21. Multiple Integrals in Curvilinear Coordinates

b. Integrating in Cylindrical Coordinates

3. Applications

Integrals in cylindrical coordinates are particularly useful when the integrand and/or the limits of integration are axially symmetric about the \(z\)-axis. We previously did Applications of Triple Integrals in rectangular coordinates. All of those applications can also be done using cylindrical coordinates.

All the examples on this page of applications of triple integrals in cylindrical coordinates will use the solid between the paraboloids given by: \[ z=10-2x^2-2y^2 \quad \text{and} \quad z=6-x^2-y^2 \] as shown in the plot. Find the bounds on the integrals in cylindrical coordinates.

The plot shows 2 paraboloids open downward. The first has z intercept at z = 10 and is steeper. The second has z intercept at z = 6 and is less steep. They intersect on a circle with radius 2 at height 2.

Both paraboloids opens downward. The first has its vertex at \((0,0,10)\). The second has its vertex at \((0,0,6)\). In cylindrical coordinates, the paraboloids are: \[ z=10-2r^2 \quad \text{and} \quad z=6-r^2 \] To find the intersection of the two paraboloids, we equate them: \[ 10-2r^2=6-r^2 \qquad \implies \qquad r^2=4 \] The solution is a circle of radius \(r=2\). So the bounds are \[ 0 \le \theta \le 2\pi \qquad 0 \le r \le 2 \qquad 6-r^2 \le z \le 10-r^2 \]

Volume

To find the volume of a 3D region \(R\) in cylindrical coordinates, the differential of volume \(dV\) must be expressed in cylindrical coordinates: \[ \text{Volume}=\iiint_R 1\,dV =\iiint_R r\,dr\,d\theta\,dz \]

Find the volume of the solid between the paraboloids.

We set up the integral as \[ V=\int_0^{2\pi}\int_0^2\int_{6-r^2}^{10-2r^2} \,r\,dz\,dr\,d\theta \] We first do the \(z\) integral and then factor the integral: \[\begin{aligned} V&=\int_0^{2\pi} \int_0^2 \left[\rule{0pt}{10pt}z\right]_{6-r^2}^{10-2r^2} r\,dr\,d\theta =\int_0^{2\pi} \int_0^2 [(10-2r^2)-(6-r^2)] r\,dr\,d\theta \\ &=\int_0^{2\pi}\,d\theta\int_0^2 (4r-r^3)\,dr =2\pi\left[2r^2-\,\dfrac{r^4}{4}\right]_0^2 =8\pi \end{aligned}\] Notice that the second step is the integral we would do to find the volume in polar coordinates.

Mass

To find the mass of a 3D region \(R\) in cylindrical coordinates, the density \(\delta\) must also be expressed in cylindrical coordinates: \[ M=\iiint_R \delta(x,y,z)\,dV =\iiint_R \delta(r,\theta,z)\,r\,dr\,d\theta\,dz \]

Find the mass of the solid between the paraboloids if its mass density is \(\delta=(x^2+y^2)z\).

In cylindrical coordinates, the density is \(\delta=r^2z\). We set up and evaluate the integral as \[ M=\iiint\delta\,dV =\int_0^{2\pi}\int_0^2\int_{6-r^2}^{10-2r^2} (r^2z)r\,dz\,dr\,d\theta \] We separate off the \(\theta\) integral and do the \(z\): \[\begin{aligned} M&=\int_0^{2\pi}\,d\theta \int_0^2\left[\dfrac{z^2}{2}\right]_{6-r^2}^{10-2r^2}r^3\,dr \\ &=(2\pi)\dfrac{1}{2}\int_0^2 ((10-2r^2)^2-(6-r^2)^2)r^3\,dr =\pi\int_0^2 (64-28r^2+3r^4)r^3\,dr \\ &=\pi\left[64\dfrac{r^4}{4}-28\dfrac{r^6}{6}+3\dfrac{r^8}{8}\right]_0^2 =\pi(64\cdot4-28\cdot\dfrac{32}{3}+3\cdot32) =\dfrac{160}{3}\pi \end{aligned}\] Notice we cannot do the mass integral as a polar integral because the density depends on \(z\).

