use the method of cylindrical shells to find the volume generated by rotation the region bounded by x=y^2+1, x=2, about y=-2. Sketch the region, the solid, and a typical shell.

Question

I know that for shells, its 2pix*f(x)*dx. i am having trouble figuring out what the "x" is for the circumference. I believe f(x), which is the height, is 2-(y^2+1), and dx is the thickness. I sketched the region already, i just need help with setting up the integral.

Answer 1

As always, draw a diagram. The parabola has vertex at (1,0) and opens to the right. The line x=2 intersects the parabola at (2,1) and (2,-1).

So, you will be revolving the sort-of-semicircular chink around the line y=-2, which is below the curve.

So, as you say,

v = ∫2πrh dy

because the shells have a horizontal axis, and thickness dy. I can understand your confusion if trying to use dx. See below.

r = y+2 because the axis of revolution is not the x-axis, but 2 units down.

h = 2-x because we're rotating an area cut off by x=2.

v = ∫2π(y+2)(2-(y^2+1)) dy
I'll let you figure the limits of integration.

If you want to use dx, you need to use discs (washers) of thickness dx. For these,

v = ∫π(R^2-r^2) dx
where R = √(x-1) and r = -√(x-1)

Answer 2

I worked it out and found the answer to be 454pi/12. this is correct right? The limits of integration i used were from 1 to 2.