Let f (x) = 6x and g(x) = x^2 + 3x, 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.
V = ∫ =
∫[g(x) - f(x)]dx
= ∫[x^2 + 3x - 6x]dx
= ∫[x^2 - 3x]dx
= ∫[x(x - 3)]dx
= 1/3x^3 - x^2 + C
V = 1/3(b^3 - a^3) - (b^2 - a^2)
where a and b are the lower and upper bounds of the region R.
using washers of thickness dx,
v = ∫[0,3] π(R^2-r^2) dx
where R = 6x and r = x^2+3x
v = ∫[0,3] π((6x)^2-(x^2+3x)^2) dx
= 729π/10
using shells of thickness dy,
v = ∫[0,18] 2πrh dy
where r = y and h = (√(y + 9/4) - 3/2)-(y/6)
v = ∫[0,18] 2πy((√(y + 9/4) - 3/2)-(y/6)) dy
= 729π/10