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

1. anonymous

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

2. anonymous

what's the equation you're dealing with?

3. anonymous

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)

4. anonymous

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

5. anonymous

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

6. anonymous

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

7. anonymous

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

8. anonymous

cool, give me one sec.

9. anonymous

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

10. anonymous

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

11. anonymous

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

12. anonymous

no prob, sure, what's the question,

13. anonymous

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...

14. anonymous

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

15. anonymous

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?

16. anonymous

yes that is right

17. anonymous

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

18. anonymous

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

19. anonymous

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.

20. anonymous

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

21. anonymous

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.

22. anonymous

awesome thanks!