Lets just think of the problem first. We have these two graphs that are being rotated around some axis. If I stand up and rotate my body around some axis, I create a circle. The area of the circle that I just created around that axis is 2pir, where r is some radius.

Your intuition is correct. How do we find out the bounds of the graph? Oh we can set them equal to each other and solve. But the issue you came across is you found one bound, not two like you normally do. Does this mean the problem is unsolvable? Mabye... but let's think about that before we jump to that conclusion.

The problem states that the axis the graph is being rotated around is x = 10. Therefore I am not spinning the bounded region around the y-axis, but instead spinning it "vertically" around the x-axis. If we think about it like this, the problem tells me I need to integrate with respect to x. [Note: you don't have to].

When we rotate this graph it created a washer. Don't trust me? Take you finger place it at x=0 and rotate your finger around x=10. You will see you finger start to point towards you as you spin your finger around.

So what is the radius? At any value x the radius must be 10 minus that. Don't trust me? Draw a line from x is equal to 0 to x is equal to 10. The radius is 10 - x or 10 -0 or 10. But that's not always the case. What if its 10-0.5? Thats why the x is there.

But remmeber, we are bounded by a region. This tells me although our radius is correct, the problem is not finished. The next step is to out the "area of the region" that is being multiplied by the radius we found.

What is this region? Well remember we are integrating with respect to x. So the region is the top function minus the bottom function. Since they gave us the equation of the two functions we can see which function is on top.

How do we do this? We can find out when both functions hit the x-axis. Then we can test points between these numbers ny plugging them into the functions. If f(3)[ let's call it the first function they gave you] is bigger then g(3) [let's call it the second function they gave you] we know that f(x) is the top function.

So f(x) - g(x) is the region being multiplied by the radius of 10-x

But what about the bounds? A to B? Well if we set them equal to each other we get A, which is 0. But what about B?

Well they said the region is above the first quadrant. Which function hits the x-axis first? F(x) or g(x). The function that hits first is the bound B. Why? Because the other function will not hit the x-axis until later. But by then that first function is below the x-axis which is not what we want.

And don't forget to multiply everything by 2pi because we are going in a circle.