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

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Answer 1 · 2018-03-29T08:18:54

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#

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#