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MarcLeclair

  • one year ago

Integral question ( finding volume): Find the volume obtained by rotating the following curves bounded by: y=sinx y=0 pi/2≤ x ≤ pi Cylindrical gave me : ∫(pi/2 to pi) 2pi (x) sinxdx but integration by part didn't give me the right answer and the answer key gave me ∫pi(x)^2 dx which I don't understand from where it comes from.

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  1. Outkast3r09
    • one year ago
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    |dw:1363165096949:dw|

  2. Outkast3r09
    • one year ago
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    |dw:1363165150768:dw|

  3. Outkast3r09
    • one year ago
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    the depends on how you're rotating it

  4. MarcLeclair
    • one year ago
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    Yeah I rotate it i get an irregularly shape object. Its upon the x axis

  5. MarcLeclair
    • one year ago
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    I have 3 methods for those, disk, cylindrical shell, and washers but I mean I used cylindrical shell in this one.

  6. Outkast3r09
    • one year ago
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    wait you did it around the x axis?

  7. MarcLeclair
    • one year ago
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    Yeah I thought I mentionned it :/

  8. Outkast3r09
    • one year ago
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    alright so what did your integral look like before you worked with it

  9. Outkast3r09
    • one year ago
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    \[2\pi\int_{\pi/2}^\pi sin(x)-0dx=2\pi\int_{\pi/2}^\pi sin(x)dx\]\]

  10. MarcLeclair
    • one year ago
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    \[\int\limits_{\pi/2}^{\pi} 2\pi(x)(sinx) dx = 2\pi \int\limits_{\pi/2}^{\pi} xsin(dx) = -(cosx)(x) - \int\limits_{\pi/2}^{\pi}-cosxdx\]

  11. MarcLeclair
    • one year ago
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    I used integration by part with cylindrical shell :/

  12. MarcLeclair
    • one year ago
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    assuming u=x dv= sinx du= dx and v= -cosx

  13. agent0smith
    • one year ago
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    Use the disk method, then you'll have sin^2 x but you can easily integrate that by using the identity for sin^2 x.

  14. MarcLeclair
    • one year ago
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    So can I ask how you differentiate between the cylindrical shell, disk and washer method?

  15. agent0smith
    • one year ago
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    from the disk method, R^2 is sin^2 x, so:\[V = \pi \int\limits_{a}^{b} \sin^2 x dx\] and \[\sin^2 x = \frac{ 1 - \cos 2x }{2 }\]

  16. MarcLeclair
    • one year ago
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    But in this case the cylindrical shell would work.

  17. agent0smith
    • one year ago
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    Use whichever method is easier... disk and washer method are one and the same. The washer method is just a disk with a hole in it, ie the inner radius is zero.

  18. MarcLeclair
    • one year ago
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    should* no?

  19. agent0smith
    • one year ago
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    In this case, cylinder method would involve changing your function into the form y = ... because the cylinders will have to be horizontal to rotate around the x-axis. You're going to have x = arcsin y and so \[V = 2 \pi \int\limits_{0}^{1} (y \arcsin y )dy\] ...not worth the hassle when the disk method is ez game.

  20. MarcLeclair
    • one year ago
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    aahhhhh Okay I get it because I have no radius I can use when my function is form of x. Thanks a lot!

  21. agent0smith
    • one year ago
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    |dw:1363166724609:dw| those would be the cylinder cross sections. What i put up there isn't 100% correct for the cylinder method, it should be modified a bit because the cylinders don't touch the y-axis... i think it's more like: \[V = 2 \pi \int\limits\limits_{0}^{1} y( \arcsin y - \pi/2)dy\]or... something... idk, too much effort to figure it out.

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