How do you find the volume of the solid y=-x+1 revolved about the x-axis?

1 Answer
Mar 29, 2018

If we restrict (0<x<1), we will obtain a cone by revolving

y=-x+1

about the x-axis, and its volume will be pi/3

Explanation:

If you were to revolve the entire function y=-x+1 about the x-axis, you would get a solid of infinite volume.

Let's instead choose to consider only the function on the interval

(0<x<1)

(the solid of revolution will be a cone)

Picture the integral by chopping the cone into a series of infinitely thin vertical slices.

Each slice is then a cylinder with its radius equal to y and thickness equal to dx

So the volume of one slice is

dV=piy^2dx

and the volume of the entire solid is the sum (integral) of all slices

int dV=int piy^2dx

V=piint y^2dx

In this problem, y=-x+1, so we write

V=piint (-x+1)^2dx

Now let's find the definite integral from 0 to 1.

piint_0^1(-x+1)^2dx

rArrpiint_0^1(x^2-2x+1)dx

rArrpi[x^3/3-x^2+x]_0^1

rArrpi[(1/3-1+1)-(0-0+0)]=pi/3

So the volume of the cone obtained by revolving

y=-x+1 for (0<x<1)

about the x-axis is

pi/3

Note: This agrees with the formula for the volume of a cone

V=pir^2h/3

Our cone has radius of 1 and a height of 1, so

V=pi(1^2)1/3=pi/3