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MarcLeclair
Group Title
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
 one year ago
MarcLeclair Group Title
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
 one year ago

This Question is Closed

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

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

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

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

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

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

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

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

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

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

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

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

agent0smith Group TitleBest ResponseYou've already chosen the best response.1
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.
 one year ago

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

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

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

agent0smith Group TitleBest ResponseYou've already chosen the best response.1
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.
 one year ago

MarcLeclair Group TitleBest ResponseYou've already chosen the best response.0
should* no?
 one year ago

agent0smith Group TitleBest ResponseYou've already chosen the best response.1
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 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.
 one year ago

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

agent0smith Group TitleBest ResponseYou've already chosen the best response.1
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 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.
 one year ago
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