the region between the graphs of y=x^2 and y=2x is rotated around the line x =2. what is the volume of the resulting solid?

Answer 1

Determine bounds

x²=2x

x²-2x=0

x(x-2)=0

x=0 and x=2

[a,b] = [0,2]

Determine height and radius functions

Height = h(x) = 2x-x² (distance between y=2x and y=x²)

Radius = r(x) = 2-x (distance between x=2 and the y-axis)

Set up integral and evaluate

A = 2π∫_a^b r(x)h(x)dx <-- Shell/Cylindrical Method about a vertical line

A = 2π∫₀² (2-x)(2x-x²)dx

A = 2π∫₀² (4x-2x²-2x²+x³)dx

A = 2π∫₀² (4x-4x²+x³)dx

A = 2π[2x²-4x³/3+x⁴/4] [0,2]

A = 2π[2(2)²-4(2)³/3+2⁴/4]

A = 2π[2(4)-4(8)/3+16/4]

A = 2π[8-32/3+4]

A = 2π[12-32/3]

A = 2π[36/3-32/3]

A = 2π[4/3]

A = 8π/3 cubic units

Hope this helped!

Answer 2

x² = 2x; x = 0 or x = 2. By shell method v = 2π∫₀²(2 - x)(2x - x²)dx = 2π∫₀²(4x - 4x² + x³)dx =

2π(2x² - 4x³/3 + x⁴/4)₀² = 2π(8 - 32/3 + 4) = 8π/3