Question

y = x^2, y = 2 − x^2; axis: y-axis
I'm finding the volume of revolution along the y-axis. How do you do this? I remember hearing that the integral had to be split up into 2 integrals in order to solve for the volume, but I'm not sure how to apply that with this problem.

Answer 1

Integrating with respect to y is always annoying, let me give it a try. just a sec

Answer 2

But for splitting just see where the curves intersect. That makes your limits

Answer 3

Answer is just \(\pi\).

Here is my work:

Integrating with respect to y:
\(\pi\int_{0}^{1}\left(\sqrt{y}\right)^{2}dy+\pi\int_{1}^{2}\left(\sqrt{-y+2}\right)^{2}dy\)

Inverse integration:
\(\pi\int_{0}^{1}\left(\sqrt{x}\right)^{2}dx+\pi\int_{1}^{2}\left(\sqrt{-x+2}\right)^{2}dx\)

Answer 4

We don't need to use washer's method even though there is a region bounded by curves. The formula for vertical disc's method is

\(V=\pi\int_{c}^{d}\left[R\left(y\right)\right]^{2}dy\)

So yes, you do need to split the integrals up here. This is because if you take a look at it graphically, the curves split into their respective area.

It's hard to figure that out without a graphical representation, so I went on Desmos to graph it out and see.

https://www.desmos.com/calculator/gr4ztydkw8

Answer 5

It's a lot like finding the area under the curve like you probably did a few lessons ago, but now you're realizing that volume is the antiderivative of area. So you're applying it to actual volume formulas.

The idea that \(V=\int_{ }^{ }A\) helps out a lot.

Answer 6

and a quick note about integrating with respect to y:
-It's just like integrating with respect to x, but your riemann height is now dy and your integrand is the same function but you're first solving for x and putting the other part in - you are not taking an inverse, just rearranging.
-a helpful trick is to use inverse integration and take the inverse of the function and integrate with respect to x.

Your limits if needed are the same between inverse and normal integration because of the definition of inverse functions