# Triple Integrals in Cylindrical and Spherical Coordinates

• By David A. Smith, Founder & CEO, Direct Knowledge
• David Smith has a B.S. and M.S. in Mathematics and has enjoyed teaching calculus, linear algebra, and number theory at both Tarrant County College and the University of Texas at Arlington. David is the Founder and current CEO of Direct Knowledge.

## Cylindrical Coordinates

Each point in three dimensions is uniquely represented in cylindrical coordinates by $(r,\theta ,z)$ using $0\leq r<\infty ,$ $0\leq \theta < 2\pi ,$ and $-\infty <z<+\infty .$ The conversion formulas involving rectangular coordinates $(x,y,z)$ and cylindrical coordinates $(r,\theta ,z)$ are \begin{align} & r=\sqrt{x^2+y^2} & \tan \theta =\frac{y}{x} \\ & z=z & x=r \cos \theta \\ & y=r \sin \theta & z=z.\end{align} A triple integral $$\iiint_Rf(x,y,z) \, dV$$ over a region $R$ can sometimes be evaluated by transforming to cylindrical coordinates if the region of integration $R$ is $z$-simple and the projection of $R$ onto the $x y$-plane is a region $D$ that can be described naturally in terms of polar coordinates.

Theorem. (Triple Integrals in Cylindrical Coordinates) Assume $R$ is a solid region with continuous upper surface $z=v(r,\theta )$ and continuous lower surface $z=u(r,\theta )$ and assume is $D$ be the projection of the solid onto the $x y$-plane expressed in polar coordinates: $$D=\{(r,\theta):\alpha \leq r \leq \beta, r_1(\theta) \leq r(\theta) \leq r_2(\theta) \}$$ where $r_1$ and $r_2$ are continuous functions of $\theta.$ If $f(x,y,z)$ is continuous on $R,$ then the triple integral of $f$ over $R$ can be evaluated as follows $$\iiint_Rf(x,y,z)dV=\int_{\alpha }^{\beta } \int_{r_1(\theta)}^{r_2 (\theta )}\int_{u(r,\theta )}^{v(r,\theta )}f(r \cos \theta ,r \sin \theta ,z) r dz dr \, d\theta .$$

Example. Find the volume of the bounded solid bounded by the paraboloid $4x^2+4y^2+z=1$ and the $x y$-plane.

Solution. The projection of the solid region onto the $x y$-plane is the region enclosed by $x^2+y^2=\frac{1}{4}.$ In cylindrical coordinates the volume is determined as, \begin{align} 4\int_0^{\pi /2}\int_0^{1/2}\int_0^{1-4r^2}r dz dr d\theta & =4\int_0^{\pi /2}\int_0^{1/2}\left(r-4 r^3\right)drd\theta \\ & =4\int_0^{\pi /2} \frac{1}{16} \, d\theta \\ & =\frac{\pi }{8}. \end{align}

Example. Evaluate the iterated integral $$I=\int_{-1}^1\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}\int_{x^2+y^2}^{\sqrt{2-x^2-y^2}}z \, dz \, dy \, dx.$$

Solution. In Cartesian coordinates the region of integration $D$ is given as $$\left\{(x,y,z) \mid x^2+y^2\leq z\leq \sqrt{2-x^2-y^2}, -\sqrt{1-x^2}\leq y\leq \sqrt{1-x^2}, -1\leq x\leq 1\right\}$$ We use cylindrical coordinates to evaluate the triple iterated integral $I$ as follows \begin{align} I& =\int_0^{2\pi }\int_0^1\int_{r^2}^{\sqrt{2-r^2}}z r dzdrd\theta \\ & =\int_0^{2\pi }\int_0^1r \left(-\frac{r^4}{2}-\frac{r^2}{2}+1\right)drd\theta \\ &=\int_0^{2\pi } \frac{7}{24} \, d\theta \\ &=\frac{7 \pi }{12}. \end{align}

Example. Evaluate the triple integral over the given region $$I=\iiint_R\left(x^4+2x^2y^2+y^4\right) \, dx \, dy \, dz$$ where $R$ is the cylindrical solid $x^2+y^2\leq a^2$ with $0\leq z\leq \frac{1}{\pi }$

Solution. We consider the region of integration as being $z$-simple by projecting onto the $x y$-plane; and in the $x y$-plane the region is bounded by $x^2+y^2=a^2.$ Notice that $$x^4+2x^2y^2+y^4=\left(x^2+y^2\right)^2.$$ In cylindrical coordinates the triple integral is evaluated as, \begin{align} I & =4\int_0^{\pi /2}\int_0^a\int_0^{1/\pi }r^4r dzdrd\theta \\ & =4\int_0^{\pi /2}\int_0^a\frac{r^5}{\pi }drd\theta \\ &=4\int_0^{\pi /2} \frac{a^6}{6 \pi } \, d\theta \\ &=\frac{a^6}{3}. \end{align}

