Use the shell method to find the volume of the solid generated by revolving the plane region
about the given line.
y=x^2, y=4x-x^2, about the x=4
Solution
when you draw the curves and the axis of rotation, you\'ll notice there are two
regions being rotated: one from x = 0 to x = 2 (from x^2 = 4x - x^2), and from x = 2 to x = 4 (the
axis of rotation). at any point x, take an infinitesimal shell of thickness dx parallel to the axis of
rotation. the height of the shell is (4x - x^2) - x^2 = 4x - 2x^2 in the first region, and 2x^2 - 4x in
the second one. the radius of the shell is 4 - x (this is the difference when rotating about the y-
axis, where the radius is just x). for the first region, dv = 2pi(4 - x)*(4x - 2x^2)dx, x from 0 to 2.
for the second region, dv = 2pi(4 - x)*(2x^2 - 4x)dx, x from 2 to 4. the next step is to perform
the integration, which i hope you already know.
about the given line.
y=x^2, y=4x-x^2, about the x=4
Solution
when you draw the curves and the axis of rotation, you\'ll notice there are two
regions being rotated: one from x = 0 to x = 2 (from x^2 = 4x - x^2), and from x = 2 to x = 4 (the
axis of rotation). at any point x, take an infinitesimal shell of thickness dx parallel to the axis of
rotation. the height of the shell is (4x - x^2) - x^2 = 4x - 2x^2 in the first region, and 2x^2 - 4x in
the second one. the radius of the shell is 4 - x (this is the difference when rotating about the y-
axis, where the radius is just x). for the first region, dv = 2pi(4 - x)*(4x - 2x^2)dx, x from 0 to 2.
for the second region, dv = 2pi(4 - x)*(2x^2 - 4x)dx, x from 2 to 4. the next step is to perform
the integration, which i hope you already know.
