Earlier in this chapter we showed how to convert a double integral in rectangular coordinates into a double integral in polar coordinates in order to deal more conveniently with problems involving circular symmetry. A similar situation occurs with triple integrals, but here we need to distinguish between cylindrical symmetry and spherical symmetry. In this section we convert triple integrals in rectangular coordinates into a triple integral in either cylindrical or spherical coordinates.
Also recall the chapter prelude, which showed the opera house l’Hemisphèric in Valencia, Spain. It has four sections with one of the sections being a theater in a five-story-high sphere (ball) under an oval roof as long as a football field. Inside is an IMAX screen that changes the sphere into a planetarium with a sky full of \(9000\) twinkling stars. Using triple integrals in spherical coordinates, we can find the volumes of different geometric shapes like these.
Review of Cylindrical Coordinates
As we have seen earlier, in two-dimensional space \(\mathbb{R}^2\) a point with rectangular coordinates \((x,y)\) can be identified with \((r,\theta)\) in polar coordinates and vice versa, where \(x = r \, \cos \theta\), \(y = r \, \sin \, \theta, \, r^2 = x^2 + y^2\) and \(\tan \, \theta = \left(\frac{y}{x}\right)\) are the relationships between the variables.
In three-dimensional space \(\mathbb{R}^3\) a point with rectangular coordinates \((x,y,z)\) can be identified with cylindrical coordinates \((r, \theta, z)\) and vice versa. We can use these same conversion relationships, adding \(z\) as the vertical distance to the point from the \((xy\)-plane as shown in \(\PageIndex{1}\).
To convert from rectangular to cylindrical coordinates, we use the conversion
\(x = r \, \cos \theta\)
\(y = r \, \sin \, \theta\)
\(z=z\)
To convert from cylindrical to rectangular coordinates, we use
\(r^2 = x^2 + y^2\) and
\(\theta = \tan^{-1} \left(\frac{y}{x}\right)\)
\(z=z\)
Note that that \(z\)-coordinate remains the same in both cases.
In the two-dimensional plane with a rectangular coordinate system, when we say \(x = k\) (constant) we mean an unbounded vertical line parallel to the \(y\)-axis and when \(y = l\) (constant) we mean an unbounded horizontal line parallel to the \(x\)-axis. With the polar coordinate system, when we say \(r = c\) (constant), we mean a circle of radius \(c\) units and when \(\theta = \alpha\) (constant) we mean an infinite ray making an angle \(\alpha\) with the positive \(x\)-axis.
Similarly, in three-dimensional space with rectangular coordinates \((x,y,z)\) the equations \(x = k, \, y = l\) and \(z = m\) where \(k, \, l\) and \(m\) are constants, represent unbounded planes parallel to the \(yz\)-plane, \(xz\)-plane and \(xy\)-plane, respectively. With cylindrical coordinates \((r, \theta, z)\), by \(r = c, \, \theta = \alpha\), and \(z = m\), where \(c, \alpha\), and \(m\) are constants, we mean an unbounded vertical cylinder with the z-axis as its radial axis; a plane making a constant angle \(\alpha\) with the \(xy\)-plane; and an unbounded horizontal plane parallel to the \(xy\)-plane, respectively.
This means that the circular cylinder \(x^2 + y^2 = c^2\) in rectangular coordinates can be represented simply as \(r = c\) in cylindrical coordinates.
(Refer to Cylindrical and Spherical Coordinates for more review.)
Integration in Cylindrical Coordinates
Triple integrals can often be more readily evaluated by using cylindrical coordinates instead of rectangular coordinates. Some common equations of surfaces in
rectangular coordinates along with corresponding equations in cylindrical coordinates are listed in Table \(\PageIndex{1}\).
These equations will become handy as we proceed with solving problems using triple integrals.
