Answer to Question #174760 in Calculus for Muhammad Ahmed

Question

With the help of triple integrals, find the volume of the sphere ρ = 2 in a) Rectangular coordianates b) Cylinderical coordiantes c) Spherical coordinates 

Answer 1

With the help of triple integrals, find the volume of the sphere ρ = 2 in a) Rectangular coordianates b) Cylinderical coordiantes c) Spherical coordinates 

Solution:

a) Rectangular coordinates:

$x^2+y^2+z^2=4$

$V=\displaystyle\int_{-2}^2dx\int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}dy\int_{-\sqrt{4-x^2-y^2}}^{\sqrt{4-x^2-y^2}}dz=$ $8\displaystyle\int_{0}^2dx\int_{0}^{\sqrt{4-x^2}}dy\int_{0}^{\sqrt{4-x^2-y^2}}dz=$ $8\displaystyle\int_{0}^2dx\int_{0}^{\sqrt{4-x^2}}\sqrt{4-x^2-y^2}dy$

Let's apply the change of variables:

$y=\sqrt{4-x^2}\cdot\sin{t}$

$dy=\sqrt{4-x^2}\cdot\cos{t}\cdot dt$

$y=0 \implies t=0$

$y=\sqrt{4-x^2}\implies t=\frac{\pi}{2}$

$V=8\displaystyle\int_{0}^2dx\int_{0}^{\pi/2}\sqrt{4-x^2-(4-x^2)\sin^2{t}}\cdot$ $\sqrt{4-x^2}\cos{t}dt=$

$8\displaystyle\int_{0}^2dx\int_{0}^{\pi/2}(4-x^2)\cos^2{t}dt=$ $8\displaystyle\int_{0}^2dx\int_{0}^{\pi/2}(4-x^2)\frac{(1+\cos{2t})}{2}dt=$ $8\displaystyle\int_{0}^2dx(4-x^2)\frac{(t+\frac{\sin{2t}}{2})}{2}|_0^{\pi/2}=$ $2\pi\displaystyle\int_{0}^2(4-x^2)dx=2\pi(4x-x^3/3)|_0^2=\frac{32\pi}{3}$

b) Cylinderical coordiantes:

$x=\rho\cos{\varphi}$

$y=\rho\sin{\varphi}$

$z=z$

$J=\rho$

The equation of the sphere:

$\rho^2+z^2=4$

$V=\displaystyle\int_0^{2\pi}d\varphi\int_0^2d\rho\int_{-\sqrt{4-\rho^2}}^{\sqrt{4-\rho^2}}\rho dz=$ $2\displaystyle\int_0^{2\pi}d\varphi\int_0^2d\rho\int_0^{\sqrt{4-\rho^2}}\rho dz=$ $4\pi\displaystyle\int_0^2\sqrt{4-\rho^2}\rho d\rho=$ $2\pi\displaystyle\int_0^2\sqrt{4-\rho^2} d\rho^2=-\frac{4\pi}{3}\displaystyle(4-\rho^2)^{3/2}|_0^2=\frac{32\pi}{3}$

c) Spherical coordinates 

$x=r\sin{\theta}\cos{\varphi}$

$y=r\sin{\theta}\sin{\varphi}$

$z=r\cos{\theta}$

$J=r^2\sin{\theta}$

$V=\displaystyle\int_0^{2\pi}d\varphi\int_0^{\pi}d\theta\int_0^2r^2\sin{\theta}dr=$ $2\pi\displaystyle\int_0^{\pi}d\theta\sin{\theta}\frac{r^3}{3}|_0^2=$ $\displaystyle\frac{16\pi}{3}\int_0^{\pi}\sin{\theta}d\theta=$ $-\displaystyle\frac{16\pi}{3}\cos{\theta}|_0^{\pi}=\frac{32\pi}{3}$

Answer: $V=\displaystyle\frac{32\pi}{3}.$