Triple Integrals in Cylindrical Coordinates

The position of a point M (x, y, z) in the xyz-space in cylindrical coordinates is defined by three numbers: ρ, φ, z, where ρ is the projection of the radius vector of the point M onto the xy-plane, φ 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).

Cylindrical system of coordinates — Figure 1.

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

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.

Region U bounded by the surface x^2+y^2=1 and the planes z=0, z=1 — Figure 2.

Projection of the region U on the xy-plane — Figure 3.

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 \(4.\)

Region U bounded by the surfaces x^2+y^2=3z and z=3 — Figure 4.

Region of integration as a circle — Figure 5.

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

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

See more problems on Page 2.