Find volume of the cone $z^2=x^2+y^2$ between the planes $z=0$ and $z=1$ using spherical coordinates

Question

I found this problem that is about calculating the volume of the cone $z^2=x^2+y^2$ between the planes $z=0$ and $z=1$. I know that the problem would be easy using cylindrical coordinates: $$\int_{0}^{2\pi}\int_{0}^{1}\int_{r}^{1}r\cdot dz\,dr\,d\theta = \frac{\pi}{3}$$ Is it possible to solve this problem using spherical coordinates? How?

Answer 1

The given region, in spherical coordinates, is the set of points

$$\left\{(\rho,\theta,\varphi)\mid0\le\rho\le\sec\varphi\land0\le\theta\le2\pi\land0\le\varphi\le\frac\pi4\right\}$$

The limits for $\rho$ follow from what I mentioned in the comment above; the limits for $\theta$ should be obvious; and the limits for $\varphi$, as I mentioned, can be found by setting $z=1$ and $\rho=\sec\varphi$ and subsequently solving for $\varphi$:

$$\begin{align} z^2=x^2+y^2\implies 1&=(\sec\varphi\cos\theta\sin\varphi)^2+(\sec\varphi\sin\theta\sin\varphi)^2\\ 1&=\tan^2\varphi\\ \tan\varphi&=1\\ \varphi&=\frac\pi4 \end{align}$$

Then the volume is the integral

$$\begin{align*} \int_0^{2\pi}\int_0^{\tfrac\pi4}\int_0^{\sec\varphi}\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\varphi\,\mathrm d\theta&=\frac{2\pi}3\int_0^{\tfrac\pi4}\sec^3\varphi\sin\varphi\,\mathrm d\varphi\\[1ex] &=\frac{2\pi}3\int_0^{\tfrac\pi4}\tan\varphi\sec^2\varphi\,\mathrm d\varphi\\[1ex] &=\frac{2\pi}3\int_0^{\tfrac\pi4}\tan\varphi\,\mathrm d(\tan\varphi)\\[1ex] &=\frac\pi3\tan^2\frac\pi4\\[1ex] &=\boxed{\frac\pi3} \end{align*}$$

as expected.