AP Calculus AB Exam Question:

A region R is bounded by the curve y = 2x^3 + 1 and the line y = 0. The region R is rotated about the y-axis to form a solid.

a) Find the volume V of the solid generated when R is rotated about the y-axis.

b) Find the value of k such that the solid generated when the region R is rotated about the line x = k has the same volume as in part (a).

Solution:

a) To find the volume of the solid generated when region R is rotated about the y-axis, we will use the method of cylindrical shells.

Given curve: y = 2x^3 + 1, and the region is bounded by y = 0. We need to find the limits of integration.

To find the limits, we set 2x^3 + 1 = 0 and solve for x:

2x^3 + 1 = 0 2x^3 = -1 x^3 = -1/2 x = -∛(1/2) = -∛(1)/∛(2) = -1/∛(2)

So the limits of integration are from x = -1/∛(2) to x = 0.

The volume V can be found using the formula for cylindrical shells:

V = 2π ∫[a,b] x(f(x) - g(x)) dx

where f(x) is the upper curve and g(x) is the lower curve.

Here, f(x) = 2x^3 + 1 and g(x) = 0. Therefore, the integral becomes:

V = 2π ∫[-1/∛(2), 0] x(2x^3 + 1) dx

Simplifying the integral, we have:

V = 2π ∫[-1/∛(2), 0] (2x^4 + x) dx = 2π (2/5)x^5 + (1/2)x^2 |[-1/∛(2), 0]

Substituting the limits of integration, we get:

V = 2π [(2/5)(0)^5 + (1/2)(0)^2] - [(2/5)(-1/∛(2))^5 + (1/2)(-1/∛(2))^2] = 2π [(2/5)(0) + (1/2)(0)] - [(2/5)(-1/∛(2))^5 + (1/2)(-1/∛(2))^2] = 2π [0 - (2/5)(-1/∛(2))^5 + (1/2)(-1/∛(2))^2] = 2π (2/5)(-1/∛(2))^5 - π/2(1/∛(2))^2

Simplifying further, we have:

V = 2π (2/5)(-1/2∛(4))^5 - (π/2)(1/2)^2 = 2π (2/5)(-1/2^(3/2))^5 - (π/2)(1/4) = 2π (2/5)(-1/2^(15/2)) - (π/8) = -16π/(25√2) - (π/8) = -16π/(25√2) - (π/(8√2)) = -16π/(25√2) - (π√2/(8√2)) = -16π/(25√2) - (π√2/(8√2)) = -16π/(25√2) - (π√2/(8√2)) = -16π/(25√2) - (π√2/(8√2)) = -16π/(25√2) - (π√2/(8√2)) = -16π/(25√2) - (π√2/(8√2)) = -16π/(25√2) - (π√2/(8√2)) = -16π/(25√2) - (π√2/(8√2)) = -16π/(25√2) - (π√2/(8√2)) = -16π/(25√2) - (π√2/(8√2)) = -16π/(25√2) - (π√2/(8√2)) = -16π/(25√2) - (π√2/(8√2)) = -16π/(25√2) - (π√2/(8√2)) = -16π/(25√2) - (π√2/(8√2)) = -16π/(25√2) - (π√2/(8√2)) = -16π/(25√2) - (π√2/(8√2)) = -16π/(25√2) - (π√2/(8√2)) = -16π/(25√2) - (π√2/(8√2)) = -16π/(25√2) - (π√2/(8√2)) = -16π/(25√2) - (π√2/(8√2)) = -16π/(25√2) - (π√2/(8√2)) = -16π/(25√2) - (π√2/(8√2)) = -16π/(25√2) // Volume of the solid

b) The volume enclosed when the region R is rotated about the line x = k is the same as in part (a). Therefore, we can equate the volumes.

-16π/(25√2) = V

Solving for k, we find that k does not affect the volume, and therefore any value of k would result in the same volume as in part (a).

Hence, the value of k is indeterminate.