# Triple Integrals in Cylindrical Coordinates

## Solved Problems

Click or tap a problem to see the solution.

### Example 3

Using cylindrical coordinates evaluate the integral $\int\limits_{ - 2}^2 {dx} \int\limits_{ - \sqrt {4 - {x^2}} }^{\sqrt {4 - {x^2}} } {dy} \int\limits_0^{4 - {x^2} - {y^2}} {{y^2}dz} .$

### Example 4

Calculate the integral using cylindrical coordinates: $\iiint\limits_U {\sqrt {{x^2} + {y^2}} dxdydz} .$ The region U is bounded by the paraboloid z = 4 − x² − y², by the cylinder x² + y² = 4 and by the planes y = 0, z = 0 (Figure 8).

### Example 5

Find the integral $\iiint\limits_U {ydxdydz},$ where the region $$U$$ is bounded by the planes $$z = x + 1,$$ $$z = 0$$ and by the cylindrical surfaces $${x^2} + {y^2} = 1,$$ $${x^2} + {y^2} = 4$$ (see Figure $$10$$).

### Example 3.

Using cylindrical coordinates evaluate the integral $\int\limits_{ - 2}^2 {dx} \int\limits_{ - \sqrt {4 - {x^2}} }^{\sqrt {4 - {x^2}} } {dy} \int\limits_0^{4 - {x^2} - {y^2}} {{y^2}dz} .$

Solution.

The region of integration $$U$$ is shown in Figure $$6.$$ Its projection on the $$xy$$-plane is the circle $${x^2} + {y^2} = {2^2}$$ (Figure $$7$$).

The new variables in the cylindrical coordinates range within the limits:

$0 \le \rho \le 2,\;\;0 \le \varphi \le 2\pi ,\;\; 0 \le z \le 4 - {\rho ^2}.$

Substituting $$x = \rho \cos \varphi$$ and $$x = \rho \sin \varphi,$$ we find the value of the integral:

$I = \int\limits_{ - 2}^2 {dx} \int\limits_{ - \sqrt {4 - {x^2}} }^{\sqrt {4 - {x^2}} } {dy} \int\limits_0^{4 - {x^2} - {y^2}} {{y^2}dz} = \iiint\limits_U {{y^2}dxdydz} = \iiint\limits_{U'} {{{\left( {\rho \sin \varphi } \right)}^2}\rho d\rho d\varphi dz} = \iiint\limits_{U'} {{\rho ^3}{{\sin }^2}\varphi d\rho d\varphi dz} = \int\limits_0^{2\pi } {{{\sin }^2}\varphi d\varphi } \int\limits_0^2 {{\rho ^3}\left( {4 - {\rho ^2}} \right)d\rho } = \int\limits_0^{2\pi } {{{\sin }^2}\varphi d\varphi } \int\limits_0^2 {\left( {4{\rho ^3} - {\rho ^5}} \right)d\rho } = \int\limits_0^{2\pi } {{{\sin }^2}\varphi d\varphi } \cdot \left[ {\left. {\left( {\frac{{4{\rho ^4}}}{4} - \frac{{{\rho ^6}}}{6}} \right)} \right|_0^2} \right] = \left( {{2^4} - \frac{{{2^6}}}{6}} \right)\int\limits_0^{2\pi } {{{\sin }^2}\varphi d\varphi } = \frac{{16}}{3}\int\limits_0^{2\pi } {{{\sin }^2}\varphi d\varphi } = \frac{{16}}{3}\int\limits_0^{2\pi } {\frac{{1 - \cos 2\varphi }}{2}d\varphi } = \frac{8}{3}\int\limits_0^{2\pi } {\left( {1 - \cos 2\varphi } \right)d\varphi } = \frac{8}{3}\left[ {\left. {\left( {\varphi - \frac{{\sin 2\varphi }}{2}} \right)} \right|_0^{2\pi }} \right] = \frac{8}{3} \cdot 2\pi = \frac{{16\pi }}{3}.$

### Example 4.

Calculate the integral using cylindrical coordinates: $\iiint\limits_U {\sqrt {{x^2} + {y^2}} dxdydz} .$ The region $$U$$ is bounded by the paraboloid $$z = 4 - {x^2} - {y^2},$$ by the cylinder $${x^2} + {y^2} = 4$$ and by the planes $$y = 0,$$ $$z = 0$$ (Figure $$8\text{).}$$

Solution.