Center of Mass

To find the center of mass of a 3D region \(R\) in cylindrical coordinates, the coordinates \(x\), \(y\) and \(z\) must also be expressed in cylindrical coordinates. So the center of mass is: \[ \bar{x}=\dfrac{M_x}{M} \quad \text{and} \quad \bar{y}=\dfrac{M_y}{M} \quad \text{and} \quad \bar{z}=\dfrac{M_z}{M} \] where the mass of the region is \[ M=\iiint_R \delta(x,y,z)\,dV =\iiint_R \delta(r,\theta,z)\,r\,dr\,d\theta\,dz \] and the \(x\), \(y\) and \(z\) \(1^\text{st}\)-moments of the mass, respectively, are \[\begin{aligned} M_x&=\iiint_R x\,\delta(x,y,z)\,dV =\iiint_R r\cos\theta\,\delta(r,\theta,z)\,r\,dr\,d\theta\,dz \\ M_y&=\iiint_R y\,\delta(x,y,z)\,dV =\iiint_R r\sin\theta\,\delta(r,\theta,z)\,r\,dr\,d\theta\,dz \\ M_z&=\iiint_R z\,\delta(x,y,z)\,dV =\iiint_R z\,\delta(r,\theta,z)\,r\,dr\,d\theta\,dz \end{aligned}\]

It is important to remember that we cannot compute \(\bar{r}=\dfrac{M_r}{M}\) and \(\bar{\theta}=\dfrac{M_\theta}{M}\) where \[\text{\Large\textcolor{red}Wrong}\qquad \begin{array}{rl} M_r&=\displaystyle\iiint_R r\,\delta(r,\theta,z)\,dV \\[8pt] M_\theta&=\displaystyle\iiint_R \theta\,\delta(r,\theta,z)\,dV \end{array} \qquad \text{\Large\textcolor{red}Wrong} \] because moments are only defined for rectangular coordinates since the balance beam derivation (\(\text{Torque} = \text{Force} \times \text{Lever Arm}\)) does not work for the \(r\) and \(\theta\) directions. If we want the cylindrical coordinates, \((\bar r,\bar\theta,\bar z)\) of the center of mass, we must first find the rectangular components and then convert to cylindrical.

Find the center of mass of the solid between the paraboloids if its mass density is \(\delta=(x^2+y^2)z\).

Before we begin, we notice that the given density function is symmetric around the \(z\)-axis (i.e. \(\delta(r,\theta,z)=r^2z\) does not depend on \(\theta\)). So we can conclude that the \(x\) and \(y\) components of the center of mass are \(\bar{x}=\bar{y}=0\). We have already computed the mass of the region as \(M=\dfrac{160}{3}\pi\). Now we compute the \(z\) moment of mass as \[\begin{aligned} M_z&=\iiint_R z\,\delta(r,\theta,z)\,dV =\int_0^{2\pi}\int_0^2\int_{6-r^2}^{10-2r^2} z(r^2z)\,r\,dz\,dr\,d\theta \\ &=2\pi\int_0^2\left[\dfrac{z^3}{3}\right]_{6-r^2}^{10-2r^2}r^3\,dr =\dfrac{2\pi}{3}\int_0^2 \left[(10-2r^2)^3-(6-r^2)^3\right]r^3\,dr \\ &=\dfrac{2\pi}{3}\int_0^2 \left[ (1000-600r^2+120r^4-8r^6)-(216-108r^2+18r^4 -r^6) \right]r^3\,dr \\ &=\dfrac{2\pi}{3}\int_0^2 \left[784-492r^2+102r^4-7r^6\right]r^3\,dr \\ &=\dfrac{2\pi}{3}\left[ 784\dfrac{r^4}{4}-492\dfrac{r^6}{6}+102\dfrac{r^8}{8}-7\dfrac{r^{10}}{10} \right]_{0}^2 \\ &=\dfrac{2\pi}{3}\left[ 784\cdot4-492\dfrac{32}{3}+102\cdot32-7\dfrac{512}{5} \right] \\ &=\dfrac{2\pi}{3}\cdot\dfrac{2176}{5} =\dfrac{4352}{15}\pi \approx 911.5 \end{aligned}\] Thus, the \(z\) component of the center of mass of the region is: \[ \bar{z}=\dfrac{M_z}{M} =\dfrac{4352\pi}{15}\dfrac{3}{160\pi} =\dfrac{136}{25}=5.44 \]

