Volume of Solid: Find Revolving Around x=-1

[science.girl]

Problem:
Find the volume of the solid generated by revolving about the line x = -1, the region bounded by the curves y = -x^2 +4x -3 and y =0.

Attempt at a Solution:
I know that subtracting y = 0 from y = -x^2 +4x -3 will give the area in 1 dimension. So, would you use shells? I'm not sure how to set this up.

Additionally, the solid is being rotated around x = -1, so it must be different than problems with solids revolving around x = 0, correct?

 

[HallsofIvy]

Problem:
Find the volume of the solid generated by revolving about the line x = -1, the region bounded by the curves y = -x^2 +4x -3 and y =0.

Attempt at a Solution:
I know that subtracting y = 0 from y = -x^2 +4x -3 will give the area in 1 dimension. So, would you use shells? I'm not sure how to set this up.

Since you are rotating around around x= -1 and are writing the formula in terms of x, yes, use "shells". "Subtracting y= 0 from y= -x^2+ 4x- 3" does NOT give area (its integral does). It give the height of the very thin cylinder forming the shell. What is the surface area? The volume of the thin cylinder is its surface area time its thickness, ds. Integrate theat.

Additionally, the solid is being rotated around x = -1, so it must be different than problems with solids revolving around x = 0, correct?

Only in that the radius of the cylindrical shells is x-(-1)= x+1, not x- 0= x.