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 and on the next page we will see that it is often useful to convert from rectangular to 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 \(z=x^2+y^2\) and \(z=8-x^2-y^2\) as shown in the plot. The first paraboloid opens upward with its vertex at the origin. The second one opens downward with its vertex at \((0,0,8)\). To find the intersection of the two paraboloids, we equate them: \[ x^2+y^2=8-x^2-y^2 \qquad \Longrightarrow \qquad x^2+y^2=4 \] The solution is a circle of radius \(r=2\). Equivalently, in cylindrical coordinates, the paraboloids are \(z=r^2\) and \(z=8-r^2\). We equate them: \[ r^2=8-r^2 \qquad \Longrightarrow \qquad r^2=4 \] Again the radius is \(r=2\). So the bounds are \[ 0 \le \theta \le 2\pi \qquad 0 \le r \le 2 \qquad r^2 \le z \le 8-r^2 \]

cylint-example

Volume

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

We set up the integral as \[ V=\int_0^{2\pi}\int_0^2\int_{r^2}^{8-r^2} \,r\,dz\,dr\,d\theta \] We factor off the \(\theta\)-integral, do the \(z\) integral and then the \(r\)-integral: \[\begin{aligned} V&=\int_0^{2\pi} \,d\theta\int_0^2 \left[z\dfrac{}{}\right]_{z=r^2}^{8-r^2} r\,dr =2\pi\int_0^2 (8-2r^2)r\,dr \\ &=2\pi\int_0^2 (8r-2r^3)\,dr =2\pi\left[4r^2-\,\dfrac{r^4}{2}\right]_0^2 =16\pi \end{aligned}\]

Mass

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

We set up and evaluate the integral as \[\begin{aligned} M&=\iiint_R \delta(r,\theta,z)\,r\,dr\,d\theta\,dz =\iiint_R (x^2+y^2)z\,r\,dr\,d\theta\,dz \\ &=\int_0^{2\pi}\int_0^2\int_{r^2}^{8-r^2} zr^3\,dz\,dr\,d\theta =\int_0^{2\pi}\,d\theta\int_0^2 \left[\dfrac{z^2}{2}\right]_{z=r^2}^{8-r^2}r^3\,dr \\ &=\dfrac{1}{2}(2\pi)\int_0^2 ((8-r^2)^2-r^4)r^3\,dr =\pi\int_0^2 (64r^3-16r^5)\,dr \\ &=\pi\left[16r^4-\,\dfrac{8}{3}r^6\right]_0^2 =\dfrac{256}{3}\pi \end{aligned}\]

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.

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 \(M=\dfrac{256}{3}\pi\). Now we compute the \(z\) moment of mass as \[\begin{aligned} M_z&=\iiint_R z\,\delta(r,\theta,z)\,dV =\int_0^{2\pi}\int_0^2\int_{r^2}^{8-r^2} z(r^2z)\,r\,dz\,dr\,d\theta \\ &=2\pi\int_0^2\int_{r^2}^{8-r^2} z^2r^3\,dz\,dr =2\pi\int_0^2 \left[\dfrac{z^3}{3}r^3\right]_{z=r^2}^{8-r^2}\,dr \\ &=\dfrac{2\pi}{3}\int_0^2 ((8-r^2)^3-r^6)(r^3)\,dr \\ &=\dfrac{2\pi}{3}\int_0^2 (-2r^9+24r^7-192r^5+512r^3)\,dr \\ &=\dfrac{2\pi}{3}\left[-\,\dfrac{1}{5}r^{10}+3r^8-32r^6+128r^4\right]_{r=0}^2 \\ &=\dfrac{2\pi}{3}\dfrac{2816}{5} =\dfrac{5632}{15}\pi \end{aligned}\] Thus, the \(z\) component of the center of mass of the region is: \[ \bar{z}=\dfrac{M_z}{M} =\dfrac{5632\pi}{15}\dfrac{3}{256\pi} =\dfrac{22}{5}=4.4 \]

So the center of mass is located at: \[ (\bar{x},\bar{y},\bar{z})=(0,0,4.4) \] The center of mass is plotted as a black dot within the paraboloidal region. It makes sense that \(\bar{z}>4\) because the density increases with \(z\) within the region.

cylapp-cm

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: \[ f_\text{ave}=\dfrac{1}{V}\iiint_R f(x,y,z)\,dV =\dfrac{1}{V}\iiint_R f(r,\theta,z)\,r\,dr\,d\theta\,dz \]