So the center of mass is located at: \[ (\bar{x},\bar{y},\bar{z})=(0,0,5.44) \] The center of mass is plotted as a blue diamond within the solid. It is OK that the center of mass is outside the solid since it is still within the convex hull from \(2\) to \(10\).

The plot shows the 2 paraboloids of the previous plot. It also shows the center of mass as a diamond at z = 5.44 which is just below the vertex of the shorter paraboloid.

Centroid

The centroid of a region \(R\) in 3D is the same as the center of mass but with constant density which we take as \(\delta=1\). Then the mass reduces to the volume and the moments of mass become moments of volume. Thus the centroid is: \[ \bar{x}=\dfrac{V_x}{V} \quad \text{and} \quad \bar{y}=\dfrac{V_y}{V} \quad \text{and} \quad \bar{z}=\dfrac{V_z}{V} \] where \(V\) is the volume and the \(x\), \(y\) and \(z\) \(1^\text{st}\)-moments of the volume, respectively, are \[\begin{aligned} V_x&=\iiint_R x\,dV =\iiint_R r\cos\theta\,r\,dr\,d\theta\,dz \\ V_y&=\iiint_R y\,dV =\iiint_R r\sin\theta\,r\,dr\,d\theta\,dz \\ V_z&=\iiint_R z\,dV =\iiint_R z\,r\,dr\,d\theta\,dz \end{aligned}\]

Find the centroid of the solid between the paraboloids.

By symmetry, we see that the \(x\) and \(y\) components of the centroid are \(\bar{x}=\bar{y}=0\). We previously computed that the volume is \(V=8\pi\). Now we compute the \(z\) moment of the volume: \[\begin{aligned} V_z&=\iiint_R z\,dV =\int_0^{2\pi}\int_0^2 \int_{6-r^2}^{10-2r^2} z\,r\,dz\,dr\,d\theta \\ &=2\pi\int_0^2 \left[\dfrac{z^2}{2}\right]_{6-r^2}^{10-2r^2} r\,dr =\pi\int_0^2 [(10-2r^2)^2-(6-r^2)^2] r\,dr \\ &=\pi\int_0^2 [(100-40r^2+4r^4)-(36-12r^2+r^4)] r\,dr \\ &=\pi\int_0^2 [64-28r^2+3r^4] r\,dr =\pi\left[32r^2-7r^4+\dfrac{r^6}{2}\right]_{r=0}^2 \\ &=\pi\left(128-112+32\right) =48\pi \end{aligned}\]

Thus, the \(z\) component of the center of mass of the region is: \[ \bar{z}=\dfrac{V_z}{V} =\dfrac{48\pi}{8\pi} =6 \] So the centroid is located at: \[ (\bar{x},\bar{y},\bar{z})=(0,0,6) \] The centroid is plotted as a green sphere within the paraboloidal region. It is OK that the centroid is right at the edge of the solid since it is still within the convex hull from \(2\) to \(10\).

The plot shows the 2 paraboloids of the previous plot. It also shows the centroid as a small circle at z = 6 which is the vertex of the shorter paraboloid.

