## Concentrationalizing one year ago Convert the triple integral into spherical coordinates.

1. Concentrationalizing

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

2. RolyPoly

I think you are integrating in this region. Not quite sure. |dw:1389713649402:dw|

3. Concentrationalizing

Thats what I get, yes. But the problem is more about knowing how to do the conversions, not the actual integration. I was able to get it into cylindrical coordinates, just didnt know how to do it properly with spherical.

4. RolyPoly

|dw:1389713913315:dw|

5. RolyPoly

Obviously, $$0\le r\le 4$$ and $$0\le \theta \le 2\pi$$

6. Concentrationalizing

Mmm....0 <= r <= 2

7. RolyPoly

Epic fail.... Yes Yes... Sorry :(

8. Concentrationalizing

No worries, lol.

9. RolyPoly

|dw:1389714352578:dw|

10. RolyPoly

So, we have $\tan \phi = \frac{2}{4}$$\phi = ... = \phi_{\text{max}}$ $0\le\phi\le\phi_{\text{max}}$

11. Concentrationalizing

Oh wow x_x I feel dumb now, lol. Im so new to these things, so not used to the thought process. For some reason I never thought to just make that triangle from the cone. Okay, so that gets me half of the answer. Now Im trying to see if I can derive the other half.

12. RolyPoly

I am soooo dumb. I consider myself as quite new to this, although not so new... Good luck!

13. Concentrationalizing

So, what Im trying to find out now is how they have two integrals in the answer. p is between 0 and 4sec(phi) for the first integral and phi is in between 0 and arctan(1/2). But then theres a 2nd integral that is added that has the range of p going from 0 to cot(phi)*csc(phi) and phi going from arctan(1/2) to pi/2. Im trying to see how to derive that part now.

14. Concentrationalizing

@RolyPoly think youd be able to figure out this last part, too?

15. RolyPoly

Why does it look so complicated?! :S

16. RolyPoly

What is the $$\phi_{\text{max}}$$ you get?

17. Concentrationalizing

Im not sure how id get a phi max here. I just know that phi would be arctan(1/2), im not sure how to find another value for it.

18. RolyPoly

$\phi_{\text{max}} = \tan^{-1}(\frac{1}{2})$ And the range of $$\phi$$ is $$0\le\phi\le\phi_{\text{max}}$$

19. Concentrationalizing

Yes, that part I have, but then there is a second integral. Here, this is what the answer says: $\int\limits_{0}^{2 \pi}\int\limits_{0}^{\arctan(1/2)}\int\limits_{0}^{4 \sec \phi} \rho^{3} \sin^{2} \phi \cos \theta d \rho d \phi d \theta + \int\limits_{0}^{2 \pi}\int\limits_{\arctan(1/2)}^{\frac{ \pi }{ 2 }}\int\limits_{0}^{\cot \phi \csc \phi} \rho^{3} \sin^{2} \phi \cos \phi d \rho d \phi d \theta$

20. RolyPoly

Must you write in this.... ugly form?

21. Concentrationalizing

Im sorry. Im just trying to answer this question......

22. RolyPoly

What is the exact question?

23. Concentrationalizing

To convert the triple integral into cylindrical and spherical coordinates and then compute the integral. I was just having trouble understanding the answer for the spherical coordinates conversion. Sure I can see the answer, but Im not sure how it was derived. The triangle on the cone earlier helped me understand the first integral, but I dont understand where the second integral comes from now.

24. RolyPoly

Actually you need not write that ugly thing, I think. Let me double check something first.

25. Concentrationalizing

alright

26. RolyPoly

Do you agree that this is the region? |dw:1389717244057:dw|

27. Concentrationalizing

I figured it was the cone portion, yes.

28. RolyPoly

(The cone part)

29. RolyPoly

Ok, then no matter in which coordinate system, the cone is the region we want to integrate, right?

30. Concentrationalizing

Yes. I think I can integrate it okay, but I just need to know how it got what it did for the spherical coordinate interpretation.

31. RolyPoly

Do you have problems in understanding why $$\rho$$ changes from 0 to 2 and $$\theta$$ changes from 0 to $$2\pi$$?

32. RolyPoly

And $$\phi$$ changes from 0 to $$\tan^{-1}(\frac{1}{2})$$ as well.

33. Concentrationalizing

I dont understand the second integral in the answer I posted. How rho goes from 0 to cot(phi)csc(phi) and how phi goes from arctan(1/2) to pi/2

34. RolyPoly

Must you use that to integrate? It is really ugly and not necessary to do that step..

35. Concentrationalizing

Its not required to integrate it, you just have to know how to derive it. The problem can be integrated using any coordinate system, but you still have to know how to convert it to all of them.

36. RolyPoly

What if I can integrating it using spherical coordinates without getting to that step?

37. RolyPoly

I mean to write that integral in spherical coordinates.

38. Concentrationalizing

I just need to know where that second integral even comes from. I mean sure, if you know how to do it another way thats fine, but Id rather understand the answer first.

39. RolyPoly

If you insist, that's fine. Are you sure that you type that correctly? $\int\limits_{0}^{2 \pi}\int\limits_{0}^{\arctan(1/2)}\int\limits_{0}^{4 \sec \phi} \rho^{3} \sin^{2} \phi \cos \theta d \rho d \phi d \theta + \int\limits_{0}^{2 \pi}\int\limits_{\arctan(1/2)}^{\frac{ \pi }{ 2 }}\int\limits_{0}^{\cot \phi \csc \phi} \rho^{3} \sin^{2} \phi \cos \phi d \rho d \phi d \theta$?

