# Triple Integrals in Cylindrical Coordinates

The position of a point $$M\left( {x,y,z} \right)$$ in the $$xyz$$-space in cylindrical coordinates is defined by three numbers: $$\rho, \varphi, z,$$ where $$\rho$$ is the projection of the radius vector of the point $$M$$ onto the $$xy$$-plane, $$\varphi$$ is the angle formed by the projection of the radius vector with the $$x$$-axis (Figure $$1$$), $$z$$ is the projection of the radius vector on the $$z$$-axis (its value is the same in Cartesian and cylindrical coordinates).

The relationship between cylindrical and Cartesian coordinates of a point is given by

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

We assume here that

$\rho \ge 0,\;\; 0 \le \varphi \le 2\pi ,\;\; - \infty \lt z \lt \infty .$

The Jacobian of transformation from Cartesian to cylindrical coordinates is

$I\left( {\rho ,\varphi ,z} \right) = \left| {\begin{array}{*{20}{c}} {\frac{{\partial x}}{{\partial \rho }}}&{\frac{{\partial x}}{{\partial \varphi }}}&{\frac{{\partial x}}{{\partial z}}}\\ {\frac{{\partial y}}{{\partial \rho }}}&{\frac{{\partial y}}{{\partial \varphi }}}&{\frac{{\partial y}}{{\partial z}}}\\ {\frac{{\partial z}}{{\partial \rho }}}&{\frac{{\partial z}}{{\partial \varphi }}}&{\frac{{\partial z}}{{\partial z}}} \end{array}} \right| = \left| {\begin{array}{*{20}{c}} {\cos \varphi }&{ - \rho \sin \varphi }&0\\ {\sin \varphi }&{\rho \cos \varphi }&0\\ 0&0&1 \end{array}} \right| = \rho \ge 0.$

Then the formula of change of variables for this transformation can be written in the form

$\iiint\limits_U {f\left( {x,y,z} \right)dxdydz} = \iiint\limits_{U'} {f\left( {\rho \cos \varphi , \rho \sin \varphi ,z} \right)\rho d\rho d\varphi dz} .$

Transition from cylindrical coordinates makes calculation of triple integrals simpler in those cases when the region of integration is formed by a cylindrical surface.

## Solved Problems

Click or tap a problem to see the solution.

### Example 1

Evaluate the integral $\iiint\limits_U {\left( {{x^4} + 2{x^2}{y^2} + {y^4}} \right)dxdydz},$ where the region $$U$$ is bounded by the surface $${x^2} + {y^2} \le 1$$ and the planes $$z = 0,$$ $$z = 1$$ (Figure $$2\text{).}$$

### Example 2

Find the integral $\iiint\limits_U {\left( {{x^2} + {y^2}} \right)dxdydz} ,$ where the region $$U$$ is bounded by the surfaces $${x^2} + {y^2} = 3z,$$ $$z = 3$$ (Figure $$4\text{).}$$

### Example 1.

Evaluate the integral $\iiint\limits_U {\left( {{x^4} + 2{x^2}{y^2} + {y^4}} \right)dxdydz},$ where the region $$U$$ is bounded by the surface $${x^2} + {y^2} \le 1$$ and the planes $$z = 0,$$ $$z = 1$$ (Figure $$2\text{).}$$

Solution.

It is more convenient to calculate this integral in cylindrical coordinates. Projection of the region of integration onto the $$xy$$-plane is the circle $${x^2} + {y^2} \le 1$$ or $$0 \le \rho \le 1$$ (Figure $$3$$).

Notice that the integrand can be written as

$\left( {{x^4} + 2{x^2}{y^2} + {y^4}} \right) = \left( {{x^2} + {y^2}} \right)^2 = \left( {{\rho ^2}} \right)^2 = \rho ^4.$

Then the integral becomes

$I = \int\limits_0^{2\pi } {d\varphi } \int\limits_0^1 {{\rho ^4}\rho d\rho } \int\limits_0^1 {dz} .$

The second integral contains the factor $$\rho$$ which is the Jacobian of transformation of the Cartesian coordinates into cylindrical coordinates. All the three integrals over each of the variables do not depend on each other. As a result the triple integral is easy to calculate as

$I = \int\limits_0^{2\pi } {d\varphi } \int\limits_0^1 {{\rho ^4}\rho d\rho } \int\limits_0^1 {dz} = 2\pi \int\limits_0^1 {{\rho ^5}d\rho } \int\limits_0^1 {dz} = 2\pi \cdot 1 \cdot \int\limits_0^1 {{\rho ^5}d\rho } = 2\pi \left. {\left( {\frac{{{\rho ^6}}}{6}} \right)} \right|_0^1 = 2\pi \cdot \frac{1}{6} = \frac{\pi }{3}.$

### Example 2.

Find the integral $\iiint\limits_U {\left( {{x^2} + {y^2}} \right)dxdydz} ,$ where the region $$U$$ is bounded by the surfaces $${x^2} + {y^2} = 3z,$$ $$z = 3$$ (Figure $$4\text{).}$$

Solution.

The region of integration is shown in Figure $$4.$$

To calculate the integral we convert it to cylindrical coordinates:

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

The differential of this transformation is

$dxdydz = \rho d\rho d\varphi dz\;\;\left( {\rho \text{ is the Jacobian}} \right).$

The equation of the parabolic surface becomes

${\rho ^2}{\cos ^2}\varphi + {\rho ^2}{\sin^2}\varphi = 3z\;\;\text{or}\;\;{\rho ^2} = 3z.$

The projection of the region of integration $$U$$ onto the $$xy$$-plane is the circle $${x^2} + {y^2} \le 9$$ with radius $$\rho = 3$$ (Figure $$5$$). The coordinate $$\rho$$ ranges from $$0$$ to $$3,$$ the angle $$\varphi$$ ranges from $$0$$ to $$2\pi,$$ and the coordinate $$z$$ ranges from $$\frac{{{\rho ^2}}}{3}$$ to $$3.$$

As a result, the integral becomes

$I = \iiint\limits_U {\left( {{x^2} + {y^2}} \right)dxdydz} = \iiint\limits_{U'} {{\rho ^2} \cdot \rho d\rho d\varphi dz} = \int\limits_0^{2\pi } {d\varphi } \int\limits_0^3 {{\rho ^3}d\rho } \int\limits_{\frac{{{\rho ^2}}}{3}}^3 {dz} = \int\limits_0^{2\pi } {d\varphi } \int\limits_0^3 {{\rho ^3}d\rho } \cdot \left[ {\left. z \right|_{\frac{{{\rho ^2}}}{3}}^3} \right] = \int\limits_0^{2\pi } {d\varphi } \int\limits_0^3 {{\rho ^3}\left( {3 - \frac{{{\rho ^2}}}{3}} \right)d\rho } = \int\limits_0^{2\pi } {d\varphi } \int\limits_0^3 {\left( {3{\rho ^3} - \frac{{{\rho ^5}}}{3}} \right)d\rho } = \int\limits_0^{2\pi } {d\varphi } \cdot \left[ {\left. {\left( {\frac{{3{\rho ^4}}}{4} - \frac{{{\rho ^6}}}{{18}}} \right)} \right|_0^3} \right] = \left( {\frac{{3 \cdot 81}}{4} - \frac{{729}}{{18}}} \right)\int\limits_0^{2\pi } {d\varphi } = \frac{{81}}{4} \cdot 2\pi = \frac{{81\pi }}{2}.$