Table \(\PageIndex{1}\): Equations of Some Common Shapes
Circular cylinder
Circular cone
Sphere
Paraboloid
Rectangular
\(x^2 + y^2 = c^2\)
\(z^2 = c^2 (x^2 + y^2)\)
\(x^2 + y^2 + z^2 = c^2\)
\(z = c(x^2 + y^2)\)
Cylindrical
\(r = c\)
\(z = cr\)
\(r^2 + z^2 = c^2\)
\(z = cr^2\)
As before, we start with the simplest bounded region \(B\) in \(\mathbb{R}^3\) to describe in cylindrical coordinates, in the form of a cylindrical box,
\(B = \{(r,\theta,z) | a \leq r \leq b, \, \alpha \leq \theta \leq \beta, \, c \leq z \leq d\}\) (Figure \(\PageIndex{2}\)).
Suppose we divide each interval into \(l, \, m\), and \(n\) subdivisions such that \(\Delta r = \frac{b \cdot a}{l}, \, \Delta \theta = \frac{\beta \cdot \alpha}{m}\), and
\(\Delta z = \frac{d \cdot c}{n}\). Then we can state the following definition for a triple integral in cylindrical coordinates.
DEFINITION: triple integral in cylindrical coordinates
Consider the cylindrical box (expressed in cylindrical coordinates)
\[B = \{(r, \theta, z)|a \leq r \leq b, \, \alpha \leq \theta \leq \beta, \, c \leq z \leq d\}.\]
If the function \(f(r, \theta, z)\) is continuous on \(B\) and if \((r_{ijk}^*, \theta_{ijk}^*, z_{ijk}^*)\) is any sample point in the cylindrical subbox
\(B_{ijk} = |r_{i-1}, r_i| \times |\theta_{j-1}, \theta_j| \times |z_{k-1}, k_i|\) (Figure \(\PageIndex{2}\)), then we can define
the triple integral in cylindrical coordinates as the limit of a triple Riemann sum, provided the following limit exists:
Note that if \(g(x,y,z)\) is the function in rectangular coordinates and the box \(B\) is expressed in rectangular coordinates, then the triple integral
\[\iiint_B g(x,y,z)dV\]
is equal to the triple integral
\[\iiint_B g(r \, \cos \theta, \, r \, \sin \, \theta, \, z) r \, dr \, d\theta \, dz\]
and we have
\[\iiint_B g(x,y,z)dV = \iiint_B g(r \, \cos \theta, \, r \, \sin \, \theta, \, z) r \, dr \, d\theta \, dz = \iiint_B f(r, \theta \, z) r \, dr \, d\theta \, dz.\]
As mentioned in the preceding section, all the properties of a double integral work well in triple integrals, whether in rectangular coordinates or cylindrical coordinates.
They also hold for iterated integrals. To reiterate, in cylindrical coordinates, Fubini’s theorem takes the following form:
Theorem: Fubini’s Theorem in Cylindrical Coordinates
Suppose that \(g(x,y,z)\) is continuous on a rectangular box \(B\) which when described in cylindrical coordinates looks like
\(B = \{(r,\theta,z) | a \leq r \leq b, \, \alpha \leq \theta \leq \beta, \, c \leq z \leq d\}\).
Then \(g(x,y,z) = g(r \, \cos \theta, r \, \sin \, \theta,z) = f(r, \theta,z)\) and
\[\iiint_B g(x,y,z)dV = \int_c^d \int_{\beta}^{\alpha} \int_a^b f(r, \theta, z) r \, dr \, d\theta \, dz.\]
The iterated integral may be replaced equivalently by any one of the other five iterated integrals obtained by integrating with respect to the three variables in other orders.
Cylindrical coordinate systems work well for solids that are symmetric around an axis, such as cylinders and cones. Let us look at some examples before we
define the triple integral in cylindrical coordinates on general cylindrical regions.
