Let f (x) = 9x and g(x) = x2 + 6x, and let r be the bounded region between the graphs of f and g. set up an integral to find the volume v of the solid generated by revolving r about the x axis and then find v.

Find where they intersect
f(x)=g(x) at

ok
9x=x^2+6x
0=x^2-3x
intersect at x=0 and x=3
which one is on 'top'
f(2)=18
g(2)=16
f(x) is on top

so it is
(we don't need to find r)
remember, the volume rotated around  x axis when f(x) is top curve and g(x) is botttom and bounds are a and b is

so
f(x)=9x
g(x)=x^2+6x
the bounds are 0 and 3
=v
=v
=v

the volume is 1344.6 cubic units