The bounds of our interval will be
<=x<=
Thus the volume of the solid obtained when the region bounded by
y=(1)/(2)x^(2)
and
y=2x
is rotated about the line
x=4
is
\int_0^4 2\pi (4-x)(2x-(1)/(2)x^(2))dx=
Try out the following questions, and if you get stuck, check out the videos below. Consider the region bounded by
y=\sqrt(x-4),y=2,x=0
, and
y=0
. Use the shell method to find the volume of the solid obtained when this region is rotated about the
x_(x)
-axis. Consider the region bounded by
y=\root(4)(x)
and
y=\root(3)((x)/(2))
. Use the shell method to find the volume of the solid obtained when this region is rotated about
y=-3
.