Calculus III 14.07 Triple Integrals in Other Coordinates

Previous	Calculus III 14.06 Triple Integrals and Applications
Next	Calculus III 14.08 Jacobian Variables

14.7 Triple Integrals in Other Coordinates

• Write and evaluate a triple integral in cylindrical coordinates.
• Write and evaluate a triple integral in spherical coordinates.

Triple Integrals in Cylindrical Coordinates

Volume for cylindrical block: \(\Delta V_{i}=r_{i} \Delta r_{i} \Delta \theta_{i} \Delta z_{i}\) Figure 14.7.1

Many common solid regions, such as spheres, ellipsoids, cones, and paraboloids, can yield difficult triple integrals in rectangular coordinates. In fact, it is precisely this difficulty that led to nonrectangular coordinate systems. This section discusses how to use cylindrical and spherical coordinates to evaluate triple integrals. Recall from Section 11.7 that the rectangular conversion equations for cylindrical coordinates are \( x\) \(=r \cos \theta \) \( y\) \(= r \sin \theta \) \( z\) \(=z.\) An easy way to remember these conversions is to note that the equations for \(x\) and \(y\) are the same as in polar coordinates and \(z\) is unchanged. In this coordinate system, the simplest solid region is a cylindrical block determined by \(r_{1} \leqslant r \leqslant r_{2} \) \(\theta_{1} \leqslant \theta \leqslant \theta_{2} \) and \(z_{1} \leqslant z \leqslant z_{2}, \) as shown in Figure 14.7.1. To obtain the cylindrical coordinates for a triple integral, consider a solid region \(Q\) whose projection \(R\) onto the \(xy\)-plane can be described in polar coordinates. Such as, \(Q=\{(x,y,z): \: (x,y) \text{ is in } R, \: h_{1}(x,y) \leqslant z \leqslant h_{2}(x,y)\}\) and \(R=\{(r,\theta): \: \theta_{1} \leqslant \theta \leqslant \theta_{2}, \: g_{1}(\theta) \leqslant r \leqslant g_{2}(\theta)\}.\) If \(f\) is a continuous function on the solid \(Q\), the triple integral for \(f\) over \(Q\) can be written as $$ \int \int_{Q} \int f(x,y,z) \: dV = \int_{R} \int \left[ \int_{h_{1}(x,y)}^{h_{2}(x,y)} f(x,y,z) \: dz \right] \: dV $$ where the double integral over \(R\) is evaluated in polar coordinates. In other words, \(R\) is a plane region that is either \(r\)-simple or \(\theta\)-simple. If \(R\) is \(r\)-simple, then the iterated form for the triple integral in cylindrical form is $$ \int \int_{Q} \int f(x,y,z) \: dV = \int_{\theta_{1}}^{\theta_{2}} \int_{g_{1}(\theta)}^{g_{2}(\theta)} \int_{h_{1}(r \cos \theta, r \sin \theta)}^{h_{2}(r \cos \theta, r \sin \theta)} f( r \cos \theta, r \sin \theta, z)r \: dz \: dr \: d\theta.$$ As in cartesian triple integration, there are six possible orders, \( \: dz \: dr \: d\theta \:\:\:\: \) \( \: dz \: d\theta \: dr \) \( \: dr \: dz \: d\theta \:\:\:\: \) \( \: dr \: d\theta \: dz \) \( \: d\theta \: dz \: dr \:\:\:\: \) \( \: d\theta \: dr \: dz. \) To visualize a particular integration order, picture the iterated integral as three sweeping motions—each adding another dimension to the solid. For example, in the order \(dr \: d\theta \: dz\), the first integration occurs in the \(r\)-direction as a point sweeps out a ray, as shown in Figure 14.7.2. Then, as \(\theta\) increases, the line sweeps out a sector, as shown in Figure 14.7.3. Finally, as \(z\) increases, the sector sweeps out a solid wedge, as shown in Figure 14.7.4. Integrate with respect to \(r\). Figure 14.7.2 Integrate with respect to \(\theta\). Figure 14.7.3 Integrate with respect to \(z\). Figure 14.7.4

