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 $\rho$.
[Math] Volume under sphere with radius 1, center (0, 0, 1) and above the cone $z = \sqrt{x^2+y^2}$
integrationmultivariable-calculusspherical coordinates
Related Solutions
First of all, the Jacobian of the spherical coordinat transformation is $\rho^2\sin\phi$ not $\rho\sin\phi$.
Also, the upper bound for $\rho$ should have been $6\cos\phi$ not $3$. The equation $\rho = 3$ describes a sphere of radius $3$ centered at the origin while the equation $\rho = 6\cos\phi$ describes a sphere of radius $3$ centered at $(x,y,z) = (0,0,3)$.
After fixing those two errors, you should get an integral which evaluates to the correct answer.
I'm adding this answer because the other answers so far answer different questions (either a different region, or a different method) than what was originally asked. I'm also going to try to explain my work in more depth to make it easier to follow along.
Note that the cone $z = \sqrt{(x^2 + y^2)}$ can be represented in spherical coordinates as $\phi = \frac{\pi}{4}$. You already calculated that $z = r \sin \phi$, and we know from the spherical coordinate transform that $z = r \cos \phi$. Combining these two equations, we get $\sin \phi = \cos \phi$ or $tan \phi = 1$, which means $\phi = \frac{\pi}{4}$. Intuitively/geometrically, you can also imagine that the cone covers the full range from $\theta = 0$ to $\theta = 2 \pi$ of rotation around the z-axis, meaning $\theta$ can be any value, and the radius can be any value as well (the cone is unbounded above), so the only parameter needed to define the cone is $\phi$.
To set up the region of integration, we know that to calculate the volume of the sphere, we would have the radius ranging from 0 to 1, with $\theta$, the angle of rotation around the z-axis, ranging from 0 to $2 \pi$, and $\phi$, the angle with the z-axis, ranging from 0 to $\pi$. In this case however, we want to exclude the region of the cone, which is the region from $\phi = 0$ to $\phi = \frac{\pi}{4}$. Thus we want to include $\phi$ only over the range from $\frac{\pi}{4}$ to $\pi$.
Thus, we have the below integral:
$$\int_{0}^{1} \int_{0}^{2 \pi} \int_{\frac{\pi}{4}}^{\pi} r^2 sin(\phi) d\phi d\theta dr$$
where $r^2 sin(\phi)$ comes from the change of base formula, and can be calculated as the absolute value of the determinant of the derivative matrix of $x$, $y$, $z$ with respect to $r$, $\theta$, and $\phi$.
The integral can be evaluated in any order, probably $\theta$ then $r$ then $\phi$ to get $2 \pi \frac{1}{3} (1 + \frac{1}{\sqrt{2}}) \approx 3.58$.
A quick intuitive check confirms that this answer makes sense: the volume of the sphere is given by $\frac{4}{3}\pi r^3$ which in this case is just $\frac{4}{3}\pi \approx 4.19 $, and we're taking out a small fraction of the top half of the sphere.
Best Answer
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 assumes 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}
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}