Find the line that divides a bounded 2d region into 2 equal areas. A region is bounded by the following equations:
$$y= x^2 -1$$
$$y=0$$
$$x=2$$
Find a value for $a$ such that the line $x=a$ divides this region into two equal areas.
Find a value for $b$ such that the line $x=b$ divides this region into two parts such that the parts will have equal volume when rotated around the y-axis. It is possible that $a=b$, but that is for you to decide.

A visual I drew of the situation described: https://imgur.com/a/vYqMK
I solved part A and got an answer of $\sqrt{3}$, but stuck on part B.  
 A: In the first part,  your region of interest looks like: 

what you may have done was set up an integral like:
$$
\int_1^a (x^2 -1) dx = \int_a^2 (x^2 - 1) dx
$$
 and solved for $a$. As a note, I quickly computed this and did not get $a =\sqrt{3}$. Double check your work, but it is possible I made a mistake. 
In the second part, we have revolved this region around the y axis to get something that looks like:

We want to draw another vertical line such that when we rotate it around, the two volumes it cuts out are equal. It might be instructive to first ask how we would find the volume of the solid in Figure 2. One way would be the so-called "shell method", whereby we find the volume by summing up the surface area of many thin cylinders. The surface area of a cylinder is $A = 2\pi \text{(radius)(height)}$. So if we wanted to find the volume of the above solid, we might set up the integral:
$$
V = \int_1^2 2 \pi \text{(area)(height)} dx = \int_1^2 2 \pi (x) (x^2-1) dx
$$
Once we have this, we can slightly adjust it (as Doug M. suggests in the comments on your post) and say that we want
$$
\int_1^b 2 \pi (x) (x^2-1) dx = \int_b^2 2 \pi (x) (x^2-1) dx
$$
Now it is a matter of solving for b.
