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 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.
 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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Outkast3r09
 one year ago
Best ResponseYou've already chosen the best response.0dw:1363165096949:dw

Outkast3r09
 one year ago
Best ResponseYou've already chosen the best response.0dw:1363165150768:dw

Outkast3r09
 one year ago
Best ResponseYou've already chosen the best response.0the depends on how you're rotating it

MarcLeclair
 one year ago
Best ResponseYou've already chosen the best response.0Yeah I rotate it i get an irregularly shape object. Its upon the x axis

MarcLeclair
 one year ago
Best ResponseYou've already chosen the best response.0I have 3 methods for those, disk, cylindrical shell, and washers but I mean I used cylindrical shell in this one.

Outkast3r09
 one year ago
Best ResponseYou've already chosen the best response.0wait you did it around the x axis?

MarcLeclair
 one year ago
Best ResponseYou've already chosen the best response.0Yeah I thought I mentionned it :/

Outkast3r09
 one year ago
Best ResponseYou've already chosen the best response.0alright so what did your integral look like before you worked with it

Outkast3r09
 one year ago
Best ResponseYou've already chosen the best response.0\[2\pi\int_{\pi/2}^\pi sin(x)0dx=2\pi\int_{\pi/2}^\pi sin(x)dx\]\]

MarcLeclair
 one year ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{\pi/2}^{\pi} 2\pi(x)(sinx) dx = 2\pi \int\limits_{\pi/2}^{\pi} xsin(dx) = (cosx)(x)  \int\limits_{\pi/2}^{\pi}cosxdx\]

MarcLeclair
 one year ago
Best ResponseYou've already chosen the best response.0I used integration by part with cylindrical shell :/

MarcLeclair
 one year ago
Best ResponseYou've already chosen the best response.0assuming u=x dv= sinx du= dx and v= cosx

agent0smith
 one year ago
Best ResponseYou've already chosen the best response.1Use the disk method, then you'll have sin^2 x but you can easily integrate that by using the identity for sin^2 x.

MarcLeclair
 one year ago
Best ResponseYou've already chosen the best response.0So can I ask how you differentiate between the cylindrical shell, disk and washer method?

agent0smith
 one year ago
Best ResponseYou've already chosen the best response.1from the disk method, R^2 is sin^2 x, so:\[V = \pi \int\limits_{a}^{b} \sin^2 x dx\] and \[\sin^2 x = \frac{ 1  \cos 2x }{2 }\]

MarcLeclair
 one year ago
Best ResponseYou've already chosen the best response.0But in this case the cylindrical shell would work.

agent0smith
 one year ago
Best ResponseYou've already chosen the best response.1Use 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.

agent0smith
 one year ago
Best ResponseYou've already chosen the best response.1In 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 xaxis. 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.

MarcLeclair
 one year ago
Best ResponseYou've already chosen the best response.0aahhhhh Okay I get it because I have no radius I can use when my function is form of x. Thanks a lot!

agent0smith
 one year ago
Best ResponseYou've already chosen the best response.1dw: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 yaxis... 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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