The volume between the paraboloids was previously found to be \(V=16\pi\). 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: \[\begin{aligned} \iiint_R &T\,dV =\int_0^{2\pi}\int_0^2\int_{r^2}^{8-r^2} (r^2\cos^2\theta)r\,dz\,dr\,d\theta \\ &=\int_0^{2\pi} \cos^2\theta\,d\theta\int_0^2 r^3\left[z\dfrac{}{}\right]_{z=r^2}^{8-r^2}\,dr \\ &=\int_0^{2\pi} \dfrac{1-\cos(2\theta)}{2}\,d\theta\int_0^2 r^3(8-2r^2)\,dr \\ &=\left[\dfrac{\theta}{2}-\,\dfrac{\sin(2\theta)}{4}\right]_0^{2\pi} \int_0^2 8r^3-2r^5\,dr \\ &=\pi\left[2r^4-\,\dfrac{1}{3}r^6\right]_{r=0}^2 =\pi\left(2^5-\,\dfrac{2^6}{3}\right) =\dfrac{32}{3}\pi \end{aligned}\] Finally, we compute the average temperature as \[ T_\text{ave}=\dfrac{1}{V}\iiint_R T\,dV =\dfrac{1}{16\pi}\dfrac{32\pi}{3}=\dfrac{2}{3} \]

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 \[\begin{aligned} V_x&=\iiint_R x\,dV =\iiint_R r\cos\theta\,r\,dr\,d\theta\,dz \\ V_y&=\iiint_R y\,dV =\iiint_R r\sin\theta\,r\,dr\,d\theta\,dz \\ V_z&=\iiint_R z\,dV =\iiint_R z\,r\,dr\,d\theta\,dz \end{aligned}\]

For the region between the paraboloids considered in the other examples, we can use symmetry to recognize that the centroid is obviously \((0,0,4)\). So in the following example we only consider the bottom half of that shape.

cylapp-bottom

By symmetric we see that the \(x\) and \(y\) components of the centroid are \(\bar{x}=\bar{y}=0\). Half of the volume we previously computed is \(V=8\pi\). Now we compute the \(z\) moment of the volume: \[\begin{aligned} V_z&=\iiint_R z\,dV =\int_0^{2\pi}\int_0^2\int_{r^2}^4 z\,r\,dz\,dr\,d\theta \\ &=2\pi\int_0^2 \left[\dfrac{z^2}{2}\right]_{z=r^2}^4 r\,dr \\ &=\pi\int_0^2 (16-r^4)r\,dr \\ &=\pi\left[8r^2-\dfrac{r^6}{6}\right]_{r=0}^2 \\ &=\pi\left(32-\dfrac{32}{3}\right) =\dfrac{64}{3}\pi \end{aligned}\] Thus, the \(z\) component of the center of mass of the region is: \[ \bar{z}=\dfrac{V_z}{V} =\dfrac{64\pi}{3}\dfrac{1}{8\pi} =\dfrac{8}{3}\approx2.67 \]

So the centroid is located at: \[ (\bar{x},\bar{y},\bar{z})=(0,0,2.67) \] The centroid is plotted as a black dot within the paraboloidal region. It makes sense that \(\bar{z}>2\) because there is more volume in the upper half of the region.

cylapp-centroid

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 4-dimensional region \(R\) between \(w=g(x,y,z)\) and \(w=f(x,y,z)\) is: \[ V_4=\iiint_R f(x,y,z)-g(x,y,z)\,dV \] In cylindrical coordinates, we have:

If \(f(r,\theta,z)\) and \(g(r,\theta,z)\) are functions on a region \(R\) in \(3\)-space and \(f(r,\theta,z) \ge g(r,\theta,z)\), then the 4-volume above the 4-dimensional region \(R\) between \(w=g(r,\theta,z)\) and \(w=f(r,\theta,z)\) is \[ V_4=\iiint_R (f(r,\theta,z)-g(r,\theta,z))\,r\,dr\,d\theta\,dz \]

The \(4\)-volume is the integral: \[ V_4=\iiint_R 3z\,dV =3\int_0^{2\pi}\int_0^2\int_{r^2}^4 z\,r\,dz\,dr\,d\theta \] This is just \(3\) times the integral we computed to find the \(z\)-moment of the volume. So the \(4\)-volume is: \[ V_4=3V_z=3\left(\dfrac{64}{3}\pi\right)=64\pi \]

PY: Add an exercise with parts for volume, mass, center of mass, centroid and average value.

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Supported in part by NSF Grant #1123255