Volume under sphere with radius 1, center (0, 0, 1) and above the cone $z = \sqrt{x^2+y^2}$

Question

so I want to find the volume of the body D defined as the region under a sphere with radius 1 with the center in (0, 0, 1) and above the cone given by $z = \sqrt{x^2+y^2}$. The answer should be $\pi$. A hint is included that you should use spherical coordinates. I've started by making a equation for the sphere, $x^2+y^2+(z-1)^2=1$. I used the transformation $(x, y, z) = (\rho\sin\phi\cos\theta, \rho\sin\phi\sin\theta, \rho\cos\phi+1)$. Now I'm struggling on defining the region D. I got that $0\leq\theta\leq2\pi$, but I can't find the bounds for $\phi$ and $\rho$.

Answer 1

If you try spherical coordinates, the following happens. For the cone, you have $$\rho\cos\phi=z=\sqrt{x^2+y^2}=\sqrt{\rho^2\sin^2\phi\cos^2\theta+\rho^2\sin^2\phi\sin^2\theta}=\rho\sin\phi.$$(this assume that you chose $\phi$ so that $\sin\phi\geq0$, that is $0\leq\phi\leq\pi$). So $\cos\phi=\sin\phi$; it follows that the equation of cone is $\phi=\pi/4$.

So, to describe the interior of your region, you will have $0\leq\phi\leq\pi/4$; simple enough. From the equation of the sphere we get $$ \rho^2=2\rho\cos\phi, $$ or $\rho=2\cos\phi$.

The volume is then \begin{align} V&=\iiint_E1\,dV=\int_0^{2\pi}\int_0^{\pi/4}\int_0^{2\cos\phi}\rho^2\sin\phi\,d\rho\,d\phi\,d\theta\\ \ \\ &=\frac{16\pi}3\,\int_0^{\pi/4}\cos^3\phi\,\sin\phi\,d\phi =-\frac{16\pi}3\,\left.\left(\frac{\cos^4\phi}4 \right)\right|_0^{\pi/4}\\ \ \\ &=\frac{4\pi}3\left(1-\frac1{4} \right)=\pi. \end{align}

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On the other hand, to work in cylindrical coordinates we have as follows. If we write $s=\sqrt{x^2+y^2}$, the intersection of the sphere and the cone happens when $s^2+(s-1)^2=1$, with solutions $s=0$ and $s=1$. As we want to be above the cone, we have to choose $s=1$. The equations of the sphere and the cone are, respectively, $z=1+\sqrt{1-r^2}$ and $z=r$. Then the volume is \begin{align} V&=\int_0^{2\pi}\int_0^1\,(1+\sqrt{1-r^2}-r)\,r\,dr\,d\theta =2\pi\,\int_0^1(r+r\sqrt{1-r^2}-r^2)\,dr\\ \ \\ &=2\pi\,\left.\left(\frac{r^2}2-\frac{\sqrt{1-r^2}}3-\frac{r^3}3\right)\right|_0^1 =2\pi\,\left(\frac12-0-\frac13+\frac13 \right)=\pi. \end{align}

Answer 2

In spherical coordinates: $$ E=\{(\rho,\theta,\phi)|0\le \theta\le 2\pi, 0\le \phi \le \pi/4, 0\le \rho \le 2\cos \phi \} $$ It follows that

$$ \boxed{ V= \iiint_E dV = \int_0^{2\pi}\int_0^{\pi/4}\int_0^{2\cos\phi}\rho^2\sin\phi \;d\rho d\phi d\theta = \pi } $$

Note: to find your bounds for $\phi$ and $\rho$:

• The upper bound for $\phi$ is precisely the cone $z=\sqrt{x^2+y^2}\; \Leftrightarrow \; \rho \cos \phi = \rho \sin \phi \; \Leftrightarrow \; \phi = \pi/4$
• The upper bound for $\rho$ is precisely the sphere $x^2+y^2+(z-1)^2=1\; \Leftrightarrow \;\rho^2\sin^2\phi+(\rho \cos\phi-1)^2=1\; \Leftrightarrow \;\rho=2\cos\phi$