Finding Volume Using the Sine Function and Disc Method

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SUMMARY

The volume generated by revolving one arch of the curve y = 5 sin(x) about the x-axis is calculated using the disc method. The correct formula for the volume is V = π∫[y]^2 dx, where y = 5 sin(x). The integral should be set up as V = π∫[5 sin(x)]^2 from 0 to π, leading to the correct evaluation of the volume. The initial mistake was neglecting to square the coefficient 5 in the radius formula.

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  • Understanding of integral calculus
  • Familiarity with the disc method for volume calculation
  • Knowledge of trigonometric functions and identities
  • Ability to perform definite integrals
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  • Practice solving integrals involving trigonometric functions
  • Learn about the application of trigonometric identities in integration
  • Explore advanced volume calculation techniques, such as the washer method
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cathy
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Homework Statement

Find the volume generated by revolving one arch of the curve y = 5 sin(x) about the x-axis.




The attempt at a solution

So I figured this would create a disc so I would have to use that the volume is ∏r^2 where r=sinx, r^2= (sin(x))^2 and that the way I should set this up is as shown:

5∏∫[sin(x)]^2 from 0 to pi. And then I would replace that with the trig identities and so on. However, this is not giving me the correct answer. Is this integral wrong?
 
Last edited:
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cathy said:
Homework Statement

Find the volume generated by revolving one arch of the curve y = 5 sin(x) about the x-axis.




The attempt at a solution

So I figured this would create a disc so I would have to use that the volume is ∏r^2 where r=sinx, r^2= (sin(x))^2 and that the way I should set this up is as shown:

5∏∫[sin(x)]^2 from 0 to pi. And then I would replace that with the trig identities and so on. However, this is not giving me the correct answer. Is this integral wrong?

r = 5sin x. You forgot to square the 5.

And not to be pedantic, but ∏r^2 is not the volume of anything. The enclosed volume of a cylinder is ##\pi r^2h##, where h is the height. Applied here the volume integral is ##\pi\int_a^by^2dx##.
 
Last edited:
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Oh! Thank you so much. I did this problem a million times, and it was such a simple mistake. :)
 

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