Question

The region is bounded by the given curves `y=x, y=4-x, 0<=x<=2` is rotated about the x-axis, how do you find the volume of the two solids of revolution?

Answer 1

See below.

Explanation:

I am reading this as the volume of the shaded area A rotated around the x axis, and the volume of shaded area B rotated around the x axis. First we find the volume of B, then we find the volume of A+B and subtract volume B from this to find the volume A

Volume of B

`V=pi int_(0)^(2)(x)^2 dx=1/3x^3`

`V=pi[1/3x^3]_(0)^(2)=[1/3(2)^3]-[1/3(0)^3]=(8pi)/3`

Volume of A + B

`(4-x)^2=16-8x+x^2`

`V=pi int_(0)^(2)(16-8x+x^2) dx=16x-4x^2+1/3x^3`

`->=[16x-4x^2+1/3x^3]_(0)^(2)`

`=[16(2)-4(2)^2+1/3(2)^3]-[16(0)-4(0)^2+1/3(0)^3]`

`V=pi[56/3]=(56pi)/3`

Volume of A = (A + B) - B.

`V=(56pi)/3-(8pi)/3=16pi`

So:

Volume of A = `color(blue)(16pi)` cubic units.

Volume of B = `color(blue)((8pi)/3)` cubic units.

Revolution of A:

Revolution of B: