calculus, find volume of the solid by rotating...?

Question

find volume of solid obtained by rotating the region bounded by the curves x=y-y^2 , x=0, rotate about the y-axis

Answer 1

First, you have to determine your bounds of integration. Remember that we're going to solve this using discs. We have to equate y - y^2 to 0, to solve for the vertical bounds of integration.

0 = y - y^2
0 = y(1 - y)
y = 0, 1
So our bounds of integration are 0 and 1.

Our volume is going to be

V = Integral (0 to 1, A(y)dy), where A(y) represents our cross section.

A(y) = pi*r^2

But we know our radius; it's X (because we want the distance from the y-axis to the graph, as we're rotating around the y-axis).

A(y) = pi*(x)^2

However, we're calculating A(y) here so we need to express x in terms of y. But x = y - y^2, so

A(y) = pi * (y - y^2)^2

Simplifying, this equals

A(y) = pi * (y^2 - y^3 - y^3 + y^4)

Therefore,

V = Integral (0 to 1, pi * (y^2 - y^3 - y^3 + y^4))

First, let's pull out the constant out of the integral.

V = pi * Integral (0 to 1, y^2 - 2y^3 + y^4)

And then solve as per the reverse power rule.

V = pi * [ (y^3 / 3) - 2(y^4 / 4) + (y^5 / 5) ] evaluated from 0 to 1.

V = pi * [1/3 - 2(1/4) + (1/5) - [0 - 0 + 0] ]
V = pi [1/3 - 1/2 + 1/5]
V = pi [10/30 - 15/30 + 6/30]
V = pi [1/30] = pi/30