# Areas of Surfaces of Revolution

• Jarvis88
In summary, the problem asks to find the area of the surface generated by revolving the curve y=√x+1, 1≤ x ≤5, about the x-axis. The attempted solution involved using the formula S= ∫ 51 2πy (√1+(dydx)2) dx, but the learner got stuck while trying to use substitution. They made an algebra error while simplifying the square root and also used the incorrect notation for the function. With the help of Chegg.com, they were able to correct their mistakes and find the correct solution.
Jarvis88

## Homework Statement

Find the area of the surface generated by revolving the curve y=√x+1, 1≤ x ≤5, about the x-axis.

I'm stuck trying to figure out how I can use substitution...if I am even able. I was trying to rewrite 1 as 4(x+1)/4(x+1) but still can't seem to get the right terms to cancel to allow me to continue further.

## Homework Equations

S= ∫ 51 2πy (√1+(dydx)2) dx

3. The Attempt at a Solution

Jarvis88 said:

## Homework Statement

Find the area of the surface generated by revolving the curve y=√x+1, 1≤ x ≤5, about the x-axis.

I'm stuck trying to figure out how I can use substitution...if I am even able. I was trying to rewrite 1 as 4(x+1)/4(x+1) but still can't seem to get the right terms to cancel to allow me to continue further.

## Homework Equations

S= ∫ 51 2πy (√1+(dydx)2) dx

3. The Attempt at a Solution

View attachment 111983
I am puzzled by what you did in the 3rd line. In the line above that you have dy/dx, but then switched in the 3rd line to dx/dy. Why?

I'm sorry, I was doing a problem previous to this one where it was dx/dy. I meant to use dy/dx throughout the problem.

If it helps, this is the answer I found on Chegg.com. However, I don't understand how they got from the 1st step to the 2nd step. I've tried it a few different ways and have yet to end up with what they have. I also don't fully trust sites like that as they are not always correct.

Jarvis88 said:
View attachment 111989

If it helps, this is the answer I found on Chegg.com. However, I don't understand how they got from the 1st step to the 2nd step.
They use ## \sqrt a \sqrt b = \sqrt{ab}##.

Ok. I'm getting tired so maybe I've made a silly algebra error somewhere, but I'm getting crazy numbers. I must not be doing an algebra manipulation correctly, but I can't seem to figure out where I've gone wrong in my simplification of the square root.
LCKurtz said:
They use ## \sqrt a \sqrt b = \sqrt{ab}##.
\

You have an algebra mistake at the end of the 2nd line on the right half of your work.

##2\pi \int_1^5\frac{\sqrt{4(x + 1) + 1}}{2}dx##
You canceled the 4 in the radical with the 2 in the denominator, leaving a factor of 2 in the radical. This is actually two mistakes.
1) ##\frac{ab + c}{a} \ne b + c## That's essentially what you did.
2) ##\frac{\sqrt{4}} 2 \ne \sqrt{2}##

BTW, your problem description threw me off:
Jarvis88 said:
Find the area of the surface generated by revolving the curve y=√x+1, 1≤ x ≤5, about the x-axis.
From your work, the function apparently is ##f(x) = \sqrt{x + 1}##. In linear form, without LaTeX, what you should have written is f(x) = √(x + 1).

Mark44 said:
BTW, your problem description threw me off:

From your work, the function apparently is ##f(x) = \sqrt{x + 1}##. In linear form, without LaTeX, what you should have written is f(x) = √(x + 1).

Honestly, I'm pretty new to all this so I'm not familiar with how to use LaTeX properly. In the problem description I was quoting the description verbatim from my assignment. Thank you for the help! I don't know how I made those silly algebra errors...

Jarvis88 said:
Honestly, I'm pretty new to all this so I'm not familiar with how to use LaTeX properly.
See our tutorial -- https://www.physicsforums.com/help/latexhelp/.

Everything you did on paper can be done using LaTeX. We prefer that members post their work directly here rather than as pasted images, as it makes it easier for helpers to identify exactly where errors are made. Also, many of the images that get posted are of poor quality and hard to read. Yours were easy to read, though.

Jarvis88
Mark44 said:
See our tutorial -- https://www.physicsforums.com/help/latexhelp/.

Everything you did on paper can be done using LaTeX. We prefer that members post their work directly here rather than as pasted images, as it makes it easier for helpers to identify exactly where errors are made. Also, many of the images that get posted are of poor quality and hard to read. Yours were easy to read, though.

Thank you! Will do!

## 1. What is a surface of revolution?

A surface of revolution is a three-dimensional shape that is created by rotating a two-dimensional curve around an axis. This axis can be any line, and the resulting shape is symmetrical around this axis.

## 2. What are some examples of surfaces of revolution?

Some common examples of surfaces of revolution include cones, spheres, cylinders, and tori (donuts). These shapes can be created by rotating a line, circle, or ellipse around an axis.

## 3. How do you calculate the area of a surface of revolution?

The formula for calculating the area of a surface of revolution depends on the type of curve being rotated. For example, the area of a sphere can be calculated using the formula A = 4πr^2, while the area of a cylinder can be calculated using the formula A = 2πrh, where r is the radius and h is the height of the cylinder.

## 4. What real-world applications use surfaces of revolution?

Surfaces of revolution have many practical applications, such as in architecture, engineering, and product design. They are also important in mathematics and physics, particularly in the study of rotational motion and fluid dynamics.

## 5. Can surfaces of revolution have irregular shapes?

Yes, surfaces of revolution can have irregular shapes as long as they are created by rotating a curve around an axis. However, these shapes may be more difficult to calculate the area of compared to regular geometric shapes.

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