This plot shows the center of mass as well as the centroid.
The center of mass at \(\bar z=-5.44\) is plotted as a blue diamond.
The centroid at \(\bar z=6\) is plotted as a green sphere.

The plot shows the 2 paraboloids of the previous plot. It also shows the center of mass as a blue diamond at z = 5.44 and the centroid as a green sphere at z = 6.

Average Value

To find the average value of a function over a 3D region in cylindrical coordinates, the function must be evaluated in cylindrical coordinates before integrating: \[ f_\text{ave}=\dfrac{1}{V}\iiint_R f(x,y,z)\,dV =\dfrac{1}{V}\iiint_R f(r,\theta,z)\,r\,dr\,d\theta\,dz \]

Compute the average of the temperature, \(T=x^2\), on the solid between the paraboloids.

The volume between the paraboloids was previously found to be \(V=8\pi\). In cylindrical coordinates, the temperature is \[ T=x^2=r^2\cos^2\theta \] We compute the integral of the temperature. We factor out the \(\theta\) integral and do the \(z\) integral before the \(r\) integral: \[\begin{aligned} \iiint_R &T\,dV =\int_0^{2\pi}\int_0^2\int_{6-r^2}^{10-2r^2} (r^2\cos^2\theta)r\,dz\,dr\,d\theta \\ &=\int_0^{2\pi} \cos^2\theta\,d\theta\int_0^2 r^3\left[\rule{0pt}{10pt}z\right]_{6-r^2}^{10-2r^2}\,dr \\ &=\int_0^{2\pi} \dfrac{1-\cos(2\theta)}{2}\,d\theta\int_0^2 r^3(4-r^2)\,dr \\ &=\left[\dfrac{\theta}{2}-\,\dfrac{\sin(2\theta)}{4}\right]_0^{2\pi} [r^4-\dfrac{r^6}{6}]_0^2\,dr \\ &=\pi\left(16-\dfrac{32}{3}\right) =\dfrac{16}{3}\pi \end{aligned}\] Finally, we compute the average temperature as \[ T_\text{ave}=\dfrac{1}{V}\iiint_R T\,dV =\dfrac{1}{8\pi}\dfrac{16\pi}{3}=\dfrac{2}{3} \]

4-Volume (Optional)

If \(f(x,y,z)\) and \(g(x,y,z)\) are functions on a region \(R\) in \(3\)-space and \(f(x,y,z) \ge g(x,y,z)\), then the 4-volume above the \(3\)-dimensional region \(R\) between \(w=g(x,y,z)\) and \(w=f(x,y,z)\) is: \[\begin{aligned} V_4&=\iiint_R f(x,y,z)-g(x,y,z)\,dV \\ &=\iiint_R (f(r,\theta,z)-g(r,\theta,z))\,r\,dr\,d\theta\,dz \end{aligned}\]

The base of a \(4\)-dimensional solid is the \(3\)-dimensional paraboloidal region discussed above. The upper surface is the graph of the function \(w=3z\). (Of course, the lower surface is \(w=0\).) Find the \(4\)-volume of this \(4\)-dimensional solid.

The \(4\)-volume is the integral: \[ V_4=\iiint_R 3z\,dV =3\int_0^{2\pi}\int_0^2\int_{6-r^2}^{10-2r^2} z\,r\,dz\,dr\,d\theta \] This is just \(3\) times the integral we computed to find the \(z\)-moment of the volume. So the \(4\)-volume is: \[ V_4=3V_z=3\left(\dfrac{16}{3}\pi\right)=16\pi \]

The exercises on the rest of this page refer to the ice cream cone below the paraboloid \(z=18-x^2-y^2\) above the cone \(z=3\sqrt{x^2+y^2}\). Find the limits of integration.

The plot shows a cone with its vertex at the bottom and a curved surface at the top which is a piece of a paraboloid.

To find the \(r\) limits, find where the paraboloid hits the cone.

