The region bounded by y1x3 y0 x2 is rotated about the line

Solution

he two curves intercept at y = x^2; x = y^2 y = y^4 y^4 - y = 0 y(y^3 - 1) = 0 y = 0 or 1, when y = 0, x = 0 when y = 1, x = 1 the volume bounded is between y = vx and y = x^2. So we need to do integral for dx from 0 to 1. V = ?0 ->1 [p ((vx)-(-3))^2 - (x^2-(-3))^2] V = ?0 ->1 [p ((x + 6vx + 9) - (x^4 + 6x^2 + 9)] V = ?0 ->1 [p ((x + 6vx + 9) - x^4 - 6x^2 - 9)] V = ?0 ->1 [p ((x + 6vx - x^4 - 6x^2)] V = [p ((1/2x^2 + 4x^(3/2) - 1/5x^5 - 2x^3 + C)] |0->1 V = [p ((1/2 + 4 - 1/5 - 2) - 0)] V = 23/10 p It is kind of long..., but you should be able to get the idea! (And hope my calculation is correct! ;-) )