5.5 Triple integrals in cylindrical and spherical coordinates (Page 7/12)

Calculus volume 3 Page 7 / 12

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.

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

A picture of l’Hemisphèric, which is a giant glass structure that is in the shape of an ellipsoid. — (credit: modification of work by Javier Yaya Tur, Wikimedia Commons)

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

0 \leq φ \leq \frac{π}{2}, 0 \leq ρ \leq r, 0 \leq θ \leq \frac{π}{2} .

Therefore,

\begin{matrix} V & = \underset{D}{∭} d x d y d z = 8 \int_{θ = 0}^{θ = π / 2} \int_{ρ = 0}^{ρ = π} \int_{φ = 0}^{φ = π / 2} ρ^{2} sin θ d φ d ρ d θ \\ = 8 \int_{φ = 0}^{φ = π / 2} d φ \int_{ρ = 0}^{ρ = r} ρ^{2} d ρ \int_{θ = 0}^{θ = π / 2} sin θ d θ \\ = 8 (\frac{π}{2}) (\frac{r^{3}}{3}) (1) \\ = \frac{4}{3} π r^{3} . \end{matrix}

This exactly matches with what we knew. So for a sphere with a radius of approximately $50$ ft, the volume is $\frac{4}{3} π {(50)}^{3} \approx 523,600 {ft}^{3} .$

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

Finding the volume of an ellipsoid

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

We again use symmetry and evaluate the volume of the ellipsoid using spherical coordinates. As before, we use the first octant $x \geq 0, y \geq 0,$ and $z \geq 0$ and then multiply the result by $8 .$

In this case the ranges of the variables are

0 \leq φ \leq \frac{π}{2}, 0 \leq ρ \leq \frac{π}{2}, 0 \leq ρ \leq 1, and 0 \leq θ \leq \frac{π}{2} .

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

x = a ρ cos φ sin θ, y = b ρ sin φ sin θ, and z = c ρ cos θ .

Then the volume of the ellipsoid becomes

\begin{matrix} V & = \underset{D}{∭} d x d y d z \\ = 8 \int_{θ = 0}^{θ = π / 2} \int_{ρ = 0}^{ρ = 1} \int_{φ = 0}^{φ = π / 2} a b c ρ^{2} sin θ d φ d ρ d θ \\ = 8 a b c \int_{φ = 0}^{φ = π / 2} d φ \int_{ρ = 0}^{ρ = 1} ρ^{2} d ρ \int_{θ = 0}^{θ = π / 2} sin θ d θ \\ = 8 a b c (\frac{π}{2}) (\frac{1}{3}) (1) \\ = \frac{4}{3} π a b c . \end{matrix}

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

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 $a = 75 ft,$ $b = 80 ft,$ and $c = 90 ft$ in the result from [link] . Hence the volume of the ellipsoid is

V_{ellipsoid} = \frac{4}{3} π (75) (80) (90) \approx 2,262,000 {ft}^{3} .

From [link] , the volume of the sphere is

V_{sphere} \approx 523,600 {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

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

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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.

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.

This figure consists of two parts, a and b. Figure a shows a representation of a hot air balloon in xyz space as a half sphere on top of a frustrum of a cone. Figure b shows the dimensions, namely, the radius of the half sphere is 28 ft, the distance from the bottom to the top of the frustrum is 44 ft, and the diameter of the circle at the top of the frustrum is 12 ft. — (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.

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, $V = \frac{4}{3} π r^{3},$ and for the volume of a cone, $V = \frac{1}{3} π r^{2} h .$
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

$T_{0} (r, θ, z) = \frac{z - r}{10} + 210 .$
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.

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

$H (r, θ, z) = −2 z - 48 .$

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

$T_{1} (r, θ, z) = \frac{z - r}{10} + 210 + (−2 z - 48),$

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

$T_{0} (r, θ, z) = \frac{z - r}{10} + 210 .$
Find the average temperature of the air in the balloon after the pilot has activated the burner for $10$ seconds.

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Source: OpenStax, Calculus volume 3. OpenStax CNX. Feb 05, 2016 Download for free at http://legacy.cnx.org/content/col11966/1.2

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