# Volume by integration

## 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

Last edited:

cepheid
Staff Emeritus
Gold Member
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.

HallsofIvy
Homework Helper
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]

cepheid
Staff Emeritus