Answer to Question #104270 in Calculus for Ajay

Question

∫∫∫cos(x^2+y^2+z^2)^(3/2) dxdydz

Answer 1

$\iiint{\cos(x^2+y^2+z^2)^{3/2} dxdydz}$

Convert $x,y,z$ to spherical coordinates $r, \varphi, \theta.$

$x=r\cos(\varphi)\sin(\theta) \\ y=r\sin(\varphi)\sin(\theta) \\ z=r\cos(\theta)$

The absolute value of Jacobian is $r^2\sin(\theta).$ Then the original integral is

$$\iiint {\cos((r^2)^{3/2})\cdot r\sin(\theta)dr d\varphi d\theta} = \iiint{r^2\cos(r^3)\sin(\theta)dr d\varphi d\theta } \\ = \int{r^2\cos(r^3)dr}\int{d\varphi}\int{\sin(\theta)d\theta} = \frac{1}{3}\int{\cos(r^3)d(r^3)}\int{d\varphi}\int{\sin(\theta)d\theta} \\ = -\frac{1}{3}\varphi\sin(r^3)\cos{(\theta)}$$

We have omitted the constants for convenience.

Return to $x,y,z.$

$r=\sqrt{x^2+y^2+z^2} \\ \varphi = \arctan{\left( \frac{y}{x}\right)} \\ \theta = \arccos{\left(\frac{z}{\sqrt{x^2+y^2+z^2}}\right)}$

Therefore, the original integral is

$$\iiint{\cos(x^2+y^2+z^2)^{3/2} dxdydz} \\ = -\frac{1}{3} \arctan{\left( \frac{y}{x}\right)}\sin((x^2+y^2+z^2)^{3/2})\cos{\left[ \arccos{\left(\frac{z}{\sqrt{x^2+y^2+z^2}}\right)} \right]} \\ = -\frac{1}{3} \arctan{\left( \frac{y}{x}\right)}\sin((x^2+y^2+z^2)^{3/2})\frac{z}{\sqrt{x^2+y^2+z^2}}$$