# Expressing a solid in spherical coordinates

I am trying to solve the following question:

The volume of the solid $$E$$ can be represented as $$\int_{-3}^3 \int_0^{\sqrt{9-x^2}} 3 - \sqrt{x^2+y^2} dydx$$ Describe the solid in spherical coordinates.

I graphed the solid and it looks like this:

So clearly $$0 \le \theta \le \pi$$ and $$0 \le \phi \le \frac{\pi}{2}.$$ Also, for each $$(\theta, \phi)$$ the value of $$\rho$$ ranges from $$0$$ until the surface of the cone. The cone is $$z = 3-\sqrt{x^2+y^2}$$. Using the substitutions \begin{align*} x &= \rho \cos \theta \sin \phi\\ y &= \rho \sin \theta \sin \phi\\ z &= \rho \cos \phi \end{align*} we get \begin{align*} \rho \cos \phi &= 3 - \sqrt{\rho^2 \sin^2 \phi}\\ \rho &= \frac {3}{\cos \phi + \sin \phi} \end{align*} So my answer is $$E_{\text{spherical}} = \{(\rho, \theta, \phi)| 0 \le \theta \le \pi, 0 \le \phi \le \frac{\pi}{2}, 0 \le \rho \le \frac {3}{\cos \phi + \sin \phi} \}$$ However, the correct answer is $$E_{\text{spherical}} = \{(\rho, \theta, \phi)| 0 \le \theta \le \pi, 0 \le \phi \le \frac{\pi}{4}, 0 \le \rho \le 3 \csc\phi \}$$

Did I make any mistake?

Thanks!

• Your answer is correct. See here: desmos.com/calculator/zpbns0uzvl Nov 9, 2021 at 15:27
• @MatthewPilling Ahh yes thanks! Nov 9, 2021 at 15:31

The solid is half of an inverted cone with vertex at $$(0, 0, 3)$$ and above $$z = 0$$.

Equation of the surface of the cone is $$~\sqrt{x^2+y^2} = 3 - z$$

That translates to $$\rho \sin\phi = 3 - \rho \cos\phi \implies \rho = \frac{3}{\cos\phi + \sin\phi}$$

Please note that $$0 \leq \phi \leq \pi/2$$ as we are above $$z = 0$$.

Also given bounds of $$x$$ and $$y$$, you can see that the projection of the solid in xy-plane is in the first and the second quadrant. That leads to $$0 \leq \theta \leq \pi$$.

$$E_{\text{spherical}} = \{(\rho, \theta, \phi)| 0 \le \rho \le \frac{3}{ \cos\phi + \sin\phi}, 0 \le \theta \le \pi, 0 \le \phi \le \frac{\pi}{2} \}$$

But to evaluate the volume, it is easier to use $$~ x = \rho \cos\theta\sin\phi, y = \rho \sin\theta \sin\phi, z = 3 - \rho \cos\phi$$

That translates the cone to $$~ \phi = \frac{\pi}{4}$$

$$\rho$$ is bound by the plane $$z = 3 - \rho \cos\phi = 0 \implies \rho = 3 \sec\phi$$

So the integral to find volume in spherical coordinates should be,

$$\displaystyle \int_0^{\pi} \int_0^{\pi/4} \int_0^{3\sec\phi} ~ \rho^2 \sin\phi ~ d\rho ~ d\phi ~ d\theta$$

• The OP did not ask about the volume or the integral that would express it... It was about describing the region in spherical coordinates. Nov 9, 2021 at 15:39
• Oh wow this makes a lot of sense, thanks! Nov 9, 2021 at 15:41
• @PierreCarre From the bounds on the integral I can see the region. Nov 9, 2021 at 15:42
• @PierreCarre sure I would add that too Nov 9, 2021 at 15:43
• @user56202 You could see the region right from the start! The bounds do not correspond to the region, although it gives you the correct value for the volume. Nov 9, 2021 at 15:43

The "correct" answer does not seem to be correct, at least not according to the change of variable you present ($$\phi = 0$$ for points in the $$z$$ axis). It should be:

$$\{(\theta, \phi, \rho): 0\leq \theta \leq \pi, 0 \leq \phi \leq \frac{\pi}{2}, 0 \leq \rho \leq \frac{3}{\cos \phi+ \sin \phi} \},$$

as you mention. So, your solution is correct.

• Thanks. But why is $\phi$ bounded below by $\pi/4$? Our solid contains points on the $z$ axis, which have a $\phi$ of $0$. Nov 9, 2021 at 15:33
• @user56202 My bad! I'll correct! Nov 9, 2021 at 15:34