# Surface Area of Revolution with Double Integration

• MHB
• Edward2022
In summary, the conversation discusses the equivalent rotation of a function over different intervals, leading to the integral $\displaystyle S = 2\pi \int_a^b f(x) \sqrt{1+[f(x)]^2} \, dx$. The conversation then goes through a series of substitutions and integration by parts to simplify the integral, resulting in a final answer of $\approx 31.27$.
Edward2022

#### Attachments

• ONE.jpg
7.3 KB · Views: 87
Last edited:
ok ... one more time

$\displaystyle S = 2\pi \int_a^b f(x) \sqrt{1+[f(x)]^2} \, dx$

rotating $f(x) = \sin{x}$ over the interval $[-\pi, 5\pi/4]$ is equivalent to rotating twice over the interval $[0,\pi]$ plus once over the interval $[0, \pi/4]$

so ...

$\displaystyle S = 4\pi \int_0^\pi \sin{x} \sqrt{1+\cos^2{x}} \, dx + 2\pi \int_0^{\pi/4} \sin{x} \sqrt{1+\cos^2{x}} \, dx$

let $u = \cos{x} \implies du = -\sin{x} \, dx$

$\displaystyle S = 4\pi \int_{-1}^1 \sqrt{1+u^2} \, du + 2\pi \int_{1/\sqrt{2}}^1 \sqrt{1+u^2} \, du$

let $u = \tan{t} \implies du = \sec^2{t} \, dt$

$\displaystyle S = 4\pi \int_{-\pi/4}^{\pi/4} \sqrt{1+\tan^2{t}} \sec^2{t} \, dt + 2\pi \int_{\arctan(1/\sqrt{2})}^{\pi/4} \sqrt{1+\tan^2{t}} \sec^2{t} \, dt$

$\displaystyle S = 4\pi \int_{-\pi/4}^{\pi/4} \sec^3{t} \, dt + 2\pi \int_{\arctan(1/\sqrt{2})}^{\pi/4} \sec^3{t} \, dt$

since the secant function is even ...

$\displaystyle S = 8\pi \int_0^{\pi/4} \sec^3{t} \, dt + 2\pi \int_{\arctan(1/\sqrt{2})}^{\pi/4} \sec^3{t} \, dt$

note integrating $\sec^3{t}$ by parts ...

$\displaystyle \int \sec^3{t} \, dt = \dfrac{1}{2} \bigg[\sec{t}\tan{t} + \ln|\sec{t}+\tan{t}| \bigg]$

so ...

$S = 4\pi \bigg[ \sec{t}\tan{t} + \ln|\sec{t}+\tan{t}| \bigg]_0^{\pi/4} + \pi \bigg[ \sec{t}\tan{t} + \ln|\sec{t}+\tan{t}| \bigg]_{\arctan(1/\sqrt{2})}^{\pi/4}$

$S = 5\pi\bigg[\sqrt{2}+\ln(\sqrt{2}+1)\bigg] - \pi \bigg[\dfrac{\sqrt{3}}{2} + \ln\left(\dfrac{\sqrt{3}+1}{\sqrt{2}}\right) \bigg] \approx 31.27$

sorry for the previous deleted posts ... too many opportunities to make careless errors, especially after consuming a few beers.

## What is the "Area of Surface of Revolution"?

The "Area of Surface of Revolution" refers to the total surface area of a three-dimensional shape created by rotating a two-dimensional curve around an axis. It is commonly used in calculus and geometry to find the surface area of objects such as cylinders, cones, and spheres.

## How do you calculate the "Area of Surface of Revolution"?

The formula for calculating the "Area of Surface of Revolution" depends on the specific shape being rotated. For example, the formula for a cylinder is A = 2πrh, where A is the surface area, r is the radius of the base, and h is the height of the cylinder. Other common formulas include A = πr² for a sphere and A = πrl for a cone, where r is the radius and l is the slant height.

## What is the difference between "Area of Surface of Revolution" and "Volume of Revolution"?

The "Area of Surface of Revolution" refers to the total surface area of a three-dimensional shape, while the "Volume of Revolution" refers to the amount of space inside the shape. In other words, the "Area of Surface of Revolution" is a 2D measurement, while the "Volume of Revolution" is a 3D measurement.

## Why is the "Area of Surface of Revolution" important in science?

The "Area of Surface of Revolution" is important in science because it allows us to calculate the surface area of complex three-dimensional shapes, which is necessary for many real-world applications. For example, in engineering, knowing the surface area of a pipe or cylinder is crucial for determining the amount of material needed for construction.

## What are some real-world applications of the "Area of Surface of Revolution"?

The "Area of Surface of Revolution" has many practical applications in fields such as engineering, architecture, and physics. It is used to calculate the surface area of objects like pipes, tanks, and containers, as well as to determine the amount of material needed for construction. In physics, it is used to calculate the surface area of objects in order to understand their properties and behavior.

• Calculus
Replies
5
Views
585
• Calculus
Replies
1
Views
2K
• Calculus
Replies
5
Views
2K
• Calculus
Replies
33
Views
4K
• Calculus
Replies
2
Views
1K
• Calculus
Replies
4
Views
1K
• Calculus
Replies
4
Views
885
• Calculus
Replies
24
Views
3K
• Calculus
Replies
2
Views
952
• Calculus
Replies
8
Views
1K