\(0 \le \theta \le 2\pi\)
\(0 \le \rule{0pt}{12pt}r \le 3\)
\(3r \le \rule{0pt}{12pt}z \le 18-r^2\)

Since the shape goes all the way around the \(z\) axis, the \(\theta\) limits are \(0 \le \theta \le 2\pi\).
In cylindrical coordinates, the paraboloid is \(z=18-r^2\) and the cone is \(z=3r\). So the \(z\) limits are given as \(3r \le z \le 12-r^2\).
To find the \(r\) limits, we equate the \(z\) equations: \[\begin{aligned} 3r=18-r^2 \quad &\implies \quad r^2+3r-18=0 \\ \implies \quad (r+6)(r-3)=0 \quad &\implies \quad r = -6,3 \end{aligned}\] Since \(r\) must be positive, the \(r\) limits are \(0 \le r \le 3\).

Find the volume of the ice cream cone.

\(V=\dfrac{135\pi}{2}\)

We use the limits found in the previous problem: \[\begin{aligned} V&=\iiint_R dV=\int_0^{2\pi} \int_0^3 \int_{3r}^{18-r^2}\,r\,dz\,dr\,d\theta \\ &=\int_0^{2\pi} \int_0^3 (18-r^2-3r)r\,dr\,d\theta \\ &=2\pi \left[9r^2-\dfrac{r^4}{4}-r^3\right]_0^3 \\ &=2\pi \dfrac{135}{4} =\dfrac{135\pi}{2} \end{aligned}\] Note: The second line is the integral for the same volume but done using polar coordinates.

mj 

Find the mass of the ice cream cone if the mass density is \(\delta=z\).

\(M=\dfrac{2673\pi}{4}\)

We use the limits found in a previous problem: \[\begin{aligned} M&=\iiint_R \delta\, dV =\int_0^{2\pi} \int_0^3 \int_{3r}^{18-r^2} z\,r\,dz\,dr\,d\theta \\ &=2\pi \int_0^3 \left[\dfrac{z^2}{2}\right]_{z=3r}^{z=18-r^2}\,r\,dr \\ &=\pi \int_0^3 ((18-r^2)^2-9r^2)\,r\,dr \\ &=\pi \int_0^3 \left(r^5-45r^3+324r\right)\,dr \\ &=\pi \left[\dfrac{r^6}{6}-\dfrac{45r^4}{4}+162r^2\right]_0^3 =\dfrac{2673\pi}{4} \end{aligned}\]

mj 

Find the center of mass of the ice cream cone if the mass density is \(\delta=z\).

The density function and the object are both symmetric about the \(z\)-axis. What does this say about the \(x\)- and \(y\)-components of the center of mass?

\((\bar x,\bar y,\bar z)=\left(0,0,\dfrac{603}{55}\right)\approx(0,0,10.96)\)

Since the shape and density are both symmetric about the \(z\)-axis, we conclude \(\bar x=\bar y=0\). We compute the \(z\)-moment of mass: \[\begin{aligned} M_z&=\iiint_R z\delta\, dV =\int_0^{2\pi} \int_0^3 \int_{3r}^{18-r^2} z^2\,r\, dz\, dr\, d\theta \\ &=2\pi \int_0^3 \left[\dfrac{z^3}{3}\right]_{z=3r}^{z=18-r^2}\,r\,dr =\dfrac{2\pi}{3} \int_0^3 \left((18-r^2)^3-27r^3\right)\,r\,dr \\ &=\dfrac{2\pi}{3} \int_0^3 \left(5832-972r^2+54r^4-r^6-27r^3\right)\,r\,dr \\ &=\dfrac{2\pi}{3} \int_0^3 \left(-r^7+54r^5-27r^4-972r^3+5832r\right)\,dr \\ &=\dfrac{2\pi}{3} \left[-\dfrac{r^8}{8}+9r^6-\dfrac{27r^5}{5}-243r^4+2916r^2\right]_0^3 \\ &=\dfrac{2\pi}{3} \dfrac{439587}{40} =\dfrac{146529\pi}{20} \end{aligned}\]

