[b][size=150]Triple integrals in cylindrical coordinates[/size][/b][br][br]Suppose we want to compute the triple integral [math]\iiint_G f(x,y,z) \ dV[/math], where [math]G[/math] is a simple xy-solid over a general polar region [math]R=\left\{(r,\theta) \ | \ c\leq \theta \leq d, \ g(\theta)\leq r \leq h(\theta)\right\}[/math]. It is more convenient to change the triple integral into cylindrical coordinates as follows:[br][br][math]\iiint_G f(x,y,z) \ dV=\int_c^d \left(\int_{g(\theta)}^{h(\theta)}\left(\int_{G(r,\theta)}^{H(r,\theta)} f(r,\theta,z) \ dz\right) \ rdr\right) \ d\theta[/math][br][br]where [math]G[/math] is bounded from below and above by [math]z=G(r,\theta)[/math] and [math]z=H(r,\theta)[/math] respectively over the polar region [math]R[/math].[br][br][br][u]Example[/u]: Find the volume of the cylinder [math]x^2+y^2=9[/math] between the planes [math]x+z=5[/math] and [math]z=1[/math].[br][br][u]Answer[/u]:[br][br]Let [math]G[/math] be the part of cylinder [math]x^2+y^2=9[/math] between [math]z=1[/math] and [math]z=5-x[/math]. Using cylindrical coordinates, the triple integral that equals the volume of [math]G[/math] can be expressed as follows:[br][br][math]\iiint_G 1 \ dV=\int_0^{2\pi}\left(\int_0^3\left(\int_1^{5-r\cos\theta} 1 \ dz\right) \ rdr\right) \ d\theta[/math] ([u]Note[/u]: The cylinder has radius 3.)[br][math]=\int_0^{2\pi}\left(\int_0^3(4-r\cos\theta) \ rdr\right) \ d\theta[/math][br][math]=\int_0^{2\pi}\left[2r^2-\frac{r^3\cos\theta}3\right]_0^3 \ d\theta[/math][br][math]=\int_0^{2\pi}(18-9\cos\theta) \ d\theta=36\pi[/math][br][br]The applet below shows that the solid [math]G[/math], which is the part of the cylinder between two planes.[br]

[b][size=150]Triple integrals in spherical coordinates[/size][/b][br][br]Sometimes we might want to express a triple integral in spherical coordinates. Suppose [math]f(x,y,z)[/math] is defined on a "spherical box" [math]B=\left\{(\rho,\theta,\phi) \ | \ a\leq \rho \leq b, \ c\leq \theta \leq d, \ p\leq \phi \leq q\right\}[/math]. We then divide each interval into [math]l,m,n[/math] subdivisions such that [math]\Delta \rho=\frac{b-a}l, \ \Delta\theta=\frac{d-c}m, \ \Delta\phi=\frac{q-p}n[/math]. Label the spherical subbox from [math]k=1[/math] to [math]k=N[/math]. Let [math](\rho_k^*,\theta_k^*,\phi_k^*)[/math] be the centre point in the [math]k^{\text{th}}[/math] spherical subbox. The volume [math]\Delta V_k[/math] of the [math]k^{\text{th}}[/math] spherical subbox equals [math](\Delta \rho_k^*)(\rho_k^*\Delta \phi)(\rho_k^*\sin(\phi_k^*) \ \Delta\theta)=(\rho_k^*)^2\sin(\phi_k^*)\Delta \rho\Delta \theta\Delta \phi[/math], as shown in the diagram below.[br][br]The triple integral in spherical coordinates is the limit of the following Riemann sum:[br][br][math]\iiint_B f(x,y,z) \ dA=\lim_{N\to\infty} \sum_{k=1}^N f(\rho_k^*,\theta_k^*,\phi_k^*)(\rho_k^*)^2\sin(\phi_k^*)\Delta\rho\Delta\theta\Delta\phi[/math][br]

Therefore, we can rewrite the triple integral as an iterated integral in spherical coordinates as follows:[br][br][math]\iiint_B f(x,y,z) \ dV=\int_p^q\left(\int_c^d\left(\int_a^b f(\rho\sin\phi\cos\theta,\rho\sin\phi\sin\theta,\rho\cos\phi)\rho^2\sin\phi \ d\rho\right) \ d\theta\right) \ d\phi[/math][br][br][br][u]Example[/u]: Find the volume of the solid [math]S[/math] bounded from above and below by the sphere [math]x^2+y^2+z^2=16[/math] and the cone [math]z=\sqrt{x^2+y^2}[/math].[br][br][u]Answer[/u]:[br][br]We first express the sphere and cone in spherical coordinates: [math]\rho=4[/math] and [math]\phi=\frac{\pi}4[/math] respectively. Therefore, [math]S=\left\{(\rho,\theta,\phi) \ | \ 0\leq \rho \leq 4, \ 0\leq \theta \leq 2\pi, \ 0 \leq \phi \leq \frac{\pi}4\right\}[/math]. Therefore, we have the following:[br][br][math]\iiint_S 1 \ dV=\int_0^{2\pi}\left(\int_0^{\frac{\pi}4}\left(\int_0^4 1 \cdot \rho^2\sin\phi \ d\rho\right) \ d\phi \right) \ d\theta[/math][br][math]=\int_0^{2\pi}\left(\int_0^{\frac{\pi}4}\left[\frac{\rho^3\sin\phi}3\right]_0^4 \ d\phi\right) \ d\theta[/math][br][math]=\int_0^{2\pi}\left(\int_0^{\frac{\pi}4}\frac{64\sin\phi}3 \ d\phi\right) \ d\theta[/math][br][math]=\int_0^{2\pi}\left(-\frac{64}3\cdot \frac{\sqrt{2}}2+\frac{64}3\right) \ d\theta[/math][br][math]=\frac{64\pi}3(2-\sqrt{2})[/math][br][br][br]The applet below shows the solid [math]S[/math].