Example 14.7.1 Finding Volume in Cylindrical Coordinates

Figure 14.7.5

Find the volume for the solid region \(Q\) cut from the sphere \(x^{2} + y^{2}+z^{2}=4\) by the cylinder \(r = 2 \sin \theta,\) as shown in Figure 14.7.5. Solution Because \(x^{2} + y^{2}+z^{2}= r^{2}+z^{2}=4\), the bounds on \(z\) are \(-\sqrt{4-r^{2}} \leqslant z \leqslant \sqrt{4-r^{2}}\). Let \(R\) be the circular projection for the solid onto the \(r\theta\)-plane. Then the bounds on \(R\) are \(0 \leqslant r \leqslant 2 \sin \theta \) and \(0 \leqslant \theta \leqslant \pi .\) The volume for \(Q\) is $$V$$ $$= \int_{0}^{\pi} \int_{0}^{2 \sin \theta} \int_{-\sqrt{4-r^{2}}}^{\sqrt{4-r^{2}}} r \: dz \: dr \: d\theta$$ $$=2 \int_{0}^{\pi/2} \int_{0}^{2 \sin \theta} \int_{-\sqrt{4-r^{2}}}^{\sqrt{4-r^{2}}} r \: dz \: dr \: d\theta $$ $$= 2 \int_{0}^{\pi/2} \int_{0}^{2 \sin \theta} 2r\sqrt{4-r^{2}} \: dr \: d\theta$$ $$= 2 \int_{0}^{\pi/2} \left. -\frac{2}{3}(4-r^{2})^{3/2} \right]_{0}^{2 \sin \theta} \: d\theta $$ $$= \frac{4}{3} \int_{0}^{\pi/2} (8-8 \cos^{3} \theta) \: d \theta$$ $$= \frac{32}{3} \int_{0}^{\pi/2} [1-(\cos \theta)(1-\sin^{2} \theta)] \: d\theta $$ $$= \frac{32}{3} \left[ \theta - \sin \theta + \frac{\sin^{3} \theta}{3} \right]_{0}^{\pi/2} $$ $$= \frac{16}{9}(3 \pi -4) \approx 9.644.$$

Example 14.7.2 Finding Mass in Cylindrical Coordinates

Figure 14.7.6

Find the mass for an ellipsoid \(Q\) given by \(4x^{2}+4y^{2}+z^{2}=16,\) lying above the \(xy\)-plane, as shown in Figure 14.7.6. The density at any point in the solid is proportional to the distance between the point and the \(xy\)-plane. Solution The density function is \( \rho(r,\theta, z) = kz,\) where \(k\) is the density constant. The ellipsoid's bounds are \(0 \leqslant z \leqslant \sqrt{16-4x^{2}-4y^{2}}=\sqrt{16-4r^{2}}, \) \(0 \leqslant r \leqslant 2, \) and \(0 \leqslant \theta \leqslant 2 \pi. \) The mass for the solid is $$m$$ $$=\int_{0}^{2\pi} \int_{0}^{2} \int_{0}^{\sqrt{16-4r^{2}}} kzr \: dz \: dr \: d\theta $$ $$= \left. \frac{1}{2} k \int_{0}^{2\pi} \int_{0}^{2} z^{2}r \right]_{0}^{\sqrt{16-4r^{2}}} \: dr \: d\theta $$ $$= \frac{1}{2} k \int_{0}^{2\pi} \int_{0}^{2} (16r-4r^{3}) \: dr \: d\theta $$ $$= \frac{1}{2} k \int_{0}^{2\pi} \left[ \vphantom{\frac{3}{4} } 8r^{2}-r^{4} \right]_{0}^{2} \: d\theta $$ $$=8k \int_{0}^{2\pi} \: d\theta = 16\pi k$$