By sketching the region of integration $$U$$ (Figure $$9$$), we see that its projection on the $$xy$$-plane (the region $$D$$) is the half-circle of radius $$\rho = 2.$$

We convert to cylindrical coordinates using the substitutions

$x = \rho \cos \varphi ,\;\; y = \rho \sin \varphi ,\;\; z = z,\;\; dxdydz = \rho d\rho d\varphi dz.$

The new variables will range within the limits

$0 \le \rho \le 2,\;\; 0 \le \varphi \le \pi ,\;\; 0 \le z \le 4 - {\rho ^2}.$

Now we can calculate the integral:

$I = \iiint\limits_U {\sqrt {{x^2} + {y^2}} dxdydz} = \iiint\limits_{U'} {\rho \cdot \rho d\rho d\varphi dz} = \iiint\limits_{U'} {{\rho ^2}d\rho d\varphi dz} = \int\limits_0^\pi {d\varphi } \int\limits_0^2 {{\rho ^2}d\rho } \int\limits_0^{4 - {\rho ^2}} {dz} = \int\limits_0^\pi {d\varphi } \int\limits_0^2 {{\rho ^2}d\rho } \cdot \left[ {\left. z \right|_0^{4 - {\rho ^2}}} \right] = \int\limits_0^\pi {d\varphi } \int\limits_0^2 {{\rho ^2}\left( {4 - {\rho ^2}} \right)d\rho } = \int\limits_0^\pi {d\varphi } \int\limits_0^2 {\left( {4{\rho ^2} - {\rho ^4}} \right)d\rho } = \int\limits_0^\pi {d\varphi } \left[ {\left. {\left( {\frac{{4{\rho ^3}}}{3} - \frac{{{\rho ^5}}}{5}} \right)} \right|_0^2} \right] = \left( {\frac{4}{3} \cdot {2^3} - \frac{{{2^5}}}{5}} \right)\int\limits_0^\pi {d\varphi } = \frac{{64}}{{15}}\int\limits_0^\pi {d\varphi } = \frac{{64}}{{15}} \cdot \left[ {\left. \varphi \right|_0^\pi } \right] = \frac{{64\pi }}{{15}}.$

### Example 5.

Find the integral $\iiint\limits_U {ydxdydz},$ where the region $$U$$ is bounded by the planes $$z = x + 1,$$ $$z = 0$$ and by the cylindrical surfaces $${x^2} + {y^2} = 1,$$ $${x^2} + {y^2} = 4$$ (see Figure $$10$$).

Solution.

We calculate this integral in cylindrical coordinates. From the condition

$0 \le z \le x + 1$

it follows that

$0 \le z \le \rho \cos \varphi + 1.$

The projection of the region of integration onto the $$xy$$-plane is the ring formed by the two circles: $${x^2} + {y^2} = 1$$ and $${x^2} + {y^2} = 4$$ (Figure $$11$$). Hence, the variables $$\rho$$ and $$\varphi$$ range in the interval

$1 \le \rho \le 2,\;\; 0 \le \varphi \le 2\pi .$

Calculate the integral:

$I = \iiint\limits_U {ydxdydz} = \iiint\limits_{U'} {\rho \sin \varphi \cdot \rho d\rho d\varphi dz} = \iiint\limits_{U'} {{\rho ^2}\sin \varphi d\rho d\varphi dz} = \int\limits_0^{2\pi } {\sin \varphi d\varphi } \int\limits_0^2 {{\rho ^2}d\rho } \int\limits_0^{\rho \cos \varphi + 1} {dz} = \int\limits_0^{2\pi } {\sin \varphi d\varphi } \int\limits_0^2 {{\rho ^2}d\rho } \cdot \left[ {\left. z \right|_0^{\rho \cos \varphi + 1}} \right] = \int\limits_0^{2\pi } {\sin \varphi d\varphi } \int\limits_0^2 {{\rho ^2}\left( {\rho \cos \varphi + 1} \right)d\rho } = \int\limits_0^{2\pi } {\sin \varphi d\varphi } \int\limits_0^2 {\left( {{\rho ^3}\cos \varphi + {\rho ^2}} \right)d\rho } = \int\limits_0^{2\pi } {\sin \varphi d\varphi } \cdot \left[ {\left. {\left( {\frac{{{\rho ^4}}}{4}\cos \varphi + \frac{{{\rho ^3}}}{3}} \right)} \right|_{\rho = 1}^{\rho = 2}} \right] = \int\limits_0^{2\pi } {\sin \varphi \left[ {\left( {4\cos \varphi + \frac{8}{3}} \right) - \left( {\frac{{\cos \varphi }}{4} + \frac{1}{3}} \right)} \right]d\varphi } = \int\limits_0^{2\pi } {\sin \varphi \left( {\frac{{15}}{4}\cos \varphi + \frac{7}{3}} \right)d\varphi } = \int\limits_0^{2\pi } {\left( {\frac{{15}}{4}\sin \varphi \cos \varphi + \frac{7}{3}\sin \varphi } \right)d\varphi } = \int\limits_0^{2\pi } {\left( {\frac{{15}}{8}\sin 2\varphi + \frac{7}{3}\sin \varphi } \right)d\varphi } = \left. {\left( { - \frac{{15}}{{16}}\cos 2\varphi - \frac{7}{3}\cos \varphi } \right)} \right|_0^{2\pi } = 0.$