Find the volume of the solid that is obtained when the region bounded by the curves x=9-y^2, x=y^2+1 and y=0 is rotated about the line y=-2. (the upper shape above the x-axis)
the curves intersect at (5,2) and (5,-2)
So, the volume can be done using shells or discs.
Using shells, consider the volume as a bunch of nested shells, each of thickness dy, radius y+2 and height the distance between the curves: (9-y^2)-(y^2+1)
v = ∫[0,2] 2πrh dy
= 2π∫[0,2] (y+2)((9-y^2)-(y^2+1)) dy
= 176π/3
Using discs (washers), we have a bunch of flat circles with a hole in them, of thickness dx. But, the bounding curves change at x=5.
v = ∫[1,5] π(R^2-2^2) dx
+ ∫[5,9] π(R^2-2^2) dx
= π∫[1,5] (2+√(x-1))^2-4 dx
+ π∫[5,9] (2+√(9-x))^2-4 dx
= 88π/3 + 88π/3
= 176π/3