ASAP in the volume of the solid generated when the region R is revolved about the line y=6.

Question

Find the volume of the solid generated when the region R is revolved about the line y=6.

Let the region R be the area enclosed by the function 

f(x)=x^1/3​+1

 and 

g(x)=1/3​x+1.

Find the volume of the solid generated when the region R is revolved about the line y=6.

 You may use a calculator and round to the nearest thousandth.

Answer 1

If you plot the two functions, there are two areas to be rotated around y = 6. The points of intersection are at x = -√27, 0, √27.

x^1/3+1 = x/3+1

x^2/3/3 = 1

x = 3^3/2 = ±√27

V₁ = π∫₀^√27(6-(x/3+1))²-(6-(x^1/3)+1))²dx

= π∫₀^√27(5-x/3)²-(5-x^1/3)²dx = 2π[x³/27-5x²/3-3x^5/3/5+15x^4/3/2]₀^√27= 57.6265

V₂= π ∫⁰_-√27(6-(x^1/3)+1))² - (6-(x/3+1))²dx

= π ∫⁰_-√27(5-x^1/3)^{2 -} (5-x/3)² dx = π[-x³/27+5x²/3+3x^5/3/5-15x^4/3/2]⁰_-√27 = 83.745

V₁+V₂ = 141.3715

Answer 2

A representative vertical strip lying in region R is perpendicular to the axis of rotation, and the region does not lie adjacent to the axis of rotation. Therefore, we will use washers: V = ∫_a^b [R_big² - R_sm²] dx

f and g intersect at x = 0 and again at x = 3√3, which are the bounds of integration.

R_big = 6 - g(x) while R_sm = 6 - f(x).