Find out the volume of the solid obtained by rotating the region bounded by x = (y - 2)^{2} and y = x around the line y = -1.
Solution : We have to first get the intersection points there.
y =( y - 2)^{2}
y = y ^{2 }- 4 y + 4
0 =y^{ 2} - 5 y + 4
0 = ( y - 4) ( y -1)
Therefore, the two curves will intersect at y = 1 & y = 4 . Following is a sketch of the bounded region and the solid.
Following are our sketches of a typical cylinder. The sketch on the left is here to illustrates some context for the sketch on the right.
Following is the cross sectional area for this cylinder.
A ( y ) = 2 ∏ ( radius ) ( width )
=2 ∏ ( y + 1) ( y - ( y - 2)^{2} )
= 2 ∏ (- y^{3} + 4 y ^{2} + y - 4)
The first cylinder will cut in the solid at y = 1 and the final cylinder will cut in at y = 4 . Then the volume is
V=∫^{d}_{c }A(y)dy
=2 ∏∫^{4}_{1 }-y^{3 }+ 4y^{2 }+ y- 4 dy
=2 ∏ (-(1/4) y^{4 }+ (4/3)y^{3}+(1/2)y^{2 }- 4y)|^{4}_{1}
=(63 ∏/2)