Volume of a Revolution

The exercise is:
"Find the volume of the solid obtained by rotating the region bounded by the given curves about the given axis."

y = x^2 , y = 4 , x = 0 , x = 2; about the y-axis

I took the inverse of the first equation, got x = sqrt(y).
Then, I figured the bounds would be at y = 0 (since that's the minimum of +sqrt(y) and -sqrt(y) ) and y = 4 (since that boundary was given).
I set up the integral: pi * the integral of sqrt(y)^2 dy from 0 to 4
I then solved it to get 4pi.

However, the book's answer is 8pi.

I set up the integral again to include the inside term as ( 2 - sqrt(y) )^2 .. Since I figured maybe I should use x = 2 as the right boundary for some reason. I got 9 1/3 pi as the answer. Still wrong.

These seem so easy to do, and I can follow the examples in the book.. yet I get them all wrong!! Can anyone please help??

Your setup is perfectly fine, but you miscalculated your integral:
V = π ∫ (√y)^2 dy (from y=0 to 4).

Since (√y)^2 = y, this becomes:
V = π ∫ (√y)^2 dy (from y=0 to 4).
= π ∫ y dy (from y=0 to 4)
= (π/2)y^2 (evaluated from y=0 to 4)
= (π/2)(16 - 0)
= 8π, which now matches with the book.

I hope this helps!

The axis of rotation is vertical, so the disk method requires us to slice horizontally, implying dy.

y = x^2 ==> x = ± sqrt(y)

However, we know that it's the positive square root, because we care about the first quadrant. We were given the limits in the question by the defining lines.

V = ∫ A(y) dy from 0 to 4

A(y) = πR^2 - πr^2 = π(R^2 - r^2)

First determine the outer radius, R. The outer radius is defined by the curve, so R = sqrt(y). The inner radius, r, is defined by the line x = 0, so r = 0. Sub those in:

A(y) = π([sqrt(y)]^2 - [0]^2)
A(y) = πy

V = ∫ πy dy from 0 to 4
V = (1/2)π y^2 from 0 to 4

Now evaluate and note that the lower limit of 0 will be 0.

V = (1/2)π (4)^2
V = 8π

Done!