40. Concentrationalizing

Yes.

41. RolyPoly

For changing the coordinates, all you need is to find the new limits and the Jacobian. I just don't understand why you insist on understand that answer instead of finding the limits by drawing a picture, which is way easier. Are you sure that it is $$\rho^3\sin^2\phi\cos\phi$$ for the second integral?

42. Concentrationalizing

The jacobian is in the section after the one Im in. And yeah, it should be cos theta for the second integral, my bad.

43. RolyPoly

How do you know you need to multiply $$\rho ^2 \sin\phi$$ to the integral then?

44. Concentrationalizing

Its in the basic conversion to spherical coordinates. There is no explanation really as to why its that, but it says to add an extra r to cylindrical and an extra p^2sin(phi) to spherical. Perhaps the jacobian explains why, but im not there yet.

45. experimentX

let $f(x) = \int_{-\sqrt{4-x^2}}^{-\sqrt{4-x^2}} \int_{x^2+y^2}^4 x dy dz$ first of all note that $$f(-x) = -f(x)$$ $f(-x) = \int_{-\sqrt{4-(-x)^2}}^{-\sqrt{4-(-x)^2}} \int_{(-x)^2+y^2}^4 (-x) dy dz = -\int_{-\sqrt{4-x^2}}^{-\sqrt{4-x^2}} \int_{x^2+y^2}^4 x dy dz = -f(x)$ for odd functions $\int_{-a}^a f(x) dx = 0$

46. Concentrationalizing

Well, thats the integral evaluated yes, but I guess all I really needed to understand was that spherical answer I posted. How that conversion was derived. In the long run it may not be important, but I at least want to understand it since it was one of my problems.

47. RolyPoly

Consider the black triangle |dw:1389718930585:dw|

48. RolyPoly

We get $\cos\phi = \frac{4}{r}$$r=\frac{4}{\cos\phi}=4 \sec\phi$

49. Concentrationalizing

Mhm, thats rho in the first integral, I managed that part, lol.

50. RolyPoly

And $$\phi$$ in the first part |dw:1389719149953:dw|

51. Concentrationalizing

Yes, arctan(1/2) because of the sides of the black triangle you drew.

52. Concentrationalizing

I gotta go :/ If either of you can explain that 2nd integral in the answer I posted, then please do and post it for me. Thanks for your time.

53. RolyPoly

@Spacelimbus Sir, sorry for tagging you here. Here's the problem: Show that $\int\limits_{-2}^{2}\int\limits_{-\sqrt{4-x^{2}}}^{\sqrt{4-x^{2}}}\int\limits_{x^{2}+y^{2}}^{4}xdzdydx$$=\int\limits_{0}^{2 \pi}\int\limits_{0}^{\arctan(1/2)}\int\limits_{0}^{4 \sec \phi} \rho^{3} \sin^{2} \phi \cos \theta d \rho d \phi d \theta + \int\limits_{0}^{2 \pi}\int\limits_{\arctan(1/2)}^{\frac{ \pi }{ 2 }}\int\limits_{0}^{\cot \phi \csc \phi} \rho^{3} \sin^{2} \phi \cos \phi d \rho d \phi d \theta$ We have problem in understanding the second integral on the right. Please give us if possible. Thanks in advance!

54. RolyPoly

Amendment to the fifth line. The fourth and the fifth line should be $\int\limits_{-2}^{2}\int\limits_{-\sqrt{4-x^{2}}}^{\sqrt{4-x^{2}}}\int\limits_{x^{2}+y^{2}}^{4}xdzdydx$$=\int\limits_{0}^{2 \pi}\int\limits_{0}^{\arctan(1/2)}\int\limits_{0}^{4 \sec \phi} \rho^{3} \sin^{2} \phi \cos \theta d \rho d \phi d \theta + \int\limits_{0}^{2 \pi}\int\limits_{\arctan(1/2)}^{\frac{ \pi }{ 2 }}\int\limits_{0}^{\cot \phi \csc \phi} \rho^{3} \sin^{2} \phi \cos \theta d \rho d \phi d \theta$

55. Spacelimbus

I will take a look into this in a few, seems to take a while to play with the limits of integration to get it in the desired form.

56. RolyPoly

I also have to go now. Will check this post again as soon as possible.

57. experimentX

|dw:1389729705792:dw|

58. experimentX

|dw:1389729892139:dw||dw:1389729934880:dw| solve $x^2 + y^2 + z^2 =r^2 \\ z = x^2 +y^2$ for 'r' and obtain $$r = \cot(\phi) \csc(\phi)$$ easy way, since it is symmetric on theta, put theta = 0. then put x=r sin(phi), z=r cos(phi) and solve.

59. Concentrationalizing

Alrighty. So why is the second integral necessary? Iwouldve thought that just the first integral wouldve covered it all, but for some reason two integrals were needed. Do you know why? @experimentX ?

60. experimentX

this is due to two different values of 'r' in two different regions. for one region you have 4sec(phi) and for other you have cot(phi)csc(phi) |dw:1389785515049:dw||dw:1389785535697:dw|

61. experimentX

note that in first region r goes from to to .. that straight line, on the second region ... r goes from 0 to that parabola.

62. Mendicant_Bias

What we're graphing is a paraboloid with a circular cross-section, right? The thing that initially confused me, looking at the first few posts, is that a cone is depicted.