## anonymous 4 years ago Find the volume of the solid generated by revolving the region bounded by the curves y =sqrt(x) and y = x - 2 and the line x = 0 around the y-axis

|dw:1327635566351:dw| The equation needs to be with respect to y The radius of the outer part of the solid of revolution is (y+2) and the radius of the inner is y^2 So the area of each cross section is $\int\limits_{-1}^{2}(\pi(y+2)^2 - \pi(y^2)^2)dy=\pi \int\limits_{-1}^{2}((y^2+4y+4) - (y^4))dy$ $\pi \int\limits_{-1}^{2}((y^2+4y+4) - y^4)dy=\pi \left[ \frac{y^3}{3} + 2y^2 + 4y\right]_{-1}^{2}$ $\pi \left[ \frac{y^3}{3} + 2y^2 + 4y\right]_{-1}^{2} = \pi \left[( \frac{8}{3} + 8 + 8)-(-\frac{1}{3}+2-4)\right]$ $\pi \left[( \frac{8}{3} + 8 + 8)-(-\frac{1}{3}+2-4)\right]=21\pi$