RaySix

Post

Volume of Solid Using Cylindrical Shells and Disks

Created by @nathanedwards

at October 31st 2023, 8:50:22 pm.

AP_Calculus_AB-Questions

#volume

#cylindrical-shells

#solid

Question:

Consider the curve given by the equation $y = \sqrt{x}$ for $0 \leq x \leq 4$ . The region enclosed by this curve and the $x$ -axis is revolved around the $y$ -axis.

(a) Use the method of cylindrical shells to find the volume of the solid generated.

(b) Use the method of disks/washers to find the volume of the solid generated.

(c) Verify your answers to parts (a) and (b) by evaluating the definite integral and finding the volume of the solid directly.

Provide your answer in terms of $\pi$ .

Answer:

(a) Method of Cylindrical Shells:

To find the volume of the solid generated by revolving the region enclosed by the curve $y = \sqrt{x}$ and the $x$ -axis around the $y$ -axis, we will use the method of cylindrical shells.

We will divide the region into infinitesimally thin vertical strip, where each strip of width $\Delta x$ is at a distance $x$ from the $y$ -axis. Therefore, the height of each cylindrical shell will be equal to the function value $y = \sqrt{x}$ , and the circumference will be given by $2\pi x$ .

The volume of each cylindrical shell can be approximated as $2\pi x \times \sqrt{x} \, \Delta x$ . Summing up these volumes from $x = 0$ to $x = 4$ , we obtain the integral:

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

To find the definite integral:

V = 2\pi \int_0^4 x^{3/2} \, dx

Calculating the integral:

V = 2\pi \left[ \frac{2}{5} x^{5/2} \right]_0^4

V = 2\pi \left( \frac{2}{5} \times 4^{5/2} - \frac{2}{5} \times 0^{5/2} \right)

V = 2\pi \left( \frac{2}{5} \times 32 - \frac{2}{5} \times 0 \right)

V = 2\pi \left( \frac{64}{5} \right)

V = \frac{128\pi}{5} \, \text{units}^3

Therefore, the volume of the solid generated by revolving the region around the $y$ -axis is $\frac{128\pi}{5} \, \text{units}^3$ .

(b) Method of Disks/Washers:

To find the volume of the solid generated by revolving the region enclosed by the curve $y = \sqrt{x}$ and the $x$ -axis around the $y$ -axis, we will use the method of disks/washers.

We will divide the region into infinitesimally thin vertical strip, where each strip of width $\Delta x$ is at a distance $x$ from the $y$ -axis. The cross-sectional area of each disk/washer will be equal to the square of the function value $y = \sqrt{x}$ .

The volume of each disk/washer can be approximated as $\pi \left( \sqrt{x} \right)^2 \, \Delta x$ . Summing up these volumes from $x = 0$ to $x = 4$ , we obtain the integral:

V = \pi \int_0^4 x \, dx

Calculating the integral:

V = \pi \left[ \frac{1}{2} x^2 \right]_0^4

V = \pi \left( \frac{1}{2} \times 4^2 - \frac{1}{2} \times 0^2 \right)

V = \pi \times 8

V = 8\pi \, \text{units}^3

Therefore, the volume of the solid generated by revolving the region around the $y$ -axis is $8\pi \, \text{units}^3$ .

(c) Verification:

To verify our previous answers, we will evaluate the definite integral directly and find the volume of the solid.

Using the definite integral:

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

Calculating the integral:

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

V = \left( \frac{2}{3} \times 4^{3/2} \right) - \left( \frac{2}{3} \times 0^{3/2} \right)

V = \frac{2}{3} \times 8

V = \frac{16}{3} \, \text{units}^3

Therefore, the volume of the solid generated by revolving the region around the $y$ -axis is $\frac{16}{3} \, \text{units}^3$ as obtained from direct evaluation of the definite integral.

Hence, the answers from parts (a) and (b) are verified.