Question

Find the volume of the solid generated by revolving the region bounded by the curves y = x^2 and y = 4x − x^2 about the line y = 6.

Answer 1

$y=x^{2} \quad ,y=4 x-x^{2} \quad ,y=6 \\[1 em] \therefore x^{2}=4 x-x^{2} \rightarrow x^{2}-2 x=0 \rightarrow x=0,2 \\[1 em] \therefore V=\pi \int_{0}^{2} r_{1}^{2}-r_{2}^{2} d x \\[1 em] r_{1}=6-x^{2} \quad, r_{2}=6-4 x+x^{2} \\[1 em] \therefore V=\pi \int_{0}^{2}\left(6-x^{2}\right)^{2}-\left(6-4 x+x^{2}\right)^{2} dx\\[1 em] =\pi \int_{0}^{2} 36-12 x^{2}+x^{4}-\left(x^{4}-8 x^{3}+28 x^{2}-48 x+36\right) d x \\[1 em] =\pi \int_{0}^{2} 36-12 x^{2}+x^{4}-x^{4}+8 x^{3}-28 x^{2}+48 x-36 d x \\[1 em] =\pi \int_{0}^{2} 8 x^{3}-40 x^{2}+48 x d x \\[1 em] =\pi\left(2 x^{4}-\frac{40}{3} x^{3}+24 x^{2}\right) \bigg|_{0}^{2} \\[1 em] \therefore V\approx 67$