Area of Surface of Revolution of ln(sec x) 0< x <(pi/6) W/A is no help here

In summary, the conversation involves a person struggling with finding the surface area obtained by rotating a curve around the x-axis. They tried using integration and Wolfram-Alpha, but were unsuccessful. The problem appeared on an exam but only required setting up the integral, not solving it. The person was told it was too difficult to solve within the given time frame. They attempted to solve it using (1/cos x), but it did not work. Another person suggests sketching the curve and using disks to find the volume element. They also ask for clarification on the response given on the exam.
  • #1
MrJones
8
0
I tried doing just the integration and I was totally stumped, and Wolfram-Alpha has been no help at all

Find the surface area obtained by rotating about the x-axis:

ln(sec x) for 0 < x < (pi/6)

This problem appeared on an exam I had, but all that was required was to set up the integral but not solve for it. When I asked why we didn't have to solve for that particular problem, I was told it was too difficult for the amount of time given for the exam.

I have been trying to solve it and initially started by using (1/cos x) to solve, but it was not helpful so far.

Anyone? I appreciate any help!

Mr J
 
Physics news on Phys.org
  • #2
MrJones said:
I tried doing just the integration and I was totally stumped, and Wolfram-Alpha has been no help at all

Find the surface area obtained by rotating about the x-axis:

ln(sec x) for 0 < x < (pi/6)

This problem appeared on an exam I had, but all that was required was to set up the integral but not solve for it. When I asked why we didn't have to solve for that particular problem, I was told it was too difficult for the amount of time given for the exam.

I have been trying to solve it and initially started by using (1/cos x) to solve, but it was not helpful so far.
Did you sketch a graph of the curve being rotated? If you have that, it's pretty easy to set up an integral that represents the volume of the solid formed by rotation. Using disks, what is the volume of the typical volume element?
 
  • #3
MrJones said:
This problem appeared on an exam I had, but all that was required was to set up the integral but not solve for it.
So what was your response on the exam -- how did you set up the integral?
 

1. What is the formula for finding the area of surface of revolution of ln(sec x)?

The formula for finding the area of surface of revolution of ln(sec x) is A = 2π∫f(x)√(1+[f'(x)]^2)dx, where f(x) is the function being revolved around the x-axis and f'(x) is its derivative.

2. How do I determine the limits of integration for ln(sec x) in the given range of 0< x <(pi/6)?

The limits of integration for ln(sec x) in the given range can be determined by setting the given values of x as the upper and lower limits. In this case, the lower limit is 0 and the upper limit is pi/6.

3. Can I use the formula A = πr^2 to find the area of surface of revolution of ln(sec x)?

No, the formula A = πr^2 is used to find the area of a circle, not the area of surface of revolution of a function. The formula for the area of surface of revolution involves integrating the function and its derivative.

4. How do I calculate the derivative of ln(sec x)?

The derivative of ln(sec x) can be calculated using the chain rule and the derivative of sec x, which is sec x * tan x. The derivative of ln(sec x) is 1/sec x * sec x * tan x, which simplifies to tan x.

5. Is there an online calculator or tool that can help me find the area of surface of revolution of ln(sec x)?

Yes, there are several online calculators and tools that can help you find the area of surface of revolution of a given function. Some popular options include Wolfram Alpha and Desmos. However, it is important to understand the formula and process behind the calculation in order to ensure accurate results.

Similar threads

  • Calculus and Beyond Homework Help
Replies
3
Views
742
  • Calculus and Beyond Homework Help
Replies
2
Views
988
  • Calculus and Beyond Homework Help
Replies
3
Views
2K
  • Calculus and Beyond Homework Help
Replies
1
Views
2K
  • Calculus and Beyond Homework Help
Replies
6
Views
853
  • Calculus and Beyond Homework Help
Replies
4
Views
1K
  • Calculus and Beyond Homework Help
Replies
9
Views
698
  • Calculus and Beyond Homework Help
Replies
5
Views
872
  • Calculus and Beyond Homework Help
Replies
13
Views
3K
  • Calculus and Beyond Homework Help
Replies
7
Views
1K
Back
Top