## Calculus: Early Transcendentals 8th Edition

$V=\frac{64 \pi }{15}$
{Step 1 of 5}: Find, find the points of intersection of $y=2x$ and $y=x^{2}$ $x^{2}=2x$ $x^{2}-2x=0$ $x \left( x-2 \right) =0$ Or $x=0$ or $x=2$ The points of Intersection are (0, 0) and (2, 4). {Step 2 of 5}: Sketch the curves. {Step 3 of 5}: Rotate the shaded region about the x-axis. Sketch the solid obtained by rotating the shaded region about the x-axis and the washer. {Step 4 of 5}: The outer radius of the washer is 2x The inner radius of the washer is x2 The cross-sectional area of the washer $A \left( x \right) = \pi \left[ \left( 2x \right) ^{2}- \left( x^{2} \right) ^{2} \right]$ $= \pi \left[ 4x^{2}-x^{4} \right]$ {Step 5 of 5}: The volume of the solid obtained by rotating the shaded region about the x-axis $V= \int _{0}^{2}A \left( x \right) dx$ $= \int _{0}^{2} \pi \left[ 4x^{2}-x^{4} \right] dx$ $= \pi \int _{0}^{2} \left( 4x^{2}-x^{4} \right) dx$ $= \pi \left[ \frac{4}{3}x^{3}-\frac{1}{5}x^{5} \right] _{0}^{2}$ [By the fundamental theorem of calculus, part 2] $= \pi \left[ \frac{32}{3}-\frac{32}{5} \right]$ $V=\frac{64 \pi }{15}$