AP Calculus AB Exam Question

A region R in the first quadrant is bounded by the x-axis, the line x = 2, and the curve y = x^3.

a) Determine the volume of the solid formed when the region R is revolved about the x-axis. b) Determine the volume of the solid formed when the region R is revolved about the line x = 4.

Answer:

a) To find the volume of the solid formed when the region R is revolved about the x-axis, we will use the method of cylindrical shells.

We start by setting up the integral for the volume of each cylindrical shell. The height of each shell will be the difference between the uppermost point on the curve y = x^3 and the x-axis, which is equal to x^3 - 0 = x^3. The radius of each shell is the distance from the axis of rotation (the x-axis) to the point on the curve, which is x. The thickness of each shell is dx.

Therefore, the formula for the volume of each cylindrical shell is given by:

dV = 2πrh*dx = 2π(x^3)(dx)

To find the total volume, we integrate this formula over the range of x = 0 to x = 2:

V = ∫[0,2] 2π(x^3) dx

Integrating with respect to x:

V = 2π ∫[0,2] (x^3) dx

Using the power rule of integration:

V = 2π [(1/4)x^4] | [0,2]

Evaluating the definite integral:

V = 2π [(1/4)(2^4)] - [(1/4)(0^4)]

Simplifying:

V = 2π [16/4]

V = 8π

Therefore, the volume of the solid formed when the region R is revolved about the x-axis is 8π cubic units.

b) To find the volume of the solid formed when region R is revolved about the line x = 4, we use the same method of cylindrical shells. However, in this case, the axis of rotation is the line x = 4, which is a distance of 4 units to the right of the x-axis.

The distance from each point on the curve to the line x = 4 is given by the difference between 4 and x. Therefore, the radius of each shell is (4 - x), and the formula for the volume of each cylindrical shell is:

dV = 2πrh*dx = 2π(4-x)(x^3)dx

To find the total volume, we integrate this formula over the range of x = 0 to x = 2:

V = ∫[0,2] 2π(4-x)(x^3) dx

Expanding and simplifying:

V = 2π ∫[0,2] (4x^3 - x^4) dx

Using the power rule of integration:

V = 2π [4(1/4)x^4 - (1/5)x^5] | [0,2]

Evaluating the definite integral:

V = 2π [(1)x^4 - (1/5)(16)] - [(1/5)(0^5)]

Simplifying:

V = 2π (16/5)

V = (32π)/5

Therefore, the volume of the solid formed when the region R is revolved about the line x = 4 is (32π)/5 cubic units.