Volume of two spheres

Question

We have two spheres.

Sphere 1: $x^2 + y^2 + z^2 = 1$ (right)

Sphere 2: $x^2 + y^2 + z^2 = 2y$ (left)

I have to calculate the volume, delimited on right by eq1 and left by eq2. What does the exercise mean by left and right? I have no textbook answer.

Answer 1

$\hskip 1.9 in$

Here is a plot of spheres. Let $S_1$ be the blue sphere and $S_2$ be red one, given: $$ \begin{align} S_1 :& x^2 + y^2 +z^2 = 1 \\ S_2 : & x^2 + y^2 +z^2 = 2y \end{align}$$

On solving $y = 1/2$ , which is the equation of blue plane. As you can see the system is symmetric about the plane, so if one can calculate the smaller red part $T$(which lies in the blue sphere), then the net volume of intersection would be just double of that. In other words, we have to find the volume of sphere $S_1$ for $1>y>1/2$.

$T$ can be subdivided into disks of radius, say $r$ with infinitesimal thickness $dy$ $$ \Rightarrow dV = \pi r^2 dy$$

If $y$ be the distance between centre of sphere and centre of disc,and $R$ be radius of $S_1$ then by pythogoreas: $$r^2 = R^2 - y^2$$ $$ \Rightarrow dV = \pi (R^2 - y^2) dy$$ $$ \Rightarrow V = \int_{1/2}^1 \pi (R^2 - y^2) dy $$

This is the volume of $T$, as stated earllier net volume is $2V$

Answer 2

Sphere 1 has the center at the origin. The equation of sphere 2 can be written as $$x^2+y^2+z^2=2y\\x^2+y^2-2y+z^2=0\\x^2+y^2-2y+1+z^2=1\\x^2+(y-1)^2+z^2=1$$ The last is the equation of the sphere with the center at $y=1$