RaySix

Post

Volume of Solid Generated by Rotation

Created by @nathanedwards

at November 1st 2023, 5:11:22 pm.

AP_Calculus_AB-Questions

#solid-of-revolution

#integral-calculus

#cylindrical-shells

Question:

Let $R$ be the region bounded by the graph of the curve $y = 2 + \sqrt{x}$ and the $x$ -axis, where $x \geq 0$ . The region $R$ is rotated about the $x$ -axis to form a solid.

a) Find the volume of the solid generated when $R$ is rotated about the $x$ -axis over the interval $0 \leq x \leq 4$ .

b) The solid generated in part (a) is sliced perpendicular to the $x$ -axis into thin disks. Find the exact value of the radius of one such disk in terms of $x$ .

c) Calculate the volume of one of the disks found in part (b).

d) Determine the exact value of the sum of the volumes of all such disks, using an integral.

Answer:

a) To find the volume of the solid generated when $R$ is rotated about the $x$ -axis, we can use the method of cylindrical shells. The volume of each shell is given by $2 \pi x f(x) \Delta x$ , where $f(x)$ is the height of the function at $x$ , and $\Delta x$ is a small width of the shell.

The height of the function $f(x)$ is given by $2 + \sqrt{x}$ .

The volume of a single shell is:

V_{\text{shell}} = 2 \pi x (2 + \sqrt{x}) \Delta x

To find the total volume, we need to integrate $V_{\text{shell}}$ over the interval $0 \leq x \leq 4$ :

V = \int_0^4 2 \pi x (2 + \sqrt{x}) \, dx

Let's evaluate this integral step by step.

First, distribute the $2 \pi x$ :

V = \int_0^4 4 \pi x + 2 \pi x \sqrt{x} \, dx

Then, integrate each term separately:

V = \left[ 2 \pi x^2 + \frac{4}{3} \pi x \sqrt{x} \right]_0^4

Evaluate the expression at the upper limit of integration:

V = \left( 2 \pi (4^2) + \frac{4}{3} \pi (4) \sqrt{4} \right) - \left( 2 \pi (0^2) + \frac{4}{3} \pi (0) \sqrt{0} \right)

This simplifies to:

V = 32 \pi + \frac{32}{3} \pi = \frac{96}{3} \pi = \boxed{32 \pi}

Therefore, the volume of the solid generated when $R$ is rotated about the $x$ -axis over the interval $0 \leq x \leq 4$ is $32 \pi$ .

b) To find the radius of one of the disks, we need to consider a horizontal slice at any given value of $x$ . The radius of the disk is equal to the height of the function $f(x)$ at that particular value of $x$ . In this case, the radius is given by $2 + \sqrt{x}$ .

Therefore, the exact value of the radius of one such disk is $\boxed{2 + \sqrt{x}}$ .

c) The volume of a disk is given by $V_{\text{disk}} = \pi r^2 \Delta x$ , where $r$ is the radius of the disk and $\Delta x$ is a small width.

Substituting the given expression for the radius, we have:

V_{\text{disk}} = \pi (2 + \sqrt{x})^2 \Delta x

Simplifying:

V_{\text{disk}} = \pi (4 + 4\sqrt{x} + x) \Delta x

Therefore, the volume of one of the disks found in part (b) is $\boxed{\pi (4 + 4\sqrt{x} + x) \Delta x}$ .

d) To determine the exact value of the sum of the volumes of all the disks, we need to integrate the expression for $V_{\text{disk}}$ over the interval $0 \leq x \leq 4$ .

Using the definite integral:

V = \int_0^4 \pi (4 + 4\sqrt{x} + x) \, dx

Let's evaluate this integral step by step.

Distribute the $\pi$ :

V = \int_0^4 4\pi + 4\pi\sqrt{x} + \pi x \, dx

Integrate each term separately:

V = \left[ 4\pi x + \frac{8}{3}\pi (\sqrt{x})^3 + \frac{1}{2} \pi x^2 \right]_0^4

Evaluate the expression at the upper limit of integration:

V = \left( 4\pi(4) + \frac{8}{3}\pi (\sqrt{4})^3 + \frac{1}{2} \pi (4)^2 \right) - \left( 4\pi(0) + \frac{8}{3}\pi (\sqrt{0})^3 + \frac{1}{2} \pi (0)^2 \right)

This simplifies to:

V = 48\pi + \frac{32}{3}\pi = \frac{144}{3}\pi + \frac{32}{3}\pi = \boxed{\frac{176}{3}\pi}

Therefore, the exact value of the sum of the volumes of all the disks is $\frac{176}{3}\pi$ .