Example \(\PageIndex{1}\): Evaluating a Triple Integral over a Cylindrical Box
Evaluate the triple integral
\[\iiint_B (zr \, \sin \, \theta) r \, dr \, d\theta \, dz\]
where the cylindrical box \(B\) is \(B = \{(r,\theta,z) |0 \leq r \leq 2, \, 0 \leq \theta \leq \pi/2, \, 0, \leq z \leq 4\}.\)
Solution
As stated in Fubini’s theorem, we can write the triple integral as the iterated integral
\[\iiint_B (zr \, \sin \, \theta) r \, dr \, d\theta \, dz = \int_{\theta=0}^{\theta=\pi/2} \int_{r=0}^{r=2} \int_{z=0}^{z=4} (zr \, \sin \, \theta) r \, dz \, dr \, d\theta.\]
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:
Evaluate the triple integral \[\int_{\theta=0}^{\theta=\pi} \int_{r=0}^{r=1} \int_{z=0}^{z=4} rz \, \sin \, \theta r \, dz \, dr \, d\theta.\]
Hint
Follow the same steps as in the previous example.
Answer
\(8\)
If the cylindrical region over which we have to integrate is a general solid, we look at the projections onto the coordinate planes. Hence the triple integral of a continuous
function \(f(r, \theta, z)\) over a general solid region \(E = \{(r, \theta, z)|(r, \theta) \in D, u_1 (r, \theta) \leq z \leq u_2 (r, \theta)\}\) in \(\mathbb{R}^3\) where \(D\) is the
projection of \(E\) onto the \(r\theta\)-plane, is
\[\iiint_E f(r, \theta, z) r \, dr \, d\theta \, dz = \iint_D \left[\int_{u_1(r,\theta)}^{u_2(r,\theta)} f(r, \theta, z) dz \right] r \, dr \, d\theta.\]
In particular, if \(D = \{(r, \theta) |G_1 (\theta) \leq r \leq g_2(\theta), \alpha \leq \theta \leq \beta \}\), then we have
\[ \iiint_E f(r,\theta, z) r \, dr \, d\theta = \int_{\theta=\alpha}^{\theta=\beta} \int_{r=g_1(\theta)}^{r=g_2(\theta)} \int_{z=u_1(r,\theta)}^{z=u_2(r,\theta)} f(r,\theta,z) r \, dz \, dr \, d\theta.\]
Similar formulas exist for projections onto the other coordinate planes. We can use polar coordinates in those planes if necessary.
Example \(\PageIndex{2}\): Setting up a Triple Integral in Cylindrical Coordinates over a General Region
Consider the region \(E\) inside the right circular cylinder with equation \(r = 2 \, \sin \, \theta\), bounded below by the \(r\theta\)-plane and bounded
above by the sphere with radius \(4\) centered at the origin (Figure 15.5.3). Set up a triple integral over this region with a function \(f(r, \theta, z)\)
in cylindrical coordinates.
Solution
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 \(z = \sqrt{16 - r^2}\). T
hen the limits for \(r\) are from \(0\) to \(r = 2 \, \sin \, \theta\). Finally, the limits for \(\theta\) are from \(0\) to \(\pi\).
Hence the region is \(E = \{(r,\theta, z)|0 \leq \theta \leq \pi, \, 0 \leq r \leq 2 \, \sin \, \theta, \, 0 \leq z \leq \sqrt{16 - r^2} \}.\)
Therefore, the triple integral is
\[\iiint_E f(r,\theta, z) r \, dz \, dr \, d\theta = \int_{\theta=0}^{\theta=\pi} \int_{r=0}^{r=2 \, \sin \, \theta} \int_{z=0}^{z=\sqrt{16-r^2}} f(r,\theta,z) r \, dz \, dr \, d\theta.\]
Exercise \(\PageIndex{2}\):
Consider the region inside the right circular cylinder with equation \(r=2 \, \sin \, \theta\) bounded below by the \(r\theta\)-plane and bounded above
by \(z = 4 - y\). Set up a triple integral with a function \(f(r,\theta,z)\) in cylindrical coordinates.
Hint
Analyze the region, and draw a sketch.
Answer
\[\iiint_E f(r,\theta, z) r \, dz \, dr \, d\theta = \int_{\theta=0}^{\theta=\pi} \int_{r=0}^{r=2 \, \sin \, \theta} \int_{z=0}^{z=4-r \, \sin \, \theta} f(r,\theta,z) r \, dz \, dr \, d\theta.\]
Example \(\PageIndex{3}\): Setting up a Triple Integral in Two Ways
Let \(E\) be the region bounded below by the cone \(z = \sqrt{x^2 + y^2}\) and above by the paraboloid \(z = 2 - x^2 - y^2\). (Figure 15.5.4).
