Find the integration limits of $\int_{0}^{3} \int_{0}^{\sqrt{9 – x^2}} \int_{0}^{\sqrt{9 – x^2 – y^2}} \frac{\sqrt{x^2 + y^2 + z^2}}{1 + (x^2 + y^2 + z^2)^2} dz dy dx$ in spherical coordinates.
So I know the integral will end up looking something like $\iiint_{}{} \frac{\rho^3}{1 + \rho^4} \sin \phi d \rho d \theta d \phi$, but I have no idea how to convert the limits. I know $0 < x < 3, \sqrt{9 – x^2} = y$ and $\sqrt{9 – x^2 – y^2} = z$, but how do I write $x, y,$ and $z$ in terms of $\rho, \theta,$ and $\phi$?
Best Answer
If you make the graph of the solid, then you see the limits more easy, first draw the region in the plane $x-y$, how $0\leq x \leq 3$ and $0\leq y \leq \sqrt{9-x^2}$ then:
Now how $0\leq z\leq \sqrt{9-x^2-y^2}$ this is a one eighth of the sphere with radius 3, in the first octant of the space $x-y-z$.
The limits are $0\leq \rho\leq 3$, $0\leq \theta\leq \pi/2$ and $0\leq \phi\leq \pi/2$