# A triple definite integral from Cartesian coordinates to Spherical coordinates. Help!

I am trying to solve evaluate this triple definite integral. The integral is given below.

$$\Lambda =\int_0^3\int_0^{\sqrt{9-x^2}}\int_0^{\sqrt{9-x^2-y^2}} \frac{\sqrt{x^2+y^2+z^2}}{1+(x^2+y^2+z^2)^2} \, dz\,dy\,dx$$

I know that this problem has to be done in spherical coordinates, since we have a triple integral, and we have the following relationship :

$$x^2+y^2+z^2$$

and I know that this equals to:

$$\rho=\sqrt{x^2+y^2+z^2}$$

I know that for spherical coordinates my x,y,z variables have the following form:

$$x = \rho\cos(\theta)\sin(\phi) \\ y = \rho\sin(\theta)\sin(\phi) \\ z = \rho\cos(\phi)$$

Once I make the substitutions my integral expression gets really complicated, and I have no idea how to proceed from there. Could anybody post a step by step solution to this triple definite integral?And also showing how the change of variables are done within the coordinates and the integral. I do believe that using spherical coordinates for this problem is the best way to solve evaluate this triple integral. Thank You!

• Did you forget the jacobian :) Nov 7 '17 at 2:31
• @IntegrateThis o yes , you are right, when changing variables from Cartesian to spherical, i should always multiple the integrand by the determinant of the jacobian matrix. Nov 7 '17 at 2:35
• The integrand is $\frac {p^3} { 1+ p^4} sin \phi$ Nov 7 '17 at 2:40
• becomes $ln |u|$, what is confusing you? The expression in $u$ have no theta or phi values, you can treat each component separately. Nov 7 '17 at 2:43
• @IntegrateThis Ok I get this, but I am confused on how you got the upper bounds for the integrational limits. Nov 7 '17 at 2:46

we need to deal the sphere in the first octant. so $\theta$ varies from $o$ to $\pi/2$, $\rho$ from $0$ to $3$, $\phi$ from $0$ to $\pi/2$. $$\Lambda =\int_0^3\int_0^{\sqrt{9-x^2}}\int_0^{\sqrt{9-x^2-y^2}} \frac{\sqrt{x^2+y^2+z^2}}{1+(x^2+y^2+z^2)^2} \, dz \, dy \, dx$$

$$=\int_0^{\pi/2}d\theta\int_{0}^{\phi=\pi/2}\sin(\phi) \, d\phi \int_0^{\rho=3} \frac{{\rho^3}}{1+(\rho)^4} \, d\rho \, d\phi$$ Here you can integrate.

• If you find any type error, please let me know.
– user464147
Nov 7 '17 at 3:02
• I think that $\theta$ varies from $0$ to $\pi/2$ Nov 7 '17 at 3:04
• no. that would be a full circle. please draw and verify.
– user464147
Nov 7 '17 at 3:11
• do you get the point?
– user464147
Nov 7 '17 at 3:12
• Now it is clear no?
– user464147
Nov 7 '17 at 3:21

You want to integrate a rotationally symmetric function over an eighth of a ball of radius $3$. Partition this ball onionlike in spherical shells of area ${1\over8}\cdot 4\pi r^2$ and thickness $dr$. In this way you obtain $$\Lambda={\pi\over8}\int_0^3{r\over 1+r^4}\>4r^2\>dr={\pi\over8}\>\log(1+r^4)\biggr|_0^3={\pi\over8}\log 82$$ Doing it all the way in terms of spherical coordinates involves computing the Jacobian of the coordinate transformation.