The region

e

enclosed by the curves

y=3x

and

y=x^(2)

is rotated about the

x

-axis. Find the volume of the resulting solid.\ Solution\ (1)\ A cross-section in the plane

P_(x)

has the shape of a washer (an annular ring) with inner radius\ and an outer radius\ , so we find the cross-sectional area by subtracting the area of the inner circle from the area of the outer circle.\ \ Therefore, we have\

V=\\\\int_0^3 A(x)dx=\\\\int_0^3 \\\\pi (9x^(2)-x^(4))dx\ =\\\\pi [|]_(0)^(3)=

The region e enclosed by the curves y=3x and y=x2 is rotated about the x-axis. Find the volume of the resulting solid. Solution (1) A cross-section in the plane Px has the shape of a washer (an annular ring) with inner radius and an outer radius , so we find the cross-sectional area by subtracting the area of the inner circle from the area of the outer circle. A(x)=9x2(x2)2=( Therefore, we have V=03A(x)dx=03(9x2x4)dx=[]03=