Volume of Rotations

Question

Calculate the volume of a solid generate by the area bounded by y=2x, y=6-x and y=0 rotated about the line x=8.

Here is what I have. I am not sure if this is correct

as y goes from 0 to 4

π ∫ R²-r² dy

π ∫ (8-y/2)² - (8-(6-y))² dy

I got 402.12 Is that correct?

Answer 1

Your answer for the Volume of revolution (V_R) is 128∏ = 402.123

The x coordinate for the center of area = xbar = ∫₀⁴∫_(y/2)^(6-y) x dxdy/Area

The Area of the triangle is (1/2)(6)(4) = 12

then xbar = 2.6666... then the distance from the line x=8 to xbar is 8 - 2.66666 = 5.33333. This is the radius around which the area is rotated to become a volume.

So, V_R = 2∏(8-xbar)(Area) = 2(3.14159)(5.333333)(12) = 128∏ = 402.123