Question

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 = ∫ =

Answer 1

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

Answer 2

∫[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.