Example 14.7.3 Finding the Inertia Moment

Figure 14.7.7

Find the inertia moment about the symmetric axis for the solid \(Q\) bounded by the paraboloid \(z=x^{2}+y^{2}\) and the plane \(z=4,\) as shown in Figure 14.7.7. The density at any point is proportional to the distance between the point and the \(z\)-axis. Solution The \(z\)-axis is symmetric to the paraboloid and the density formula is, \(\rho (x,y,z) = k \sqrt{x^{2}+y^{2}}\). Therefore, the inertia moment about the \(z\)-axis is $$I_{z} = \int \int_{Q} \int k (x^{2}+y^{2})\sqrt{x^{2}+y^{2}} \: dV.$$ Integration in cylindrical coordinates is useful when factors involving \(x^{2}+y^{2}\) appear in the integrand so that will be done here. The bounds in cylindrical coordinates are \( 0 \leqslant r \leqslant \sqrt{x^{2}+y^{2}} = \sqrt{z}\). This produces $$I_{z}$$ $$=k \int_{0}^{4} \int_{0}^{2\pi} \int_{0}^{\sqrt{z}} r^{2}(r)r \: dr \: d\theta \: dz $$ $$= \left. k \int_{0}^{4} \int_{0}^{2\pi} \frac{1}{5} r^{5} \right]_{0}^{\sqrt{z}} \: d\theta \: dz $$ $$= k \int_{0}^{4} \int_{0}^{2\pi} \frac{1}{5} z^{5/2} \: d\theta \: dz$$ $$= \frac{1}{5} k \int_{0}^{4} z^{5/2} (2\pi) \: dz $$ $$= \frac{2\pi k}{5} \left[ \frac{2}{7} z^{7/2} \right]_{0}^{4}= \frac{512k \pi}{35}. $$

Triple Integrals in Spherical Coordinates

Figure 14.7.8

Triple integrals involving spheres or cones are often easier to evaluate by converting to spherical coordinates. Recall from Section 11.7 that the cartesian conversion equations for spherical coordinates are \(x\) \(=\rho \sin \phi \cos \theta \) \(y\) \(=\rho \sin \phi \sin \theta \) \(z\) \(=\rho \cos \phi. \) In spherical coordinates the simplest region is a spherical block determines by \(\{(\rho, \theta, \phi): \: \rho_{1} \leqslant \rho \leqslant \rho_{2}, \: \theta_{1} \leqslant \theta \leqslant \theta_{2}, \: \phi_{1} \leqslant \phi \leqslant \phi_{2} \}\) where \(\rho_{1} \geqslant 0, \: \theta_{2} - \theta_{1} \leqslant 2 \pi\), and \(0 \leqslant \phi_{1} \leqslant \phi_{2} \leqslant \pi\), as shown in Figure 14.7.8. If \((\rho, \theta, \phi)\) is a point inside a spherical block, then the block's volume is approximated by \(\Delta V_{i} \approx \rho_{i}^{2} \sin \phi_{i} \Delta \rho_{i} \Delta \phi_{i} \Delta \theta_{i}\).

Definition 14.7.1 Volume In Spherical Coordinates

If \(f\) is integrable over a solid region \(Q\) in spherical coordinates. The formula for the volume is

$$ \int \int_{Q} \int f(x,y,z) \: dV = \int_{\theta_{1}}^{\theta_{2}} \int_{\phi_{1}}^{\phi_{2}} \int_{\rho_{2}}^{\rho_{2}} f(\rho \sin \phi \cos \theta, \rho \sin \phi \sin \theta, \rho \cos \phi) \rho^{2} \sin \phi \: d\rho \: d \phi \: d \theta $$

Like triple integrals in cylindrical coordinates, triple integrals in spherical coordinates are evaluated with iterated integrals. As with cylindrical coordinates, the integration order can be viewed as three sweeping motions—each adding another dimension to the solid. For instance, the iterated integral

$$ \int_{0}^{2 \pi} \int_{0}^{\pi/4} \int_{0}^{3} \rho^{2} \sin \phi \: d\rho \: d \phi \: d \theta,$$

which is shown in Figure 14.7.9, Figure 14.7.10, and Figure 14.7.11.

