## Calculus: Early Transcendentals 8th Edition

a) A=$\frac{5}{12}$ b)V = $\frac{41 \pi }{105}$ c) V= $\frac{13 \pi }{30}$
{Step 1 of 9} A) First, find the points of intersection of the curves y=x3 and y=2x-x2 $x^3=2x-x^2$ $x^3+x^2-2x=0$ $x(x^2+x-2)=0$ $x=0$ or $x2+x-2=0$ $x=0$ or $x2+2x-x-2=0$ $x=0$ or $(x+2)(x-1)=0$ $x=0$ or $x=-2$ or $x=1$ The points of intersection are (0,0), (-2,-8), and (1,1). Consider the two points (0,0) and (1,1), which are in the first quadrant. {Step 2 of 9} Sketch the curves. {Step 3 of 9} Find the area of the shaded region, which is labelled R. A=$\int _{0}^{1} \left[ \left( 2x-x^{2} \right) -x^{3} \right] dx$ =$\left[ x^{2}-\frac{1}{3}x^{3}-\frac{1}{4}x^{4} \right] _{0}^{1}$ =1-$\frac{1}{3}$ -$\frac{1}{4}$ A=$\frac{5}{12}$ {Step 4 of 9} B) Rotate this region about the x-axis. Use the slicing method. Consider a strip in this region parallel to the y-axis. Rotate this region about the x-axis to produce a washer. {Step 5 of 9} The outer radius of the washer is $2x-x^2$ The inner radius of the washer is $x^3$ The cross-sectional area of the washer A(x) =$\pi \left[ \left( 2x-x^{2} \right) ^{2}- \left( x^{3} \right) ^{2} \right]$ =$\pi \left[ 4x^{2}+x^{4}-4x^{3}-x^{6} \right]$ {Step 6 of 9} Then the volume of the solid V=$\int _{0}^{1}A \left( x \right) dx$ =$\pi \int _{0}^{1} \left[ 4x^{2}+x^{4}-4x^{3}-x^{6} \right]$dx =$\pi \left[ \frac{4}{3}x^{3}+\frac{1}{5}x^{5}-x^{4}-\frac{1}{7}x^{7} \right] _{0}^{1}$ =$\pi \left[ \frac{4}{3}+\frac{1}{5}-1-\frac{1}{7} \right]$ V = $\frac{41 \pi }{105}$ {Step 7 of 9} ( C ) rotate region R about the y-axis and use the cylindrical shell method. Consider a vertical strip in region R. rotate the region about the y-axis to produce a cylindrical shell with a radius of x and a height of $(2x-x^2-x^3)$. {Step 8 of 9} Draw a Figure {Step 9 of 9} The circumference of the shell is 2$\pi x$ The volume of the solid V = $\int _{0}^{1} \left( 2 \pi x \right) \left( 2x-x^{2}-x^{3} \right) dx$ V= 2$\pi \int _{0}^{1} \left( 2x^{2}-x^{3}-x^{4} \right) dx$ =2$\pi \left[ \frac{2}{3}x^{3}-\frac{1}{4}x^{4}-\frac{1}{5}x^{5} \right] _{0}^{1}$ =2$\pi \left[ \frac{2}{3}-\frac{1}{4}-\frac{1}{5} \right]$ V= $\frac{13 \pi }{30}$