find volume of solid obtained by rotating the region bounded by the y=x^2+1, and y=9-X^2,rotate about y=-1?

Question

i appreciate help^^ i worked on this problem but cant seem to get the right answer ><

Answer 1

There's a few tricky things in this one, I'll point them out for you and then you'll be able to get the right answer.
You have two parabolas, one facing up and one facing down, and if you solve the system you find that the limits of integration are from x = -2 to x = 2. These limits may give a problem cancelling each other out, just in case they do, let's integrate from x = 0 to x = 2, to get half of the solid, and then double the result. We can do this because the figure is symetric to the y axis. Since there's a hole in the middle of this figure, we'll use the washer method, which states that the volume is pi integral (limits) [R(x)]^2 - [r(x)]^2 dx
Now, R(x) is the outer radius, and since you're rotating it around the line y = -1 rather than the x axis, R(x) = 9 -(-1) - x^2 = 10 - x^2. Also, r(x) = x^2 + 1 - (-1) = x^2 + 2 So you have
2 pi integral 0 to 2 [(10-x^2)^2 - (x^2 + 2)^2] dx
Multiply that out, integrate it between 0 and 2, and double the result, and that oughta do it.