planes.
Set up an integral for the volume of the region bounded by the cone [latex] = <(+)>>>[/latex] and the hemisphere [latex] = -->>[/latex] (see the figure below).
Figure 4. A region bounded below by a cone and above by a hemisphere.
Show SolutionUsing the conversion formulas from rectangular coordinates to spherical coordinates, we have:
Set up a triple integral for the volume of the solid region bounded above by the sphere [latex] = [/latex] and bounded below by the cone [latex] = <<\pi>/>[/latex].
Show Solution[latex]V(E)=\displaystyle\int_^<\theta=2\pi>\displaystyle\int_<\phi=0>^<\varphi=\pi/3>\displaystyle\int_^\rho^2\sin\varphi \ d\rho \ d\varphi \ d\theta.[/latex]
Let [latex]E[/latex] be the region bounded below by the cone [latex] = +>>[/latex] and above by the sphere [latex] = + + [/latex] ( Figure 5 ). Set up a triple integral in spherical coordinates and find the volume of the region using the following orders of integration:
Figure 5. A region bounded below by a cone and above by a sphere.
Show Solution1. Use the conversion formulas to write the equations of the sphere and cone in spherical coordinates.
For the sphere:
Hence the integral for the volume of the solid region [latex]E[/latex] becomes
2. Consider the [latex][/latex]-plane. Note that the ranges for [latex][/latex] and [latex][/latex] (from part a.) are
The curve [latex] = [/latex] meets the line [latex] = <<\pi>/>[/latex] at the point [latex](<<\pi>/>,>/>)[/latex]. Thus, to change the order of integration, we need to use two pieces:
[latex]\hspace\begin&0\leq\rho\leq\sqrt2/2 \\&0\leq\varphi\leq\pi/4\end[/latex] [latex]\text< and >[/latex] [latex]\begin&\sqrt2/2\leq\rho\leq1 \\&0\leq\varphi\leq\cos^\rho\end[/latex]
Hence the integral for the volume of the solid region [latex]E[/latex] becomes
In each case, the integration results in [latex]> = <\frac<\pi>>[/latex].
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.
Convert the following integral into cylindrical coordinates:
Show SolutionThe ranges of the variables are
The first two inequalities describe the right half of a circle of radius 1. Therefore, the ranges for [latex][/latex] and [latex]r[/latex] are
The limits of [latex]z[/latex] are [latex]^>[/latex], hence
Convert the following integral into spherical coordinates:
Show SolutionThe ranges of the variables are
The first two ranges of variables describe a quarter disk in the first quadrant of the [latex]xy[/latex]-plane. Hence the range for [latex][/latex] is [latex] \ <\leq>\ \ <\leq>\ <\frac<\pi>>[/latex].
The lower bound [latex] = ^> + ^>>>[/latex] is the upper half of a cone and the upper bound [latex] = - ^> - ^>>>[/latex] is the upper half of a sphere. Therefore, we have [latex] \ <\leq>\ \ <\leq>\ >[/latex], which is [latex] \ <\leq>\ \ <\leq>\ <3<\sqrt>>[/latex].
For the ranges of [latex][/latex], we need to find where the cone and the sphere intersect, so solve the equation
[latex]\hspace\begin 3\sqrt2\cos\varphi&=3 \\ \cos\varphi&=\frac1 \\ \varphi&=\frac<\pi>4. \end[/latex]
Putting this together, we obtain
Use rectangular, cylindrical, and spherical coordinates to set up triple integrals for finding the volume of the region inside the sphere [latex]x^2+y^2+z^2=4[/latex] but outside the cylinder [latex]x^2+y^2=1[/latex].
Cylindrical: [latex]\displaystyle\int_^<\theta=2\pi>\displaystyle\int_^\displaystyle\int_>^>r \ dz \ dr \ d\theta.[/latex]
Watch the following video to see the worked solution to the above Try It
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.
Find the volume of the spherical planetarium in l’Hemisphèric in Valencia, Spain, which is five stories tall and has a radius of approximately [latex]50[/latex] ft, using the equation [latex]x^2+y^2+z^2=r^2[/latex].
