anonymous
  • anonymous
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.
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
|dw:1363165096949:dw|
anonymous
  • anonymous
|dw:1363165150768:dw|
anonymous
  • anonymous
the depends on how you're rotating it

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anonymous
  • anonymous
Yeah I rotate it i get an irregularly shape object. Its upon the x axis
anonymous
  • anonymous
I have 3 methods for those, disk, cylindrical shell, and washers but I mean I used cylindrical shell in this one.
anonymous
  • anonymous
wait you did it around the x axis?
anonymous
  • anonymous
Yeah I thought I mentionned it :/
anonymous
  • anonymous
alright so what did your integral look like before you worked with it
anonymous
  • anonymous
\[2\pi\int_{\pi/2}^\pi sin(x)-0dx=2\pi\int_{\pi/2}^\pi sin(x)dx\]\]
anonymous
  • anonymous
\[\int\limits_{\pi/2}^{\pi} 2\pi(x)(sinx) dx = 2\pi \int\limits_{\pi/2}^{\pi} xsin(dx) = -(cosx)(x) - \int\limits_{\pi/2}^{\pi}-cosxdx\]
anonymous
  • anonymous
I used integration by part with cylindrical shell :/
anonymous
  • anonymous
assuming u=x dv= sinx du= dx and v= -cosx
agent0smith
  • agent0smith
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.
anonymous
  • anonymous
So can I ask how you differentiate between the cylindrical shell, disk and washer method?
agent0smith
  • agent0smith
from the disk method, R^2 is sin^2 x, so:\[V = \pi \int\limits_{a}^{b} \sin^2 x dx\] and \[\sin^2 x = \frac{ 1 - \cos 2x }{2 }\]
anonymous
  • anonymous
But in this case the cylindrical shell would work.
agent0smith
  • agent0smith
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.
anonymous
  • anonymous
should* no?
agent0smith
  • agent0smith
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.
anonymous
  • anonymous
aahhhhh Okay I get it because I have no radius I can use when my function is form of x. Thanks a lot!
agent0smith
  • agent0smith
|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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