The area bounded by the curve y = x^3 and the line y = 4x, in x > 0, is rota....?

Question

The area bounded by the curve y = x^3 and the line y = 4x, in x > 0, is rotated 2π radians
about the x-axis. Calculate the volume generated

Answer 1

You need to integrate π*int((4x)^2 - (x^3)^2) dx) with limits of integration from 0 to 2. The limits are 0 to 2 because 0 and 2 are the x-values where x^3 = 4x for x > 0. This is called disk integration because it returns the infinitesimal volumes of a series of annular "disks" that combine to form the solid. The integrated function (4x)^2 - (x^3)^2, and taking into account the factor of pi outside the integration, is the difference between the area of the circle defined by the outer perimeter and the area of the hole defined by the inner perimeter, because the area of a circle with radius r is π*r^2. 4x is greater than x^3 on this interval, so 4x is the outer perimeter and x^3 is the inner perimeter.

π*int((4x)^2 - (x^3)^2) dx) = π*int(16x^2 - x^6) dx), which you should be able to calculate easily enough using simple polynomial integration. Remember, the limits are 0 to 2.

Answer 2

You'll probably want to use the disk method.
First find the points where the two functions meet.
x^3 = 4x
x^3 - 4x = 0
x (x^2 - 4) = 0
x = 0 or x = 2

Note that the radius of the outer disk is going to be 4x and the radius of the inner disk is x^3, since 4x is above x^3. Now we just have to integrate pi * (r1^2-r2^2) over 0 to 2.

pi * Int (0 to 2) (4x)^2 - (x^3)^2 dx
= pi * Int (0 to 2) (16x^2 - x^6) dx
= pi * ( (16/3)x^3 - (1/7)x^7) evaluated at (0 to 2)
= pi * ( (128/3 - 128/7) )
= (512/21)*pi