Trig integrals and finding volume.

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Homework Help Overview

The problem involves finding the volume obtained by rotating the curve defined by y=sin(x) around the x-axis, within the bounds of x from π/2 to π. The original poster indicates the use of the cylindrical disk method for this calculation.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • The original poster attempts to apply the cylindrical disk method and has performed integration by parts but is confused by discrepancies with the answer key. Other participants question the axis of rotation and clarify the formula for the disk method.

Discussion Status

Participants are exploring different methods for calculating the volume, discussing the appropriateness of the disk versus cylindrical shell methods. There is an ongoing examination of the conditions under which each method is applicable, with no clear consensus reached yet.

Contextual Notes

There is some confusion regarding the choice between the disk and cylindrical shell methods, as well as the implications of the axis of rotation on the volume calculation. The original poster expresses uncertainty about how to approach the problem differently based on the answer key's differing work.

MarcL
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Homework Statement



Find the volume obtained by rotating the following curves bounded by:

y=sinx y=0 pi/2≤ x ≤ pi

Homework Equations



I know I have to use the cylindrial disk method so ∫2pi(x)f(x)dx.

The Attempt at a Solution



I did the following ∫(pi/2 to pi) 2pi (x) sinxdx
I factor out the 2pi and I am left with ∫(x)(sinx)(dx) always from pi/2 to pi
I then used integration by part with u=x du=dx and dv=sinx v=-cosx

therefore uv -∫vdu --> -(xcos(x)) (from pi/2 to pi) - ∫-cos(x)dx from pi/2 to pi

However my answer key shows a completely different work / answer.. I don't know how to approach the question differently
 
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Which axis are you rotating about?
Also if this is about the x-axis, all you have to do is use:
[tex]\int_a^b \pi f(x)^{2}dx[/tex]
 
Last edited:
Wait but I thought a cylindrical would be 2pi(x) f(x) dx . X for the radius and f(x) for the height. And yes it was about the x-axis
 
What's the area of a disk?
 
It would be pi (x)^2. And this was a stupid question. I meant cylindrical shell... So my teacher used the disk method. How would I be able to decide between cylindrical shell, the disk and washer method :/ ( I mean washer there would have to be a "hole" of some sort no?)
 
Nope you can use the shell or disk method anytime, sometimes (like exams) there is only one way to do it because the other way will make the integral really nasty.
 

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