set the limits of integration of the spherical coordinates between two paraboloids and a plane

Question

Find the volume of the solid $\mathcal{S}$ enclosed laterally by the paraboloids $\mathcal{P}_1$ of equation $z = x^2 + y^2$ and $\mathcal{P}_2$ of equation $z = 3(x^2 + y^2)$ and from above by the plane $z = 1$ using spherical Coordinates. How one may set the limits of integration of the spherical coordinates?

Now, for a solid bounded below by paraboloid $z=a^2(x^2+y^2)$ for $a>0$, and above by plane $z=1$, the volume can be obtained and it is equal to $\dfrac{\pi}{2a^2}$. Using this result we can get the volume of the solid $\mathcal{S}$ and obtain that $$ V(\mathcal{S})=V_{\mathcal{P}_1}-V_{\mathcal{P}_2}=\dfrac{\pi}{2 \times 1}-\dfrac{\pi}{2\times 3}=\dfrac{\pi}{3}. $$ But, how one can obtain the volume of $S$ using spherical coordinates?

Answer 1

There are two components of the integral that must be considered:

1. When $\rho$ varies between the inner and outer paraboloid

2. When $\rho$ varies between the inner paraboloid and the plane $z=1$.

We consider these cases separately.

First, we transform the boundary surfaces into spherical coordinates. $z=1$ becomes $\rho \cos\phi = 1$, so $\rho = \sec \phi$. For $z = x^2 + y^2$, we have $\rho \cos \phi = \rho^2 \sin^2\phi$, so $\rho = \cot\phi \csc\phi$. Similarly, for $z = 3(x^2 + y^2)$, we have $\rho = \frac{1}{3}\cot\phi \csc\phi$. These will form the bounds on $\rho$ in our triple integrals.

We now need to find the bounds on $\phi$. The plane intersects the inner paraboloid at the point $(r,z) = (1/\sqrt{3},1)$, where $r^2 = x^2+y^2$. Recalling that $z = \rho \cos \phi$ and $r = \rho \sin \phi$, we then have $\phi = \tan^{-1}(1/\sqrt{3}) = \frac{\pi}{6}$. Using a similar calculation for the outer paraboloid, we find this intersection at $\phi = \frac{\pi}{4}$. This range corresponds to case (2) and case (1) is given by $\phi\in[\pi/4,\pi/2]$ where we vary from the plane intersection with the outer paraboloid all the way down to the $xy$-plane.

We then have

$$ V = \int_0^{2\pi}\int_{\frac{\pi}{6}}^{\frac{\pi}{4}}\int_{\frac{1}{3}\cot\phi \csc\phi}^{\sec \phi} \rho^2\sin\phi~\mathrm{d}\rho \mathrm{d}\phi \mathrm{d}\theta + \int_0^{2\pi}\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\int_{\frac{1}{3}\cot\phi \csc\phi}^{\cot\phi \csc\phi} \rho^2\sin\phi~\mathrm{d}\rho \mathrm{d}\phi \mathrm{d}\theta. $$

Answer 2

Let's firstly consider our figure only in first quadrant and at the end will multiply result on $4$.

Let me consider spherical coordinates $$\begin{cases}x=r\sin\phi\cos\theta & \\ y= r \sin\phi\sin\theta & \\ z=r\cos\theta \end{cases}$$ Accordingly we have $0\leqslant r < \infty$, $\phi \in [0,\pi]$ and $\theta \in (0, 2\pi]$.

Paraboloid $z=x^2+y^2$ gives $r=\frac{\cos\phi }{\sin^2\phi}$ and $z=3(x^2+y^2)$ gives $r=\frac{\cos\phi }{3\sin^2\phi}$. As expected they are not dependent on $\theta$, so, it takes all range in first quadrant $\left[0, \frac{\pi}{2} \right]$. Plane $z=1$, obviously become $r=\frac{1}{\cos\phi}$.

With respect to $\phi$ volume is divided in two parts: from $\frac{\pi}{6}$ up to $\frac{\pi}{4}$ where ray from origin firstly meet $z=3(x^2+y^2)$ and then plane $z=1$ i.e. we are between $r=\frac{\cos\phi }{3\sin^2\phi}$ and $r=\frac{1}{\cos\phi}$. And from $\frac{\pi}{4}$ up to $\frac{\pi}{2}$ volume is between $r=\frac{\cos\phi }{3\sin^2\phi}$ and $r=\frac{\cos\phi }{\sin^2\phi}$. So we obtain:

$$\int\limits_{0}^{\frac{\pi}{2}}d\theta\int\limits_{\frac{\pi}{6}}^{\frac{\pi}{4}}d\phi\int\limits_{\frac{\cos\phi }{3\sin^2\phi}}^{\frac{1}{\cos\phi}}r^2\sin\phi dr+\int\limits_{0}^{\frac{\pi}{2}}d\theta\int\limits_{\frac{\pi}{4}}^{\frac{\pi}{2}}d\phi\int\limits_{\frac{\cos\phi }{3\sin^2\phi}}^{\frac{\cos\phi }{\sin^2\phi}}r^2\sin\phi dr = \frac{\pi}{12}$$

multiplying it on $4$, as mentioned, gives $\frac{\pi}{3}$.

Let me say, that, perhaps, most natural way to calculate this volume is consider its projection, for example, on $OZY$ and consider body of rotation around an axis $OZ$.

(Having prepared my answer, I found someone else’s already given, but still decided to leave my own, firstly, so that the questioner knew that I did not leave it without attention and, secondly, the person who wished could choose the answer he liked more.)