# How to solve $\int_0^2 \int_0^\sqrt{4-x^{2}} \int_0^\sqrt{4-x^2 -y^2} z \sqrt{4-x^2 -y^2} \, dz \, dy \, dx$ in spherical coordinate

$$\int_0^2 \int_0^\sqrt{4-x^{2}} \int_0^\sqrt{4-x^2 -y^2} z \sqrt{4-x^2 -y^2} \, dz \, dy \, dx$$

The task is to solve this integral using spherical coordinate. After I tried to change the variable, I got $$\int _0^{\frac{\pi }{2}}\int _0^{\frac{\pi }{2}}\int _0^2\left(\rho \:\cos\left(\phi \right)\sqrt{4-\rho ^2\left(\sin\left(\phi \right)\right)^2}\right)\:\rho ^2\sin\left(\phi \right)d\rho \:d\theta \:d\phi$$

Which I think pretty ugly with $$\sqrt{4-\rho ^2\left(\sin\left(\phi \right)\right)^2}$$ . Is there anything I did wrong on the variable changing process? If it's not, what are the approaches to solve this integral?

• Do the $\phi$ integral first. – Ninad Munshi May 20 at 9:24
• What are the rules to change the integration order on spherical coordinate system? – Patrick Herp May 20 at 9:27
• The bounds are constants, there is no need to do anything special. – Ninad Munshi May 20 at 9:27

$$\int_0^{2}\int_0^{\frac{\pi}{2}}\int_0^{\frac{\pi}{2}}\rho^3\sin\phi\cos\phi\sqrt{4-\rho^2\sin^2\phi}\:d\theta\:d\phi\:d\rho = \frac{\pi}{4}\int_0^{2}\int_0^{\frac{\pi}{2}} \rho\sqrt{4-\rho^2\sin^2\phi} \:d(\rho^2\sin^2\phi)\:d\rho$$
$$= \frac{\pi}{6}\int_0^2 -\rho \left[4-\rho^2\sin^2\phi\right]^{\frac{3}{2}}\biggr|_0^{\frac{\pi}{2}}\:d\rho = \frac{\pi}{6}\int_0^2 8\rho-\rho(4-\rho^2)^{\frac{3}{2}}\:d\rho = \frac{8\pi}{5}$$
• How exactly did you change into differential $\:d(\rho^2\sin^2\phi)$ ? – Patrick Herp May 20 at 10:04
• @PatrickHerp that's just a basic u substitution without writing $du=\cdots$. Like for example $$\int_0^\pi \sin x\cos x\:dx = \int_0^\pi \sin x \:d(\sin x) = \frac{1}{2}\sin^2 x \Bigr|_0^\pi$$ – Ninad Munshi May 20 at 10:08
• And the differential doesn't effect the remaining $\rho$ variable? – Patrick Herp May 20 at 10:14