Set up a triple integral in cylindrical coordinates to find the volume of the region, using the following orders of integration:
a. \(dz \, dr \, d\theta\)
b. \(dr \, dz \, d\theta\)
Solution
a. The cone is of radius 1 where it meets the paraboloid. Since \(z = 2 - x^2 - y^2 = 2 - r^2\) and \(z = \sqrt{x^2 + y^2} = r^2\)
(assuming \(r\) is nonnegative), we have \(2 - r^2 = r\). Solving, we have \(r^2 + r - 2 = (r + 2)(r - 1) = 0\). Since \(r \geq 0\),
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 \(E = \{(r, \theta, z) |0 \leq \theta \leq 2\pi, \, 0 \leq r \leq 1, \, r \leq z \leq 2 - r^2 \}\).
Hence the integral for the volume is
\[V = \int_{\theta=0}^{\theta=2\pi} \int_{r=0}^{r=1} \int_{z=r}^{z=2-r^2} r \, dz \, dr \, d\theta.\]
b. 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 \[E = \{(r,\theta,z)|0 \leq \theta \leq 2\pi, \, 0 \leq z \leq 1, \, 0 \leq r \leq z\} \cup \{(r,\theta,z)|0 \leq \theta \leq 2\pi, \, 1 \leq z \leq 2, \, 0 \leq r \leq \sqrt{2 - z}\}.\]
Now the integral for the volume becomes
\[V = \int_{\theta=0}^{\theta=2\pi} \int_{z=0}^{z=1} \int_{r=0}^{r=z} r \, dr \, dz \, d\theta + \int_{\theta=0}^{\theta=2\pi} \int_{z=1}^{z=2} \int_{r=0}^{r=\sqrt{2-z}} r \, dr \, dz \, d\theta.\]
Exercise \(\PageIndex{3}\):
Redo the previous example with the order of integration \(d\theta \, dz \, dr\).
Hint
Note that \(\theta\) is independent of \(r\) and \(z\).
Answer
\(E = \{(r,\theta,z)|0 \leq \theta \leq 2\pi, \, 0 \leq z \leq 1, \, 0 \leq r \leq 2 - z^2\}\) and \[V = \int_{r=0}^{r=1} \int_{z=0}^{z=2 - r^2} \int_{\theta=0}^{\theta=2\pi} r \, d\theta \, dz \, dr.\]
Example \(\PageIndex{4}\): Finding a Volume with Triple Integrals in Two Ways
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
\(x^2 + y^2 = 1\) (Figure 15.5.5). 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:
\[\begin{align} V (E) &= \int_{\theta=0}^{\theta=2\pi} \int_{r=0}^{r=1} \int_{z=0}^{z=\sqrt{4-r^2}} r \, dz \, dr \, d\theta \\ &= \int_{\theta=0}^{\theta=2\pi} \int_{r=0}^{r=1} \left[ \left>rz\right|_{z=0}^{z=\sqrt{4-r^2}}\right] dr \, d\theta = \int_{\theta=0}^{\theta=2\pi} \int_{r=0}^{r=1} \left(r\sqrt{4 - r^2}\right) dr \, d\theta\\&= \int_0^{2\pi} \left(\frac{8}{3} - \sqrt{3} \right) d\theta = 2\pi \left(\frac{8}{3} - \sqrt{3} \right) \, \text{cubic units.} \end{align}\]
b. 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 \((1,\sqrt{3})\) in the \(rz\)-plane
\[E_1 = \{ (r,\theta,z) | 0 \leq r \leq \sqrt{4 - r^2}, \, \sqrt{3} \leq z \leq 2, \, 0 \leq \theta \leq 2\pi\}\] and
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 \(\PageIndex{6}\)). Note that \(\rho > 0\) 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.
Recall the relationships that connect rectangular coordinates with spherical coordinates.
