Calculus Integral - volume of the solid obtained by rotating R around the line?

Question

The region R is bounded by the graphs of x-2y and x=y^2 . Set up (but do not evaluate) the integral that gives the volume of the solid obtained by rotating R around the line x = -1.

Answer 1

x=2y and x=y^2 intersect at (0,0) and (4,2). Imagine using cylindrical shells rotating around the line x=-1. The height of the cylinders will be the difference of the y's: sqrt(x)-x/2. The radii of the shells will be x-(-1)=x+1.

Answer 2

This is tough to answer just through internet, but I will give you the guidelines in order to solve it.

1. Determine the points of intersection between x=2y and x=y^2. They should be y=0 and y=2.

2. Build a cylinder slice at coordinate y of thickness Dy that goes to x=2y (the outer boundary of region R) and has x=-1 as its axis of rotation and symmetry ; the cylinder will be hollow at its center.

3. The volume of the slice will be DV = A * Dy, where A is the surface area of the slice.

4. A = PI*(1+2y)^2 - PI*(1+y^2)^2: this is the area of the hollow slice.

5. Then the volume V is approximately = SUM DV = SUM ( A * Dy ). When the thickness Dy goes towards zero, then: Dy = dy and the SUM becomes an integral:

V is exactly: V = INTEGRAL (A * dy). The limits of the integral should be y=0, y=2. The numerical result is:

V = 6.93. You should check the numerical calc. as I did it really rapidly.

Hope this helps.

Answer 3

Try doing your own homework, especially when it's impossible to illustrate an explanation like that on a site like this...