# Let R be a region bounded by y=x^2, y=2x, and y=3. Solid S is obtained by rotating region R about x=-1.?

1. Write the integral for the volume of S using the washer method.(no need to compute it).

2. Find the volume of S using the shell method.

Can someone help me with this problem please? So for the first part of the problem, I am getting V = (pi) * integral(sqrt(y) - (-1)) dy over [0,3]. I'm not sure if I did it right.

For the second part, I got the radius is x+1 and the height is 3.

Then V = (2pi) * integral(3*(x+1))dx (I'm not sure what the bounds are).

I am having trouble with this problem. Please help me out. Thank you.

The region R is the region bounded below the line y=3.

### 2 Answers

- PopeLv 71 month ago
Have you tried sketching the three boundaries? They bound two finite regions and several infinite regions. As it is, region R is not well defined, so we cannot even get a good start on the problem.

The limits on your first integral suggest that you may be using the lower of the two finite regions, but neither of the integrals is correct, or anywhere close.

Someone might be willing to provide guidance here, but I would not touch it until the problem is first clearly stated.

- RockItLv 71 month ago
Important, using washers, you will integrate with respect to dy, whose limits of integration are y=0 and y=3. So, your inner and outer radius for the "washer" which are given in terms of x in your problem statement need to be expressed in terms of y instead.

You need to visualize this. The area between dy and inner and outer radius which form a bounding "rectangle". Thats what you are summing up.