Let R be the region in the first quadrant bounded by the x-axis and the curve y=2x-x^2. The volume produced when R is revolved about the x-axis is

Question

Let R be the region in the first quadrant bounded by the x-axis and the curve y=2x-x^2. The volume produced when R is revolved about the x-axis is

Answer 1

I'm going to assume you mean the bounded part of the area etween the curve 2x-x² and the x-axis, the bounded piece lies between x=0 and x=2. The volume upon revolution can be split into small cylinders each of which has for radius the height of the curve and for width dx, the differential volume of one cylinder is:

dV = dx π * (2x - x²)² = π (4x² - 4x³ + x⁴) dx

the volume of the figure is the sum of all the differential volumes:

V = π ( 4/3 x³ - x⁴ + x⁵/5) |₀² = π ( 32/3 - 32 + 64/5) = 16π/15