Triple Integrals in Cylindrical Coordinates
Solved Problems
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 \(7\)).
Figure 6. 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 \(\rho = 2.\)
Figure 8. Figure 9. 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
\[1 \le \rho \le 2,\;\; 0 \le \varphi \le 2\pi .\]
Figure 10. Figure 11. Calculate the integral:
\[
I = \iiint\limits_U {ydxdydz}
= \iiint\limits_{U'} {\rho \sin \varphi \cdot \rho d\rho d\varphi dz}
= \iiint\limits_{U'} {{\rho ^2}\sin \varphi d\rho d\varphi dz}
= \int\limits_0^{2\pi } {\sin \varphi d\varphi } \int\limits_0^2 {{\rho ^2}d\rho } \int\limits_0^{\rho \cos \varphi + 1} {dz}
= \int\limits_0^{2\pi } {\sin \varphi d\varphi } \int\limits_0^2 {{\rho ^2}d\rho } \cdot \left[ {\left. z \right|_0^{\rho \cos \varphi + 1}} \right]
= \int\limits_0^{2\pi } {\sin \varphi d\varphi } \int\limits_0^2 {{\rho ^2}\left( {\rho \cos \varphi + 1} \right)d\rho }
= \int\limits_0^{2\pi } {\sin \varphi d\varphi } \int\limits_0^2 {\left( {{\rho ^3}\cos \varphi + {\rho ^2}} \right)d\rho }
= \int\limits_0^{2\pi } {\sin \varphi d\varphi } \cdot \left[ {\left. {\left( {\frac{{{\rho ^4}}}{4}\cos \varphi + \frac{{{\rho ^3}}}{3}} \right)} \right|_{\rho = 1}^{\rho = 2}} \right]
= \int\limits_0^{2\pi } {\sin \varphi \left[ {\left( {4\cos \varphi + \frac{8}{3}} \right) - \left( {\frac{{\cos \varphi }}{4} + \frac{1}{3}} \right)} \right]d\varphi }
= \int\limits_0^{2\pi } {\sin \varphi \left( {\frac{{15}}{4}\cos \varphi + \frac{7}{3}} \right)d\varphi }
= \int\limits_0^{2\pi } {\left( {\frac{{15}}{4}\sin \varphi \cos \varphi + \frac{7}{3}\sin \varphi } \right)d\varphi }
= \int\limits_0^{2\pi } {\left( {\frac{{15}}{8}\sin 2\varphi + \frac{7}{3}\sin \varphi } \right)d\varphi }
= \left. {\left( { - \frac{{15}}{{16}}\cos 2\varphi - \frac{7}{3}\cos \varphi } \right)} \right|_0^{2\pi } = 0.\]
