Answer to Question #143563 in Calculus for Promise Omiponle

Question

Evaluate the following integral by converting to spherical coordinates: ∫(0 to 1)∫(0 to sqrt(1-x^2))∫(sqrt(x^2+y^2) to sqrt(2-x^2-y^2)) xy dz dy dx

Answer 1

$\displaystyle\int_0^1 \int_0^{\sqrt{1 - x^2}}\int_{\sqrt{x^2 + y^2}}^{\sqrt{2 - x^2 - y^2}} \,xy \mathrm{d}z\mathrm{d}y\mathrm{d}x\\ \, \mathrm{d}z\, \mathrm{d}y\, \mathrm{d}x = \rho^2\sin{\theta}\mathrm{d}\varphi\, \mathrm{d}\theta\, \mathrm{d}\rho\\ x = \rho\cos{\varphi}\sin{\theta}, y = \rho\sin{\varphi}\sin{\theta}, z = \rho\cos{\theta}\\ \sqrt{x^2 + y^2} = z\\ x^2 + y^2 + z^2 = 2z^2\\ \rho^2 = 2\rho^2\cos{\theta}\\ \cos{\theta} = \frac{1}{\sqrt{2}}, \theta = \frac{\pi}{4}\\ \sqrt{2 - x^2 - y^2} = z\\ 2 - x^2 - y^2 = z^2\\ 2 = x^2 + y^2 + z^2\\ \rho^2 = 2, \rho = \sqrt{2}\\ \sqrt{1 - x^2}= y\\ 1 = x^2 + y^2\\ x^2 + y^2 = \rho^2\sin^2{\theta}(\cos^2{\varphi} + \sin^2{\varphi}) = \rho^2\sin^2{\theta}\\ 1 = \rho^2\sin^2\theta\\ \sin^2\theta = \frac{1}{2}, \theta = \frac{\pi}{4}\\ x = 0, \rho\cos\varphi\cos\theta = 0, \varphi = \frac{\pi}{2}\\ \rho\sin\varphi\cos\theta = 1\\ \sqrt{2}\sin\varphi\sin\left(\frac{\pi}{4}\right) = 1\\ \varphi = \frac{\pi}{2}\\ \begin{aligned} \int_0^1 \int_0^{\sqrt{1 - x^2}}\int_{\sqrt{x^2 + y^2}}^{\sqrt{2 - x^2 - y^2}}xy \,\mathrm{d}z\mathrm{d}y\mathrm{d}x &= \int_0^{\frac{\pi}{2}} \int_0^{\frac{\pi}{4}}\int_0^{\sqrt{2}} (\rho^2\cos\varphi\sin\varphi\sin^2\theta\cdot\rho^2\sin\theta)\, \mathrm{d}\rho\,\mathrm{d}\theta\,\mathrm{d}\varphi \\&=\int_0^{\sqrt{2}} \rho^4\, \mathrm{d}\rho \int_0^{\frac{\pi}{2}} \frac{\sin(2\varphi)}{2}\,\mathrm{d}\varphi \int_0^{\frac{\pi}{4}}\sin^3{\theta}\, \mathrm{d}\theta \\&=-\frac{\rho^5}{5}\biggr\vert_0^{\sqrt{2}} \cdot\frac{\cos(2\varphi)}{4}\biggr\vert_0^{\frac{\pi}{2}} \cdot\frac{-3\cos{\theta} + \frac{\cos(3\theta)}{3}}{4}\biggr\vert_0^{\frac{\pi}{4}} \\&=-\frac{\rho^5}{5}\biggr\vert_0^{\sqrt{2}} \cdot\frac{\cos(2\varphi)}{4}\biggr\vert_0^{\frac{\pi}{2}} \cdot\frac{-9\cos{\theta} + \cos(3\theta)}{12}\biggr\vert_0^{\frac{\pi}{4}} \\&=\frac{4\sqrt{2}}{5} \cdot\frac{1}{4} \cdot\left(-\frac{3\sqrt{2}}{8} - \frac{\sqrt{2}}{24} + \frac{9}{12} - \frac{1}{12}\right) \\&=\frac{\sqrt{2}}{5}\cdot\left(\frac{2}{3}-\frac{3\sqrt{2}}{8} - \frac{\sqrt{2}}{24}\right) \\&=\frac{1}{5}\cdot\left(\frac{2\sqrt{2}}{3} - \frac{5}{6}\right) \\&=\frac{2\sqrt{2}}{15} - \frac{1}{6} \end{aligned}$