Find the volume bounded by the $xy$ plane, cylinder $x^2 + y^2 = 1$ and sphere $x^2 + y^2 +z^2 = 4$

Question

Find the volume bounded by the $xy$ plane, cylinder $x^2 + y^2 = 1$ and sphere $x^2 + y^2 +z^2 = 4$.

I am struggling with setting up the bounds of integration.
First, I will calculate the 'first-quadrant' piece of the volume.
$z$ will traverse from $0$ to $2$.
$x$ should start from the cylinder and go to the edge of the current circle of the sphere: $$\sqrt{1-y^2} \le x \le \sqrt{4-x^2-z^2}$$ However, the same applies to $y$: (I am only calculating half of the volume right now, where the smaller circle is the lower bound):
$$\sqrt{1-x^2} \le y \le \sqrt{4-y^2-z^2}$$ However, this cannot work as both $x$ and $y$ are dependent.
What is the error?

Answer 1

Let $f(x, y) = \sqrt{4-(x^2 + y^2)}$ be the function which for each point closer than $4$ to the origin of the $xy$-plane gives the height of the top half of $x^2 + y^2 + z^2 = 4$ over that point. Then what you want is the integral of this function, over the unit disc.

Expressed in polar coordinates, this is given by $$ \int_0^{2\pi}d\theta\int_0^1r\sqrt{4-r^2}dr $$

Answer 2

There are several ways to calculate the triple integral $$ V=\iiint_B\,dxdydz $$ where $B$ is a tower (cylinder) with a roof (sphere).

One way as you do is to split the integration as $$ V=\int_0^2\Big(\iint_{B_z}\,dxdy\Big)dz. $$ Here $B_z$ is the intersection of the body $B$ with the plane of constant $z$ (horizontal slices). It is troublesome as the body should be split into two parts (one for the tower and one for the roof). It is more convenient here to do the vertical splitting $$ V=\iint_{x^2+y^2\le 1}\Big(\int_0^{\text{top of the sphere}}dz\Big)dxdy. $$