Example. Evaluate the iterated integral $$\int_0^{\pi }\int_0^2\int_0^{\sqrt{4-r^2}}r \sin \theta \,dz \, dr \,d\theta .$$

Solution. We find \begin{align} \int_0^{\pi }\int_0^2\int_0^{\sqrt{4-r^2}}r \sin \theta \, dz \, dr \, d\theta & =\int_0^{\pi }\int_0^2r \sqrt{4-r^2} \sin \theta \, dr \, d\theta \\ & =\int_0^{\pi } \frac{8 \sin \theta }{3} \, d\theta \\ &=\frac{16}{3} \end{align}

Example. Evaluate the iterated integral $$\int_0^{\pi /4}\int_0^1\int_0^{\sqrt{r}}r^2 \sin \theta \, dz \, dr \, d\theta.$$

Solution. We find \begin{align} \int_0^{\pi /4}\int_0^1\int_0^{\sqrt{r}}r^2 \sin \theta \, dz \, dr \, d\theta & =\int_0^{\pi /4}\int_0^1r^{5/2} \sin \theta \, dr \, d\theta \\ & =\int_0^{\pi /4} \frac{2 \sin \theta }{7} \, d\theta \\ &=\frac{2}{7} \left(1-\frac{1}{\sqrt{2}}\right) \\ &=\frac{2-\sqrt{2}}{7} \end{align}

## Spherical Coordinates

Each point in three dimensions is uniquely represented in spherical coordinates by $(\rho ,\theta ,\phi )$ using $0\leq \rho <\infty ,$ $0\leq \theta < 2\pi ,$ and $0\leq \phi \leq \pi .$ The conversion formulas from rectangular coordinates $(x,y,z)$ to spherical coordinates $(\rho ,\theta ,\phi )$ are \begin{equation*}\begin{array}{lllll} \rho =\sqrt{x^2+y^2+z^2} & & \theta = \tan^{-1} \left( \frac{y}{x}\right) & \qquad &\phi =\cos ^{-1}\left(\frac{z}{\sqrt{x^2+y^2+z^2}}\right) \\ x=\rho \sin \phi \cos \theta & & y =\rho \sin \phi \sin \theta & & z=\rho \cos \phi \end{array}\end{equation*} Using Jacobians, we can show that the element of volume in spherical coordinates is $$dV=\rho ^2 \sin \phi \,d\rho \, d\theta \, d\phi.$$

Theorem. (Triple Integrals in Spherical Coordinates) If $f(x,y,z)$ is continuous on the closed bounded region $R$, then the triple integral of $f$ over $R$ is given by \begin{align} & \iiint_Rf(x,y,z)dV \\ & = \iiint_S f(\rho \sin \phi \cos \theta ,\rho \sin \phi \sin \theta ,\rho \cos \phi ) \, \rho ^2 \sin \phi \, d\rho \, d\phi \, d\theta \end{align} where $S$ is the region $R$ expressed in spherical coordinates.

Example. Find the volume of the bounded solid $S$ inside the sphere of radius $a.$

Solution. Since an equation of the sphere is $\rho =a$ for $0\leq \theta <2\pi$ and $0\leq \phi \leq \pi ,$ the volume is determined by evaluating a triple iterated integral in spherical coordinates \begin{align} \iiint_S \, dV & =\int_0^{2\pi }\int_0^{\pi }\int_0^a\rho ^2 \sin \phi \, d\rho \, d\phi \, d\theta \\ & = \int_0^{2\pi }\int_0^{\pi }\frac{1}{3} a^3 \sin \phi \, d\phi \, d\theta =\int_0^{2\pi } \frac{2 a^3}{3} \, d\theta \\ & =\frac{4 a^3 \pi }{3}. \end{align}

Example. Evaluate the triple integral over the given region $$I=\iiint_D z^2 \, dx \, dy \, dz$$ where $D$ is the solid hemisphere $x^2+y^2+z^2\leq 1$ and $z\geq 0$

Solution. Using spherical coordinates to evaluate the triple integral \begin{align} I=&4\int_0^{\pi /2}\int_0^{\pi /2}\int_0^1 \rho ^4 \cos ^2 \phi \sin \phi \ d\rho \, d\theta \, d\phi \\ & =4\int_0^{\pi /2}\int_0^{\pi /2}\frac{1}{5} \cos ^2 \phi \sin \phi \ d\theta \, d\phi \\ &=4\int_0^{\pi /2} \frac{1}{10} \pi \cos ^2 \phi \sin \phi \, \ d\phi \\ & =\frac{4\pi }{10}\left(-\frac{\cos \phi }{4}-\frac{1}{12} \cos (3 \phi )\right) \\ & =\frac{2 \pi }{15}. \end{align}

