# Find the integral of e^(x^2+y^2+z^2)^3/2 using spherical coordinates.

### 1 Answer | Add Yours

`int int int_E` `e^((x^2+y^2+z^2)^(3/2)) = int_0^(2pi) int_0^pi int_0^1 e^(rho^2)^(3/2) rho^2 sin phi d rho d phi d theta`

`2pi int_0^pi int_0^1 e^(rho^3) rho^2 sin phi d rho d phi`

`2pi int_0^1 (1/3)e^(rho^3) sin phi d phi`

`2pi int_0^pi (1/3) (e-1) sin d phi = (2pi/3)(e-1)(1 + 1)`

`2pi int_0^pi (1/3) (e-1) sin d phi = (4pi/3)(e-1)`

**Hence, evaluating the triple integral using spherical coordinates yields `int int int_E e^(x^2+y^2+z^2)^3/2 = (4pi/3)(e-1).` **