Question regading integration (Volume generated)

  • Thread starter Sanosuke Sagara
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In summary, the question asks how to calculate the volume generated by revolving the region S in the first quadrant, bounded by the coordinate axes, the line x=3 and the curve y=\sqrt{1+x2} around the y-axis. The volume is calculated using LateX equation, and the error in the calculation is corrected.
  • #1
Sanosuke Sagara
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I have my question,solution,doubt in the attachment that followed.I hope that anyone will help me figure out this problem.Thanks for anybody that spend some time on this question.
 

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  • #2
Help, I really need somebody to look out this question for me whether I have done wrong or not.I have my solution to the question in the attachment.
 
  • #3
I might be slightly paranoid, but why would I open a Word document from a complete stranger? Besides, there's a perfectly good way to write (mostly) anything you need here on the forum, using LaTeX.
 
  • #4
I don't know how to write equation in LateX form , so I post up the question with attachment.Trust me ,the ataachment don't have virus and won't affected your computer.
 
  • #5
You crashed Word!

--J

(PS. Major accomplishment, but can't help ya' since I can't see it, sorry.)
 
  • #6
can you ask your question in plain old words?
 
  • #7
The question asks to evaluate the volume generated by revolving the region S in the first quadrant, bounded by the coordinate axes , the line x=3 and the curve [itex]y=\sqrt{1+x2}[/itex] around the y-axis.

Check out this thread for LateX:
https://www.physicsforums.com/showthread.php?t=8997&highlight=latex
You can also click the latex equations from other posters to see the exact code used. It's fast and easy to learn this way. I advise making access to the problems easier if you want people to help you.

Anyway, you seem to have a made a slight mistake in calculating the 'total volume'
The volume generated by rotating the rectangle bounded by x=0, y=0, x=3 and y=[itex]\sqrt{10}[/itex] is [itex]3^2\pi \sqrt{10}[/itex], instead of [itex]3 \pi \sqrt{10}[/itex].

The volume use have to subtract is:
[tex]\pi \int_1^{\sqrt{10}}(y^2-1)dy=\pi\left[\frac{10\sqrt{10}}{3}-\sqrt{10}+2/3\right][/tex]
You got that part right.

If you subtract the above from [itex]\pi 9\sqrt{10}[/itex] you get the right answer.
 
Last edited:
  • #8
[itex]\pi 9\sqrt{10}[/itex]
 
  • #9
Soory,I was just trying with the LateX equation and thanks for Galileo for seeking out the question for me and correct my errors.
 

FAQ: Question regading integration (Volume generated)

What is integration?

Integration is a mathematical process used to find the area under a curve or the volume of a three-dimensional shape.

What is the purpose of finding the volume generated by integration?

The purpose of finding the volume generated by integration is to calculate the total amount of space enclosed by a three-dimensional shape, such as a solid object or a region bounded by a curve.

How do you calculate the volume generated using integration?

To calculate the volume generated using integration, you need to first set up an integral that represents the cross-sectional area of the shape, and then integrate it over the desired limits of integration.

What are the different methods of integration?

The different methods of integration include the Fundamental Theorem of Calculus, u-substitution, integration by parts, trigonometric substitution, and partial fractions.

What are some real-world applications of calculating the volume generated by integration?

Calculating the volume generated by integration has many real-world applications, such as calculating the volume of a container, finding the amount of water in a reservoir, and determining the capacity of a fuel tank. It is also used in fields such as engineering, physics, and economics.

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