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!
Best Answer
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$