From spherical coordinates to rectangular 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.
Theorem: Fubini’s Theorem for Spherical Coordinates
If \(f(\rho,\theta, \varphi)\) is continuous on a spherical solid box \(B = [a,b] \times [\alpha,\beta] \times [\gamma , \psi]\), then
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 w
consider triple integrals in spherical coordinates on general spherical regions.
Example \(\PageIndex{5}\): Evaluating a Triple Integral in Spherical Coordinates
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:
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
Set up a triple integral for the volume of the solid region bounded above by the sphere \(\rho = 2\) and bounded below by the cone \(\varphi = \pi/3\).
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 \(\PageIndex{8}\): Converting from Rectangular Coordinates to Cylindrical Coordinates
Convert the following integral into cylindrical coordinates:
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 \(x^2 + y^2 = 1\).
Answer: Rectangular
\[\int_{x=-2}^{x=2} \int_{y=-\sqrt{4-x^2}}^{y=\sqrt{4-x^2}} \int_{z=-\sqrt{4-x^2-y^2}}^{z=\sqrt{4-x^2-y^2}} dz \, dy \, dx - \int_{x=-1}^{x=1}
\int_{y=-\sqrt{1-x^2}}^{y=\sqrt{1-x^2}} \int_{z=-\sqrt{4-x^2-y^2}}^{z=\sqrt{4-x^2-y^2}} dz \, dy \, dx.\]
Answer: Cylindrical
\[\int_{\theta=0}^{\theta=2\pi} \int_{r=1}^{r=2} \int_{z=-\sqrt{4-r^2}}^{z=\sqrt{4-r^2}} r \, dz \, dr \, d\theta.\]
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 \(\PageIndex{10}\): 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 \(x^2 + y^2 + z^2 = r^2\).
Solution
We calculate the volume of the ball in the first octant, where \(x \leq 0, \, y \leq 0\), and \(z \leq 0\), 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
Example \(\PageIndex{12}\): 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
\(x^2 + y^2 + z^2 = 50^2\).
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 substract.
First we find the volume of the ellipsoid using \(a = 75\) ft, \(b = 80\) ft, and \(c = 90\) ft in the result from Example. Hence the volume of the ellipsoid
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 \(28\) feet. A graph of our balloon model and a cross-sectional diagram showing the dimensions are shown in the following figure.
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.
a. 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.)
b. Verify the answer using the formulas for the volume of a sphere, \(V = \frac{4}{3}\pi r^3\), and for the volume of a cone,
\(V = \frac{1}{3} \pi r^2 h\).
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 \[T_0 (r,\theta,z) = \frac{z - r}{10} + 210.\]
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
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
\[H(r,\theta,z) = -2z - 48.\]
Then the temperature of the air in the column is given by \[T_1(r,\theta,z) = \frac{z - r}{10} + 210 + (-2z - 48),\]
while the temperature in the remainder of the balloon is still given by \[T_0(r,\theta,z) = \frac{z - r}{10} + 210.\]
3. Find the average temperature of the air in the balloon after the pilot has activated the burner for \(10\) seconds.
Key Concepts
To evaluate a triple integral in cylindrical coordinates, use the iterated integral
the limit of a triple Riemann sum, provided the following limit exists:
\[lim_{l,m,n\rightarrow\infty} \sum_{i=1}^l \sum_{j=1}^m \sum_{k=1}^n f(r_{ijk}^*, \theta_{ijk}^*, s_{ijk}^*) r_{ijk}^* \Delta r \Delta \theta \Delta z \nonumber\]
triple integral in spherical coordinates
the limit of a triple Riemann sum, provided the following limit exists:\[lim_{l,m,n\rightarrow\infty} \sum_{i=1}^l \sum_{j=1}^m \sum_{k=1}^n f(\rho_{ijk}^*, \theta_{ijk}^*, \varphi_{ijk}^*) (\rho_{ijk}^*)^2 \sin \, \varphi \Delta \rho \Delta \theta \Delta \varphi \nonumber\]
Contributors and Attributions
Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org.