Since we previously computed \(M=\dfrac{2673\pi}{4}\), we conclude: \[ \bar z=\dfrac{M_z}{M}=\dfrac{146529\pi}{20}\dfrac{4}{2673\pi} =\dfrac{603}{55}\approx10.96 \] Thus the center of mass is at: \[ (\bar x,\bar y,\bar z)=\left(0,0,\dfrac{603}{55}\right)\approx(0,0,10.96) \] The center of mass is a blue diamond in the plot.
Notice the center of mass is above the center plane \(z=9\) since there is more mass toward the top.

The plot shows the same ice cream cone of the previous plot. It also shows the center of mass as a blue diamond at z = 10.96.

mj 

Find the centroid of the ice cream cone.

The object is symmetric about the \(z\)-axis. What does this say about the \(x\)- and \(y\)-components of the centroid?

\((\bar x,\bar y,\bar z)=\left(0,0,\dfrac{99}{10}\right)=(0,0,9.9)\)

Again, the object is symmetric about the \(z\)-axis, so \(\bar x=\bar y=0\). We compute the \(z\)-moment of volume: \[ V_z=\iiint_R z\, dV=\int_0^{2\pi} \int_0^3 \int_{3r}^{18-r^2} z\,r\,dz\,dr\,d\theta \] This is the same integral as the mass. So \[ V_z=M=\dfrac{2673\pi}{4} \]

Since the volume is \(V=\dfrac{135\pi}{2}\), we conclude: \[ \bar z=\dfrac{V_z}{V}=\dfrac{2673\pi}{4}\dfrac{2}{135\pi}=\dfrac{99}{10}=9.9 \] Thus the centroid is located at: \[ (\bar x,\bar y,\bar z)=\left(0,0,\dfrac{99}{10}\right)=(0,0,9.9) \] The centroid is a green sphere in the plot.
Notice the centroid is above the center plane \(z=9\) since there is more volume toward the top.

The plot shows the same ice cream cone of the previous plot. It also shows the centroid as a green sphere at z = 9.9.

mj 

This plot shows the center of mass as well as the centroid.
The center of mass is plotted as a blue diamond.
The centroid at is plotted as a green sphere.
Notice that the center of mass is above the centroid because the density, \(\delta=z\), is bigger toward the top.

The plot shows the ice cream cone with the center of mass as a blue diamond at z = 10.96 and the centroid as a green sphere at z = 9.9.

Find the average value of the the function \(g(x,y,z)=y^2\) on the ice cream cone.

\(g_{\text{ave}}=\dfrac{36}{25}=1.44\)

In cylindrical coordinates, \(g(r,\theta,z)=y^2=r^2\sin^2\theta\). Thus: \[\begin{aligned} \iiint_R g\, dV&=\int_0^{2\pi} \int_0^3 \int_{3r}^{18-r^2} r^2\sin^2\theta\, r\, dz\, dr\, d\theta \\ &=\int_0^{2\pi} \int_0^3 (18-r^2-3r)\,r^3\sin^2\theta\, dr\, d\theta \\ &=\int_0^3 \left(18r^3-r^5-3r^4\right)\, dr \int_0^{2\pi}\dfrac{1-\cos(2\theta)}{2}\, d\theta \\ &=\left[\dfrac{9r^4}{2}-\dfrac{r^6}{6}-\dfrac{3r^5}{5}\right]_0^3 \left[\dfrac{\theta}{2}-\dfrac{\sin(2\theta)}{4}\right]_0^{2\pi} \\ &=\dfrac{486}{5}\pi \end{aligned}\] Finally, \[ g_{\text{ave}}=\dfrac{1}{V}\iiint_R g\, dV =\dfrac{2}{135\pi}\dfrac{486\pi}{5}=\dfrac{36}{25}=1.44 \]

mj 

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