AP Calculus AB Exam Question:

A solid is formed by rotating the region bounded by the curve defined by the equation y = x^2 + 1 and the x-axis, where x is between -1 and 3, about the y-axis.

(a) Find the volume of the solid using the method of disks.

(b) Find the volume of the solid using the method of washers.

(c) Verify your answers to parts (a) and (b) by evaluating the definite integral for the same region.

Step-by-step Solution:

(a) Using the Method of Disks:

To find the volume of the solid using the method of disks, we integrate the cross-sectional areas of the disks formed by revolving the curve around the y-axis.

The equation of the curve is given as y = x^2 + 1. We need to find the limits of integration. Since x is between -1 and 3, we can rewrite the equation as x = sqrt(y - 1).

The radius of each disk is the distance between the y-axis and the curve, which is x. Therefore, the radius of each disk is sqrt(y - 1).

The differential thickness of each disk, dy, is used to find the corresponding area of the disk.

The volume of each disk, dV, is given by dV = π(radius)^2 dy.

To get the total volume, we integrate the volume function from the lower limit to the upper limit:

V = ∫[a,b] dV = ∫[a,b] π(sqrt(y - 1))^2 dy

V = π∫[a,b] (y-1) dy

Now, we need to determine the limits of integration. Since x is between -1 and 3, y is between 0 and 10.

V = π∫[0,10] (y-1) dy

V = π [(y^2/2 - y)] |[0,10]

V = π [(10^2/2 - 10) - (0^2/2 - 0)]

V = π (50 - 0)

Answer: The volume of the solid using the method of disks is 50π cubic units.

(b) Using the Method of Washers:

To find the volume of the solid using the method of washers, we integrate the volumes of the washers formed by revolving the curve around the y-axis.

The equation of the curve is given as y = x^2 + 1. We need to find the limits of integration. Since x is between -1 and 3, we can rewrite the equation as x = sqrt(y - 1).

The outer radius of each washer is the distance between the y-axis and the curve, which is x. Therefore, the outer radius of each washer is sqrt(y - 1).

The inner radius of each washer is the distance between the y-axis and the x-axis, which is simply the y-axis itself. Therefore, the inner radius of each washer is 0.

The differential thickness of each washer, dy, is used to find the corresponding volume of the washer.

The volume of each washer, dV, is given by dV = π(outer radius)^2 - π(inner radius)^2 dy.

To get the total volume, we integrate the volume function from the lower limit to the upper limit:

V = ∫[a,b] dV = ∫[a,b] π(outer radius)^2 - π(inner radius)^2 dy

V = ∫[a,b] π(sqrt(y - 1))^2 - π(0)^2 dy

V = ∫[a,b] π (y-1) dy

Using the same limits of integration as in part (a), the integral becomes:

V = π∫[0,10] (y-1) dy

V = π [(y^2/2 - y)] |[0,10]

V = π [(10^2/2 - 10) - (0^2/2 - 0)]

V = π (50 - 0)

Answer: The volume of the solid using the method of washers is 50π cubic units.

(c) Verifying with Definite Integral:

To verify our answers from part (a) and (b), we can evaluate the definite integral for the same region.

The definite integral for the region bounded by the curve is:

V = ∫[a,b] A(x) dx,

where A(x) is the area function.

The area function, A(x), is given by A(x) = π(y)^2.

Using the equation y = x^2 + 1, we have A(x) = π(x^2 + 1)^2.

Since x is between -1 and 3, the limits of integration are -1 and 3.

V = ∫[-1,3] π(x^2 + 1)^2 dx

V = π ∫[-1,3] (x^4 + 2x^2 + 1) dx

V = π [(1/5)x^5 + (2/3)x^3 + x] |[-1,3]

V = π [(1/5)(3^5) + (2/3)(3^3) + 3] - π [(1/5)(-1^5) + (2/3)(-1^3) - 1]

V = π [(243/5) + (54/3) + 3] - π [(-1/5) + (-2/3) - 1]

V = π (48.6 + 36 + 3) - π (-0.2 - 0.666 - 1)

V = π (87.6) - π (-1.866)

V ≈ 87.6π + 1.866π

V ≈ 89.466π

Answer: The volume of the solid, as verified by the definite integral, is approximately 89.466π cubic units.