Question

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

Answer 1

Umm. I can draw......but i don't know if i can draw triple integrals... sorry

Answer 2

what's the equation you're dealing with?

Answer 3

one 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)

Answer 4

sorry hope that's understandable, i can't type out a lot of those symbols

Answer 5

is it \[ \int\limits_{0}^{2\pi}\int\limits_{0}^{\sqrt{3}}\int\limits_{3}^{r^2}rdzdrd \theta\]

Answer 6

if not try using the equation button on the bottowm of the chat window

Answer 7

yep everything's right except the last one is (3-r^2) to 0, sorry i forgot to insert the 0

Answer 8

cool, give me one sec.

Answer 9

so 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 (3-r^2) for z. Zero cancels out so now it's\[ \int\limits_{0}^{2\pi}\int\limits_{0}^{\sqrt{3}}3r-r^3drd \Theta\]. Then we integrate wrt r, and sub in root 3

Answer 10

\[\int\limits_{0}^{2\pi}(3/2)r^2-(1/4)r^4d \Theta =\int\limits_{0}^{2\pi}(9/2)-(9/4)d \Theta = \int\limits(9/4)d \Theta\], this = 9pi/2. It's not something you can really draw because it is a hypervolume, but that's the answer though.

Answer 11

awesome thanks so much! glad to know that i actually got the right answer! haha. are you able to help me with one more?

Answer 12

no prob, sure, what's the question,

Answer 13

sweet 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...

Answer 14

\[\int\limits_{-2}^{2} \int\limits_{-\sqrt{4-x ^{2}}}^{\sqrt{4-x ^{2}}}\int\limits_{x ^{2} +y ^{2}}^4 xdzdydx\]

Answer 15

I don't have some of my notes with me, but I think x =rcos(theta) y= rsin(theta) and z=z in cylindrical right?

Answer 16

yes that is right

Answer 17

Then I think there is an extra factor or something like that when integrating too right? like an extra r or something like that?

Answer 18

yes there is an extra r, sorry the box wasn't letting me type anything for awhile

Answer 19

I'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.

Answer 20

that's alright, thanks for your help i really appreciate it!

Answer 21

I 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.

Answer 22

awesome thanks!