Example. Evaluate the triple integral over the given region $$I=\iiint_D \frac{dx\, dydz}{\sqrt{x^2+y^2+z^2}}$$ where $D$ is the solid sphere $x^2+y^2+z^2\leq 3.$

Solution. Since the integrand is symmetric about the origin we can use symmetry. In spherical coordinates the triple integral $I$ evaluates to, \begin{align} I & =8\int_0^{\pi /2}\int_0^{\pi /2}\int_0^{\sqrt{3}} \rho \sin \phi \, d\rho \, d\theta \, d\phi \\ & =8\int_0^{\pi /2}\int_0^{\pi /2}\frac{3 \sin \phi }{2} \, d\theta \, d\phi \\ & =8\int_0^{\pi /2} \frac{3}{4} \pi \sin (\phi ) \, d\phi \\ &=6 \pi. \end{align}

Example. Evaluate the iterated integral $$I =\int_0^{\pi /2}\int_0^{2\pi }\int_0^2 \cos \phi \sin \phi \,d\rho \,d\theta d\phi.$$

Solution. We find \begin{align} I & =\int_0^{\pi /2}\int_0^{2\pi }2 \cos \phi \sin \phi \, d\theta \, d\phi \\ & = \int_0^{\pi /2} 4 \pi \cos \phi \sin \phi \, d\phi =2 \pi. \end{align}

Example. Evaluate the iterated integral $$I=\int_0^{\pi /2}\int_0^{\pi /4}\int_0^{\cos \phi }\rho ^2 \sin \phi \, d\rho \, d\theta \, d\phi.$$

Solution. We find \begin{align} I& =\int_0^{\pi /2}\int_0^{\pi /4}\frac{1}{3} \cos ^3 \phi \sin \phi \, d\theta \, d\phi \\ & =\int_0^{\pi /2} \frac{1}{12} \pi \cos ^3(\phi ) \sin \phi \, d\phi \\ & =\frac{\pi }{48}. \end{align}

Example. Find the volume of the bounded solid in the spherical solid $\rho \leq 4$ after the solid cone $\phi \leq \pi /6$ has been removed.

Solution. We will use spherical coordinates and evaluate a triple iterated integral to find the volume \begin{align} V& =\int_0^{2\pi }\int_{\pi /6}^{\pi }\int_0^4\rho ^2 \sin \phi \, d\rho \, d\phi \, d\theta \\ & =\int_0^{2\pi }\int_{\pi /6}^{\pi }\frac{64 \sin \phi }{3} \, d\phi \, d\theta \\ & =\int_0^{2\pi } \frac{64}{3} \left(1+\frac{\sqrt{3}}{2}\right) \, d\theta \\ & =\frac{64\pi }{3} \left(2+\sqrt{3}\right). \end{align}

Example. Evaluate the triple iterated integral $$I=\int_0^3\int_0^{\sqrt{9-y^2}}\int_{\sqrt{x^2+y^2}}^{\sqrt{18-x^2-y^2}} \left(x^2+y^2+z^2\right) \, dz \, dx \,dy.$$

Solution. Converting to spherical coordinates we evaluate $I$ as follows, \begin{align} I&=\int_0^{\pi /4}\int_0^{\pi /2}\int_0^{3\sqrt{2}}\rho ^4 \sin \phi \, d\rho \, d\theta \, d\phi \\ & =\int_0^{\pi /4}\int_0^{\pi /2}\frac{972}{5} \sqrt{2} \sin \phi \, d\theta \, d\phi \\ &=\int_0^{\pi /4} \frac{486}{5} \sqrt{2} \pi \sin \phi \, d\phi \\ &=\frac{486\pi }{5} \left(\sqrt{2} -1\right). \end{align}

## Exercises on Triple Integrals in Cylindrical and Spherical Coordinates

Exercise. Evaluate the following iterated integrals.

$(1) \quad \int_0^{\pi }\int_0^2\int_0^{\sqrt{4-r^2}}r \sin \theta \, dz \, dr \, d\theta$

$(2) \quad \int_0^{\pi /4}\int_0^1\int_0^{\sqrt{r}}r^2 \sin \theta \, dz \, dr \, d\theta$

$(3) \quad \int_0^{2\pi }\int_0^4\int_0^1z r \, dz \, dr \, d\theta$

$(4) \quad \int_{-\pi /4}^{\pi /3}\int_0^{\sin \theta }\int_0^{4 \cos \theta } r \, dz \, dr \, d\theta$

$(5) \quad \int_0^{\pi /2}\int_0^{\pi /4}\int_0^{\cos \phi }\rho ^2 \sin \phi \, d\rho \, d\theta \, d\phi$

$(6) \quad \int_0^{\pi /2}\int_0^{2\pi }\int_0^2 \cos \phi \sin \phi \, d \rho \, d\theta \, d\phi$