R is the region bounded above by the graph of f(x)=2e^−x^2 and below by the x-axis over the interval [1,2]. Find the volume of the solid of revolution formed by revolving R around the y-axis.

Answer 1

using shells of thickness dx,
v = ∫[1,2] 2πrh dx
where r = x and h = y
v = ∫[1,2] 2πx*2e^(-x^2) dx = 2π(e^3-1)*e^-4
do this by letting u = x^2 and you have ∫[1,4] 2πe^-u du

Answer 2

or, using discs (washers) of thickness dy,
v = ∫[2/e^4,2/e] π(R^2-r^2) dy
where r = 1 and R^2 = ln(2/y)
v = ∫[2/e^4,2/e] π*(ln(2/y) - 1) dy = 2π(e^3-1)*e^-4