Find the volume formed by rotating the region enclosed by: y = 0.9x and y^3 = x with y ≥ 0 about the x -axis

Answer 1

Begin by rewriting y^3 = x as y = x^(1/3)

Then we can calculate where the two curves meet by setting the two equations equal to each other:

0.9x = x^(1/3)

0.9x^(2/3) = 1

x^(2/3) = 10/9

x = 1.17

The volume is given by ∫ (π y₂² - π y₁²) dx

y₂ is the curve on top and y₁ is the curve on the bottom: V = ∫ π (x^2/3 - 0.81x²) dx

Integrate this from x = 0 to x = 1.17 (the boundaries of the region in the first quadrant)