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Volume of Solid: Revolving Region About Line

Created by @nathanedwards

at November 1st 2023, 2:14:50 pm.

AP_Calculus_AB-Questions

#volume

#ap-calculus-ab

#cylindrical-shells

AP Calculus AB Exam Question:

Let $R$ be the region enclosed by the curve $y = x^2 + 1$ and the line $y = 3x - 2$ . The region is revolved about the line $y = 4$ . Find the volume of the solid generated.

Answer:

To find the volume of the solid generated by revolving the region $R$ about the line $y = 4$ , we can use the method of cylindrical shells.

Step 1: Determine the limits of integration. To find the limits of integration, we need to determine the x-values where the two curves intersect. Set $x^2 + 1 = 3x - 2$ and solve for $x$ :

x^2 + 1 = 3x - 2

x^2 - 3x + 3 = 0

This quadratic equation does not have real solutions, which means that the curves $y = x^2 + 1$ and $y = 3x - 2$ do not intersect. However, we can still determine the limits of integration based on the x-values where the curves would intersect if they did.

The parabola $y = x^2 + 1$ intersects the line $y = 4$ at $x = 3$ and $x = -1$ . Therefore, the limits of integration are -1 and 3.

Step 2: Set up the volume integral using the cylindrical shell method. The volume of a cylindrical shell can be calculated using the formula $V = 2\pi rh \Delta x$ , where $r$ is the distance from the axis of rotation to the shell, $h$ is the height of the shell, and $\Delta x$ is the thickness of the shell.

Since the region is revolved about the line $y = 4$ , the distance from the axis of rotation to the shell is $4 - (x^2 + 1)$ . The height of the shell is the difference between the y-values of the curves at the x-coordinate, which is $3x - 2 - (x^2 + 1)$ . Finally, the thickness of the shell is $\Delta x$ .

The volume integral can be set up as follows:

V = \int_{-1}^{3} 2\pi (4 - (x^2 + 1))(3x - 2 - (x^2 + 1)) dx

Step 3: Evaluate the integral. Simplify the integrand:

V = \int_{-1}^{3} 2\pi (3x - x^2 - 3)(3x - x^2 - 3) dx

V = \int_{-1}^{3} 2\pi (9x^2 - 6x^3 + x^4 - 9x + 6x^2 - 3x^3 + 3x - 2) dx

V = \int_{-1}^{3} 2\pi (-9x^3 +4x^4 - 3x^2 - 6x + 2) dx

Integrate term by term:

V = 2\pi \left[-\dfrac{9}{4}x^4 + \dfrac{4}{5}x^5 - x^3 - 3x^2 + 2x\right]_{-1}^{3}

Substitute the limits of integration:

V = 2\pi \left(-\dfrac{9}{4}(3)^4 + \dfrac{4}{5}(3)^5 - (3)^3 - 3(3)^2 + 2(3)\right) - 2\pi \left(-\dfrac{9}{4}(-1)^4 + \dfrac{4}{5}(-1)^5 - (-1)^3 - 3(-1)^2 + 2(-1)\right)

Simplify the expression:

V = 2\pi \left(-\dfrac{9}{4}(81) + \dfrac{4}{5}(243) - 27 - 27 + 6\right) - 2\pi \left(-\dfrac{9}{4} + \dfrac{4}{5} + 1 - 3 + (-2)\right)

V = 2\pi \left(-\dfrac{243}{4} + \dfrac{972}{5} - 54 + 6\right) - 2\pi \left(-\dfrac{9}{4} + \dfrac{4}{5} - 2\right)

V = 2\pi \left(-\dfrac{972}{4} + \dfrac{1944}{5} - 216 + 12\right) - 2\pi \left(-\dfrac{45}{4} + \dfrac{20}{5} - 10\right)

V = 2\pi \left(-243 + \dfrac{3888}{5} - 216 + 12\right) - 2\pi \left(-\dfrac{45}{4} + 4 - 10\right)

V = 2\pi \left(-243 + \dfrac{3888}{5} - 216 + 12\right) - 2\pi \left(-\dfrac{45}{4} + 4 - 10\right)

V = 2\pi \left(\dfrac{864}{5}\right) - 2\pi \left(-\dfrac{57}{4}\right)

Evaluate the expression:

V = 2\pi \left(\dfrac{864}{5}\right) + 2\pi \left(\dfrac{57}{4}\right)

V = \dfrac{5}{2}\cdot 2\pi \left(\dfrac{864}{5}\right) + \dfrac{4}{2}\cdot 2\pi \left(\dfrac{57}{4}\right)

Simplify the expression:

V = 2\pi \cdot 864 + \pi \cdot 57

V = 1728\pi + 57\pi

V = \boxed{1785\pi}

Thus, the volume of the solid generated by revolving the region $R$ about the line $y = 4$ is $1785\pi$ .