multivariable-calculus
Write a triple integral in cylindrical coordinates for the volume of the solid cut from a ball of radius 2 by a cylinder of radius 1, one of whose rulings is a diameter of the ball.
I am unable to understand how to get $z=f(r,\theta)$ and the limits.
Any help would be appreciated
Here, I evaluate the forth part of the region so after evaluating the triple integral you should multiply the result to $4$. According to Fig below, we have a rectangle triangular $OCD$. So $\sin(\phi_0)=1/2$ and so $\phi_0=\pi/6$. It means that $\phi\in [\pi/6,\pi/2]$. As you found out the rang of $\theta$ is $[0,\pi/2]$. But about the $\rho$:
You see that our region is starting to shape from the intersection till we meet $z=0$. For finding the range of $\rho$, make an arrow arbitrary to cut the cylinder at $A$ and next cut the sphere at $B$. What is $\rho$ at $A$ and what is at $B$? We know that in a spherical coordinates: $$x=\rho\sin(\phi)\cos(\theta)\\y=\rho\sin(\phi)\sin(\theta)\\z=\rho\cos(\phi)$$ then the equation of our cylinder $x^2+y^2=1$ be converted to $\rho^2\sin^2(\phi)=1$ or $\rho=\frac{1}{\sin(\phi)}$. This is the first coordinate $\rho$ at $A$. Clearly, $\rho$ at $B$ is $2$. Hence you have: $$\int_0^{\pi/2}d\theta\int_{\pi/6}^{\pi/2}\sin(\phi)d\phi\int_{\frac{1}{\sin(\phi)}}^{2}\rho^2d\rho$$.
You know the equation of such part of the sphere is $$z^2=4-(x^2+y^2),x\in[0..2],y\in[0..2]$$ But $r^2=x^2+y^2$ and then $z=\sqrt{4-r^2}$. The ranges of our new variables are : $$\theta|_0^{\pi/2}, r|_0^2, z|_0^{\sqrt{4-r^2}}$$ So we have to evaluate $$\int_0^{\pi/2}\int_0^2\int_0^{\sqrt{4-r^2}}dv$$
Best Answer
Hint:
The better way to solve this problems to use cylindrical coordinates.
The sphere of center $(0,0,0)$ and radius $2$ has equation: $r^2+z^2=4$ and, without loss of generality, we can assume that cylinder has the axis parallel to the $z$ axis , and center at $(0,1,0)$ and radius $1$ so it has equation: $r=2\sin \theta$. this means that the limits of the volume limited by the sphere and the cylinder are: $$ 0\le \theta \le 2\pi \qquad 0\le r \le 2 \sin \theta \qquad-\sqrt{4-r^2}\le z \le \sqrt{4-r^2} $$
so, using the symmetries of the figure, the volume is: $$ V=4 \int_0^{\frac{\pi}{2}}\int_0^{2\sin \theta}\int_0^{\sqrt{4-r^2}}rdzdrd\theta $$ where $rdzdrd\theta$ is the volume element in cylindrical coordinates.
can you do from this?