posted by Taeyeon .
2. Let R be the region of the first quadrant bounded by the x-axis and the cuve y=2X-X^2
a. Find the volume produced when R is revolved around the x-axis
b. Find the volume produced when R is revolved around the y-axis
the x-intercepts are 0 and 2
a) Vol = π ∫ (2x-x^2)^2 dx from 0 to 2
= π∫(4x^2 - 4x^3 + x^4) dx from 0 to 2
the rest is straighforward
b) this is is a little harder.
Solve the function for x
x^2 - 2x + y = 0
x = (2 ± √(4 - 4y) )/2
so the radius = 2 + √(4-4y) - (2 - √(4-4y))
so radius^2 = 4(4-4y) = 16 - 16y
volume = π∫(16-16y) dy from 0 to 1 , (the max height of the parabola is 1)
again , the rest is easy.