# Volume by integration

In summary, the homework statement is asking if it is possible to find the volume of a revolution about the x-axis between y=1 and y=2 using the washer method. It seems that it is possible to do so, but the problem is that the integral is not split up into two parts and the inner radius is not constant.f

## Homework Statement

is it possible to find the of revolution about x-axis volume of x=y^2+1 ,x=0 between y=1 and y=2
using washer method?

## Homework Equations

V=pi integral from a to b R^2-r^2 thickness

## The Attempt at a Solution

the problem is when I made the integral i had two inner radii first the line x =1 till the point x=1 because the parabola only starts there so I am confused do I just sum two integrals?
Edit I meant the first radius is line y =1 not x =1 sorry

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I don't remember the washer method exactly, but since this area goes right up to the y-axis, isn't the inner radius zero? In other words, can't you just integrate over *cylinders* of infinitesimal thickness dy and radius x(y) from y=1 to y=2?

but I am rotating around the x-axis not the y one and it is from y=1 to y=2

also the question doesn't want the volume itself it just wants to know which method is possible to use and to setup the integral for them.

The "washer" method is exactly the same doing the disc method, using the two boundaries as radii and then subtracting. It looks to me like it would be simplest to break this into to parts.

First, y= 1 and $x= y^2+ 1$ intersect at x= 2 so for x from 0 to 2, you just have the difference of two cylinders, of raidii 1 and 2, and height 2. That would be the same as
$$\pi\int_{x=0}^2 (4-1)dx$$

y= 2 and $x= y^2+ 1$ intersect at x= 5 so from x= 2 to x= 5, you have an upper bound of y= 2 and a lower bound of $y= \sqrt{x- 1}$. The volume of that, rotated around the x-axis will be
[tex]\pi\int_{x=2}^5 (4- (x-1)) dx= \pi\int_{x=2}^5 4 dx- \pi\int_{x=2}^5 (x-1)dx[/itex]

Oh. Okay, so maybe you do have to split it up into two integrals. From x = 0 to the value of x for which y = 1 intersects x(y) (i.e. x(1)) , the inner radius is constant at 1, and the outer radius is constant at 2. From x = x(1) to x = x(2), the inner radius is given by x(y) and the outer radius is still constant and equal to 2. Do you agree?

EDIT: Beaten by HallsofIvy!

oh I see thanks I never got a problem when we had to add volumes before so it kind of confused me .