Figure 7. (credit: modification of work by Javier Yaya Tur, Wikimedia Commons)
Show SolutionWe calculate the volume of the ball in the first octant, where [latex]x\geq0[/latex], [latex]y\geq0[/latex], and [latex]z\geq0[/latex], using spherical coordinates, and then multiply the result by [latex]8[/latex] for symmetry. Since we consider the region [latex]D[/latex] as the first octant in the integral, the ranges of the variables are
[latex]\hspace\begin V&=\underset<\displaystyle\iiint>dx \ dy \ dz = 8\displaystyle\int_^<\theta=\pi/2>\displaystyle\int_^<\rho=\pi>\displaystyle\int_^<\varphi=\pi/2>\rho^2\sin\theta \ d\varphi \ d\rho \ d\theta \\ &=8\displaystyle\int_^<\varphi=\pi/2>d\varphi\displaystyle\int_^<\rho=\pi>\rho^2 \ d\rho\displaystyle\int_^<\theta=\pi/2>\sin\theta \ d\theta \\ &=8\left(\frac<\pi>2\right)\left(\frac3\right)(1) \\ &=\frac43\pi^3 \end[/latex]
This exactly matches with what we knew. So for a sphere with a radius of approximately [latex]50[/latex] ft, the volume is [latex]\frac43\pi^3\approx523,000\text< ft.>^3[/latex].
For the next example we find the volume of an ellipsoid.
Find the volume of the ellipsoid [latex]\frac+\frac+\frac=1[/latex].
Show SolutionWe again use symmetry and evaluate the volume of the ellipsoid using spherical coordinates. As before, we use the first octant [latex]x\geq0[/latex], [latex]y\geq0[/latex], and [latex]z\geq0[/latex] and then multiply the result by 8.
In this case the ranges of the variables are
[latex]\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>[/latex]
Also, we need to change the rectangular to spherical coordinates in this way:
Then the volume of the ellipsoid becomes
[latex]\hspace\begin V&=\underset<\displaystyle\iiint>dx \ dy \ dz \\ &=8\displaystyle\int_^<\theta=\pi/2>\displaystyle\int_^\displaystyle\int_^<\varphi=\pi/2>abc\rho^2\sin\theta \ d\varphi \ d\rho \ d\theta \\ &=8abc\displaystyle\int_^<\varphi=\pi/2>d\varphi\displaystyle\int_^\rho^2 \ d\rho\displaystyle\int_^<\theta=\pi/2>\sin\theta \ d\theta \\ &=8abc\left(\frac<\pi>2\right)\left(\frac13\right)(1) \\ &=\frac43\pi \end[/latex]
Find the volume of the space inside the ellipsoid [latex]\frac+\frac+\frac=1[/latex] and outside the sphere [latex]x^2+y^2+z^2=50^2[/latex].
Show SolutionThis 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 [latex]a=75[/latex] ft, [latex]b=80[/latex] ft, and [latex]c=90[/latex] ft in the result from Example “Finding the Volume of an Ellipsoid”. Hence the volume of the ellipsoid is
From Example “Chapter Opener: Finding the Volume of l’Hemisphèric”, the volume of the sphere is
Therefore, the volume of the space inside the ellipsoid [latex]\frac+\frac+\frac=1[/latex] and outside the sphere [latex]x^2+y^2+z^2=50^2[/latex] is approximately
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 [latex]500[/latex] 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 [latex]28[/latex] 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 [latex]28[/latex] feet and the radius of the small end of the frustum is [latex]6[/latex] 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.
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
the dimensions of the balloon and the hot air, namely, the radius of the half sphere is 28 ft, the distance from the bottom to the top of the frustrum is 44 ft, the diameter of the circle at the top of the frustrum is 12 ft, and the inner column of hot air has height 20 ft and diameter 12 ft." width="385" height="385" />
Assume that after the pilot activates the burner for [latex]10[/latex] seconds, the temperature of the air in the column described above increases according to the formula
