Mumbai University > First Year Engineering > sem 2 > Applied Maths 2

Marks : 6

Year : DEC 2014

$$Let \space I =\int\int\int\dfrac {dx\space dy\space dz}{x^2+y^2+z^2}$$

Using spherical polar co-ordinate for sphere $x^2 + y^2 + z^2 =a^2 $ center $(0, 0, 0)$ and radius a

Put $x = r \sin θ \cos\phi $

$Y = r \sin θ \sin \phi \\ Z = r \cos θ \\ Dx \space dy \space dz = x^2 \sin θ \space dr \space dθ \space d\phi \\x^2+y^2+z^2= (r\sin \theta\cos\phi)^2+(r\sin\theta\sin\phi)^2+(r\cos\theta)^2 \\ =r^2\sin^2\theta[\cos^2\phi+\sin^2\phi+r^2\cos^2\theta]\\ =r^2\sin^2\theta +r^2\cos^2\theta\\ =r^2$

Limits of r is 0 to a

Limits of $\theta$ is 0 to $\pi$

Limits of $\phi$ is $0$ to $2\pi$ $\\$ $$\therefore I=\int\limits_{\phi -0}^{2\pi}\int\limits_{\theta=0}^{\pi}\int\limits^a_{r=0}\dfrac 1{r^2}r^2\sin\theta dr\space d\theta \space d\phi$$

$ =\int\limits_0^{2\pi}\int\limits_0^{\pi}\sin\theta[r]_0^a\space d\theta \space d\phi \\ =\int\limits_0^{2\pi}\int\limits_0^{\pi} a\sin\theta \space d\theta\space d\phi\\ =\int\limits_0^{2\pi}a[-\cos\theta]_0^{\pi}d\phi \\ =\int\limits_0^{2\pi}a[-\cos\pi +\cos 0] d\phi \\ =\int\limits_0^{2\pi}a[1+1]d\theta\\ =2a[\phi]_0^{2\pi}\\ =4\pi a$