Question

Evaluate the following integral by converting it to cylindrical coordinates: ∫(-3 to 3)∫(0 to sqrt(9-x^2))∫(0 to 9-x^2-y^2) sqrt(x^2+y^2) dz dy dx

Answer 1

$I = \int_{-3}^{3}\int_0^{\sqrt{9 - x^2}}\int_0^{\sqrt{9 - x^2 - y^2}} \sqrt{x^2 + y^2}\, \mathrm{d}z\, \mathrm{d}y\, \mathrm{d}x\\ \textsf{Converting to cylindrical coordinates.}\\ \, \mathrm{d}z\, \mathrm{d}y\, \mathrm{d}x = \rho\mathrm{d}z\, \mathrm{d}\rho\, \mathrm{d}\theta\\ x = \rho\cos{\theta}, y = \rho\sin{\theta}\\ x^2 + y^2 = \rho^2(\cos^2{\theta} + \sin^2{\theta}) = \rho^2\\ y = 0, \sin{\theta} = 0, \theta = 2\pi\\ y = \sqrt{9 - x^2}\\ y = \sqrt{9 - \rho^2\cos^2{\theta}}\\ x= -3, \rho\cos{\theta} = -3\\ x= -3, \rho\cos(2\pi) = \rho = -3\\ x= 3, \rho\cos(2\pi) = \rho = 3\\ \therefore y = 3\sin{\theta} = \sqrt{9 - 9\cos^2{2\pi}} = \sqrt{9 - 9} = 0, \theta = 0\\ z = \sqrt{9 - x^2 - y^2} = \sqrt{9 - \rho^2}\\ \begin{aligned} I = \int_{-3}^{3}\int_0^{\sqrt{9 - x^2}}\int_0^{\sqrt{9 - x^2 - y^2}} \sqrt{x^2 + y^2}\, \mathrm{d}z\, \mathrm{d}y\, \mathrm{d}x &= \int_{-3}^{3}\int_0^{2\pi} \int_0^{\sqrt{9 - \rho^2}} \sqrt{\rho^2} \cdot \rho\mathrm{d}z\, \mathrm{d}\theta\, \mathrm{d}\rho \\&=\int_{-3}^{3}\int_0^{2\pi} \int_0^{\sqrt{9 - \rho^2}} \rho^2\,\mathrm{d}z\, \mathrm{d}\theta\, \mathrm{d}\rho \\&=\int_{-3}^{3}\rho^2\,\mathrm{d}\rho\int_0^{\sqrt{9 - \rho^2}} \mathrm{d}z \int_0^{2\pi}\, \mathrm{d}\theta \\&= 2\pi\int_{-3}^{3}\rho^2\,\mathrm{d}\rho \cdot1 \vert_0^{\sqrt{9 - \rho^2}} \\&= 2\pi\int_{-3}^{3}\rho^2\sqrt{9 - \rho^2}\,\mathrm{d}\rho \end{aligned}\\ \textsf{Substitute}\, \rho = 3\sin\alpha\\ \begin{aligned} \therefore I &= 2\pi\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}9\sin^2{\alpha}\sqrt{9 - 9\sin^2\alpha}\cdot3\cos{\alpha}\mathrm{d}\alpha \\&= 2\pi\int_{0}^{\frac{\pi}{2}}9\sin^2{\alpha}\sqrt{9 - 9\sin^2\alpha}\cdot3\cos{\alpha}\mathrm{d}\alpha \\&= 324\pi\int_{0}^{\frac{\pi}{2}}\sin^2{\alpha}\cos^2{\alpha}\,\mathrm{d}\alpha \\&= 324\pi\left(\frac{\alpha}{8} + \frac{\sin{\alpha}\cos{\alpha}}{8}- \frac{\sin{\alpha}\cos^{3}{\alpha}}{4}\right)\vert_{0}^{\frac{\pi}{2}} \\&= 324\left(\frac{\pi}{16}\right) = \frac{81\pi^2}{4} \end{aligned}\\ \textsf{Note:}\, \int_{-a}^{a} f(x)\, \mathrm{d}x = 2\int_{0}^{a} f(x)\, \mathrm{d}x, \,\,\textsf{if }\,f(x) \textsf{is even.}$