How to find limits of an integral in spherical and cylindrical coordinates if you transform it from cartesian coordinates

Question

If you have a volume integral in Cartesian coordinates with given limits of x,y and z and you want to transfer it to another coordinate system like spherical and cylindrical coordinates. I can easily determine how the integral will look like but how to find the limits of the integral(for example, limits of $r,\theta$ and $\phi$ in spherical system )?

For example :

To calculate a triple integral in spherical coordinates for the volume inside $z^{2}=x^{2}+y^{2}$ between the planes z=1 and z=2. Now I have the integral in Cartesian with limits : $$\int_{z=1}^{2}\int_{x=-z}^{z} \int_{y=-\sqrt{z^2-x^2}}^{\sqrt{z^2-x^2}} d x d y d z$$ Now if I turned it into a spherical coordinate [$x=r\sin\theta\cos\phi$,$y=r\sin\theta\sin\phi$,$z=r\cos\theta$].
The integral will be: $$\int\int\int r^{2}\sin\theta dr d\theta d\phi$$ But how do I calculate the limits of $r$,$\theta$ and $\phi$?

I'm not asking how to solve this particular question, I want to know a general method to find limits when changing coordinates like this without using any computer programs like graph plotters.

Answer 1

There is no general formula. You must have a good understanding of how spherical coordinate works if you are converting to it. You also should have a rough visualization of the surface in your mind. Even a rough $2D$ sketch helps. Then use the limits in cartesian coordinates to find the limits in spherical coordinates.

Now if we work through the example you have written down,

As you mentioned, $x = r \cos \phi \sin \theta, y = r \sin \phi \sin \theta, z = r \cos \theta$.

Now for the cone $z^2 \geq x^2 + y^2$, $r^2 \cos^2 \theta \geq r^2 \sin^2 \theta \implies \theta \leq \frac{\pi}{4}$.

Now bounds of $z$ is $1 \leq z \leq 2$. Given $z = r \cos \theta$, we obtain,

$\sec \theta \leq r \leq 2 \sec \theta$.

Limits of $\phi$ is simply $0$ to $2 \pi$.

So the integral becomes,

$\displaystyle \int_0^{2\pi} \int_0^{\pi/4} \int_{\sec \theta}^{2 \sec \theta} r^2 \sin \theta \ dr \ d\theta \ d\phi$.

Answer 2

That region is what you get rotating the red region from the figure below arounf the $z$ axis (the vertical axis). So, $\varphi$ can take any value from $0$ to $2\pi$. Furthermore, $\varphi$ can take any value from $0$ to $\frac\pi4$. Finally, you have $z=r\cos\theta$, and so $1\leqslant z\leqslant2\iff\frac1{\cos\theta}\leqslant r\leqslant\frac2{\cos\theta}$. So, compute$$\int_0^{2\pi}\int_0^{\pi/4}\int_{1/\cos(\theta)}^{2/\cos(\theta)}r^2\sin(\theta)\,\mathrm dr\,\mathrm d\theta\,\mathrm d\varphi.$$