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anonymous
 5 years ago
hey! anyone out there good at drawing triple integrals or good in dzdr(dtheta) plane?
anonymous
 5 years ago
hey! anyone out there good at drawing triple integrals or good in dzdr(dtheta) plane?

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Umm. I can draw......but i don't know if i can draw triple integrals... sorry

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what's the equation you're dealing with?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0one is the triple integral of (2pi to 0), (square root of 3 to 0), then (3 r squared) with the inside of the integral as rdzdr(dtheta)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sorry hope that's understandable, i can't type out a lot of those symbols

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0is it \[ \int\limits_{0}^{2\pi}\int\limits_{0}^{\sqrt{3}}\int\limits_{3}^{r^2}rdzdrd \theta\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if not try using the equation button on the bottowm of the chat window

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yep everything's right except the last one is (3r^2) to 0, sorry i forgot to insert the 0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0cool, give me one sec.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so first we integrate with respect to z, so now it's going to be\[\int\limits_{0}^{2\pi}\int\limits_{0}^{\sqrt{3}}rzdrd \Theta\], then we sub in (3r^2) for z. Zero cancels out so now it's\[ \int\limits_{0}^{2\pi}\int\limits_{0}^{\sqrt{3}}3rr^3drd \Theta\]. Then we integrate wrt r, and sub in root 3

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{0}^{2\pi}(3/2)r^2(1/4)r^4d \Theta =\int\limits_{0}^{2\pi}(9/2)(9/4)d \Theta = \int\limits(9/4)d \Theta\], this = 9pi/2. It's not something you can really draw because it is a hypervolume, but that's the answer though.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0awesome thanks so much! glad to know that i actually got the right answer! haha. are you able to help me with one more?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no prob, sure, what's the question,

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sweet it has to do with converting a triple integral with rectangular coordinates to cylindrical coordinates, i'll try to type it out in the equation thing...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{2}^{2} \int\limits_{\sqrt{4x ^{2}}}^{\sqrt{4x ^{2}}}\int\limits_{x ^{2} +y ^{2}}^4 xdzdydx\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I don't have some of my notes with me, but I think x =rcos(theta) y= rsin(theta) and z=z in cylindrical right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Then I think there is an extra factor or something like that when integrating too right? like an extra r or something like that?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes there is an extra r, sorry the box wasn't letting me type anything for awhile

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'm looking at it, and I don't want to give you bad advice If I had my notes with me, I could help, but the thing that has got me is the integrating operators, I don't think we can just put r^2 sin(thetat)cos(theta)*r d(theta) in there. and I know we need a dr, but without my notes I'm drawing it up. Sorry about that one.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0that's alright, thanks for your help i really appreciate it!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I think I've got it, sorry, it's looks like you have to sub rcos(thetat) for x, then times the extra r = r^2cos(theta)dzdrd(theta). your bounds for dz change from r^2 to 4, then dr 2, to 2, and d(theta) from 0 to 2pi, to complete the circle the volume is over. Then it's pretty much like before.
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