\(\rho\) varies from 0 to 3 with \(\phi\) and \(\theta\) held constant. Figure 14.7.9

\(phi\) varies from 0 to \(\pi/4\) with \(\theta\) held constant. Figure 14.7.10

\(\theta\) varies from 0 to \(2\pi\). Figure 14.7.11

Example 14.7.4 Finding Volume in Spherical Coordinates

Figure 14.7.12

Find the volume for the solid region \(Q\) bounded below by the upper nappe from the cone, \(z^{2}=x^{2}+y^{2}\), and above by the sphere, \(x^{2}+y^{2}+z^{2}=9\), as shown in Figure 14.7.12. Solution In spherical coordinates, the sphere's equation is \( \rho^{2} = x^{2}+y^{2}+z^{2}=9,\) where \(\rho =3.\) The sphere and cone intersect when \((x^{2}+y^{2})+z^{2}=z^{2}+z^{2} = 9,\) where $$ z = \frac{3}{\sqrt{2}}.$$ Because \(z= \rho \cos \phi\), \(\phi \) is $$ \left( \frac{3}{\sqrt{2}} \right)\left( \frac{1}{3}\right) = \cos \phi \rightarrow \phi = \frac{\pi}{4}.$$ The integration order \(d\rho \: d\phi \: d\theta\), where \(0 \leqslant \rho \leqslant 3 \), \(0 \leqslant \phi \leqslant \pi/42 \pi \), and \(0 \leqslant \pi \leqslant 2\pi, \) can be used. $$\int \int_{Q} \int \: dV $$ $$=\int_{0}^{2\pi} \int_{0}^{\pi/4} \int_{0}^{3} \rho^{2} \sin \phi \: d\rho \: d \phi \: d \theta $$ $$= \int_{0}^{2\pi} \int_{0}^{\pi/4} 9 \sin \phi \: d\rho \: d \phi \: d \theta $$ $$=9 \left. \int_{0}^{2\pi} -\cos \phi \right]_{0}^{\pi/4} \: d \theta $$ $$= 9 \int_{0}^{2\pi} \left( 1 - \frac{\sqrt{2}}{2}\right) \: d\theta $$ \(= 9\pi(2-\sqrt{2}) \approx 16.563\)

Example 14.7.5 Finding the Mass Center for a Solid Region

Find the mass center for the solid region \(Q\) with uniform density, bounded below by the upper nappe from the cone \(z^{2}=x^{2}+y^{2}\), and above by the sphere, \(x^{2}+y^{2}+z^{2}=9\), as shown in Figure 14.7.12.
Solution Because the density is uniform, the density at any point \((x,y,z)\) is \(k\). By symmetry, the mass center lies on the \(z\)-axis. Thus, only

$$\bar{z}=\frac{M_{xy}}{m},$$

where \(m=kV=9\pi(2-\sqrt{2})\) from Example 14.7.4 need be calculated. Since \(z= \rho \cos \phi\),

$$\frac{M_{xy}}{m}=\int \int_{Q} \int kz\: dV $$	$$= k \int_{0}^{3}\int_{0}^{2\pi} \int_{0}^{\pi/4} (\rho \cos \phi)\rho^{2} \sin \phi \: d \phi \: d \theta \: d\rho$$
	$$= k \int_{0}^{3}\int_{0}^{2\pi} \rho^{3} \left. \frac{\sin^{2} \phi}{2} \right]_{0}^{\pi/4} \: d \theta \: d\rho $$
	$$= \frac{1}{4} k \int_{0}^{3} \int_{0}^{2\pi} \rho^{3} \: d \theta \: d\rho $$
	$$= \frac{1}{2} (k)(\pi) \int_{0}^{3} \rho^{3} \: d\rho = \frac{81k\pi}{8} $$

Therefore, \(\bar{z}\) is

$$\bar{z}$$	$$=\frac{M_{xy}}{m} $$
	$$= \frac{81k\pi/8}{9\pi(2-\sqrt{2})} $$
	$$= \frac{9(2+\sqrt{2})}{16} \approx 1.920 $$

and the mass center is approximately \((0,0,1.92)\).

Internal Links

Parent Article: Calculus III 14 Multiple Integration