Challenge of the Day!?

Question

Didn't give many challenges. because school work is too stressful, but now here's another one.

Find the volume of x = e^y - 1 after it is revolved around the y-axis (The volume between y = 0 and y = 3) .

Answer 1

V = int(pi*x(y)^2*dy) from 0 to 3

V = pi*int[(e^y - 1)^2] from 0 to 3

V = pi*int[e^y*(e^y - 1) - (e^y-1)] from 0 to 3

V = pi*[1/2*(e^y-1)^2 - (e^y - y)] from 0 to 3

V = pi*[ 1/2*(e^3-1)^2 - (e^3 - 3)] - pi*[1/2*(e^0 - 1)^2 - (e^0 - 0)]

V = pi*[ 1/2*(e^3-1)^2 - (e^3 - 3)] + pi= 521.6405

Answer 2

The curve divides the first quadrant bounded by (0,0) and (e^3-1,3) into two areas: A, the area below the curve and A' the area above the curve. But A' is

A' = integral(xdy) from y = 0 to y=3
= (e^y -y) y=0,3
= e^3-4
the volume is 2 pi * A' = 2pi(e^3-4)

Answer 3

integral(pi*x(y)^2*dy,y,0,3) = 521.640483