Calculus: Early Transcendentals 8th Edition

a) $V=\frac{2 \pi }{15}$ b) $V=\frac{ \pi }{6}$ c) $V=\frac{8 \pi }{15}$
{Step 1 of 10} First, find the points of intersection of the curves $y=x$ and $y=x^{2}$ For this, $x=x^{2}$ $x^{2}-x=0$ $x \left( x-1 \right) =0 or \left( x=0 or x=1 \right)$ The point of intersection are $\left( 0,0 \right)$ and $\left( 1,1 \right)$ . {Step 2 of 10} Sketch the curves. {Step 3 of 10} Rotate this shaded region about the x-axis. Consider a strip parallel to the y-axis in this shaded region. Rotation about the x-axis produces a washer. The outer radius of the washer is $x$ The inner radius of the washer is $x^{2}$ {Step 4 of 10} The cross-sectional area of the washer $A \left( x \right) = \pi \left[ x^{2}- \left( x^{2} \right) ^{2} \right]$ $= \pi \left[ x^{2}-x^{4} \right]$ The volume of the solid $V= \int _{0}^{1} \pi \left( x^{2}-x^{4} \right) dx$ $V= \pi \int _{0}^{1} \left( x^{2}-x^{4} \right) dx$ $= \pi \left[ \frac{x^{3}}{3}-\frac{x^{5}}{5} \right] _{0}^{1}$ $= \pi \left[ \frac{1}{3}-\frac{1}{5} \right]$ $V=\frac{2 \pi }{15}$ {Step 5 of 10} Consider the solid obtained by rotating this region about the y-axis. It is easier to use the cylindrical shell method to find the volume of this solid. Consider a strip at a distance of x from the origin. Rotation produces a cylindrical shell with radius x. {Step 6 of 10} Draw the Figure {Step 7 of 10} The circumference of the shell is $2 \pi x$, and the height of the shell is $x-x^{2}$ The volume of the solid $V= \int _{0}^{1} \left( 2 \pi x \right) \left( x-x^{2} \right) dx$ $=2 \pi \int _{0}^{1} \left( x^{2}-x^{3} \right) dx$ $=2 \pi \left[ \frac{1}{3}x^{3}-\frac{1}{4}x^{4} \right] _{0}^{1}$ $=2 \pi \left[ \frac{1}{3}-\frac{1}{4} \right]$ $V=\frac{ \pi }{6}$ {Step 8 of 10} Rotate this shaded region about $y=2$Use the slicing method {Step 9 of 10} The outer radius of the washer is $2-x^{2}$ The inner radius of the washer is $2-x$ The cross-sectional area of washer is $A \left( x \right) = \pi \left[ \left( 2-x^{2} \right) ^{2}- \left( 2-x \right) ^{2} \right]$ $= \pi \left[ 4+x^{4}-4x^{2}-4-x^{2}+4x \right]$ $= \pi \left[ x^{4}-5x^{2}+4x \right]$ {Step 10 of 10} The volume of the solid obtained by rotating the bounded region about $y=2$ $V= \int _{0}^{1}A \left( x \right) dx$ $= \int _{0}^{1} \pi \left[ x^{4}-5x^{2}+4x \right] dx$ $= \pi \int _{0}^{1} \left( x^{4}-5x^{2}+4x \right) dx$ $= \pi \left[ \frac{1}{5}x^{5}-\frac{5}{3}x^{3}+4\frac{x^{2}}{2} \right] _{0}^{1}$ $= \pi \left[ \frac{1}{5}-\frac{5}{3}+2 \right]$ $V=\frac{8 \pi }{15}$