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Concentrationalizing
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Convert the triple integral into spherical coordinates.
 10 months ago
 10 months ago
Concentrationalizing Group Title
Convert the triple integral into spherical coordinates.
 10 months ago
 10 months ago

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Concentrationalizing Group TitleBest ResponseYou've already chosen the best response.0
\[\int\limits_{2}^{2}\int\limits_{\sqrt{4x^{2}}}^{\sqrt{4x^{2}}}\int\limits_{x^{2}+y^{2}}^{4}xdzdydx\]
 10 months ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.1
I think you are integrating in this region. Not quite sure. dw:1389713649402:dw
 10 months ago

Concentrationalizing Group TitleBest ResponseYou've already chosen the best response.0
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.
 10 months ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.1
dw:1389713913315:dw
 10 months ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.1
Obviously, \(0\le r\le 4\) and \(0\le \theta \le 2\pi\)
 10 months ago

Concentrationalizing Group TitleBest ResponseYou've already chosen the best response.0
Mmm....0 <= r <= 2
 10 months ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.1
Epic fail.... Yes Yes... Sorry :(
 10 months ago

Concentrationalizing Group TitleBest ResponseYou've already chosen the best response.0
No worries, lol.
 10 months ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.1
dw:1389714352578:dw
 10 months ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.1
So, we have \[\tan \phi = \frac{2}{4}\]\[\phi = ... = \phi_{\text{max}}\] \[0\le\phi\le\phi_{\text{max}}\]
 10 months ago

Concentrationalizing Group TitleBest ResponseYou've already chosen the best response.0
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.
 10 months ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.1
I am soooo dumb. I consider myself as quite new to this, although not so new... Good luck!
 10 months ago

Concentrationalizing Group TitleBest ResponseYou've already chosen the best response.0
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.
 10 months ago

Concentrationalizing Group TitleBest ResponseYou've already chosen the best response.0
@RolyPoly think youd be able to figure out this last part, too?
 10 months ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.1
Why does it look so complicated?! :S
 10 months ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.1
What is the \(\phi_{\text{max}}\) you get?
 10 months ago

Concentrationalizing Group TitleBest ResponseYou've already chosen the best response.0
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.
 10 months ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.1
\[\phi_{\text{max}} = \tan^{1}(\frac{1}{2})\] And the range of \(\phi\) is \(0\le\phi\le\phi_{\text{max}}\)
 10 months ago

Concentrationalizing Group TitleBest ResponseYou've already chosen the best response.0
Yes, that part I have, but then there is a second integral. Here, this is what the answer says: \[\int\limits_{0}^{2 \pi}\int\limits_{0}^{\arctan(1/2)}\int\limits_{0}^{4 \sec \phi} \rho^{3} \sin^{2} \phi \cos \theta d \rho d \phi d \theta + \int\limits_{0}^{2 \pi}\int\limits_{\arctan(1/2)}^{\frac{ \pi }{ 2 }}\int\limits_{0}^{\cot \phi \csc \phi} \rho^{3} \sin^{2} \phi \cos \phi d \rho d \phi d \theta\]
 10 months ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.1
Must you write in this.... ugly form?
 10 months ago

Concentrationalizing Group TitleBest ResponseYou've already chosen the best response.0
Im sorry. Im just trying to answer this question......
 10 months ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.1
What is the exact question?
 10 months ago

Concentrationalizing Group TitleBest ResponseYou've already chosen the best response.0
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.
 10 months ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.1
Actually you need not write that ugly thing, I think. Let me double check something first.
 10 months ago

Concentrationalizing Group TitleBest ResponseYou've already chosen the best response.0
alright
 10 months ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.1
Do you agree that this is the region? dw:1389717244057:dw
 10 months ago

Concentrationalizing Group TitleBest ResponseYou've already chosen the best response.0
I figured it was the cone portion, yes.
 10 months ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.1
(The cone part)
 10 months ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.1
Ok, then no matter in which coordinate system, the cone is the region we want to integrate, right?
 10 months ago

Concentrationalizing Group TitleBest ResponseYou've already chosen the best response.0
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.
 10 months ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.1
Do you have problems in understanding why \(\rho\) changes from 0 to 2 and \(\theta\) changes from 0 to \(2\pi\)?
 10 months ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.1
And \(\phi\) changes from 0 to \(\tan^{1}(\frac{1}{2})\) as well.
 10 months ago

Concentrationalizing Group TitleBest ResponseYou've already chosen the best response.0
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
 10 months ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.1
Must you use that to integrate? It is really ugly and not necessary to do that step..
 10 months ago

Concentrationalizing Group TitleBest ResponseYou've already chosen the best response.0
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.
 10 months ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.1
What if I can integrating it using spherical coordinates without getting to that step?
 10 months ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.1
I mean to write that integral in spherical coordinates.
 10 months ago

Concentrationalizing Group TitleBest ResponseYou've already chosen the best response.0
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.
 10 months ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.1
If you insist, that's fine. Are you sure that you type that correctly? \[\int\limits_{0}^{2 \pi}\int\limits_{0}^{\arctan(1/2)}\int\limits_{0}^{4 \sec \phi} \rho^{3} \sin^{2} \phi \cos \theta d \rho d \phi d \theta + \int\limits_{0}^{2 \pi}\int\limits_{\arctan(1/2)}^{\frac{ \pi }{ 2 }}\int\limits_{0}^{\cot \phi \csc \phi} \rho^{3} \sin^{2} \phi \cos \phi d \rho d \phi d \theta\]?
 10 months ago

Concentrationalizing Group TitleBest ResponseYou've already chosen the best response.0
Yes.
 10 months ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.1
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?
 10 months ago

Concentrationalizing Group TitleBest ResponseYou've already chosen the best response.0
The jacobian is in the section after the one Im in. And yeah, it should be cos theta for the second integral, my bad.
 10 months ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.1
How do you know you need to multiply \(\rho ^2 \sin\phi\) to the integral then?
 10 months ago

Concentrationalizing Group TitleBest ResponseYou've already chosen the best response.0
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.
 10 months ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
let \[ f(x) = \int_{\sqrt{4x^2}}^{\sqrt{4x^2}} \int_{x^2+y^2}^4 x dy dz\] first of all note that \( f(x) = f(x) \) \[ f(x) = \int_{\sqrt{4(x)^2}}^{\sqrt{4(x)^2}} \int_{(x)^2+y^2}^4 (x) dy dz = \int_{\sqrt{4x^2}}^{\sqrt{4x^2}} \int_{x^2+y^2}^4 x dy dz = f(x)\] for odd functions \[ \int_{a}^a f(x) dx = 0\]
 10 months ago

Concentrationalizing Group TitleBest ResponseYou've already chosen the best response.0
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.
 10 months ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.1
Consider the black triangle dw:1389718930585:dw
 10 months ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.1
We get \[\cos\phi = \frac{4}{r}\]\[r=\frac{4}{\cos\phi}=4 \sec\phi\]
 10 months ago

Concentrationalizing Group TitleBest ResponseYou've already chosen the best response.0
Mhm, thats rho in the first integral, I managed that part, lol.
 10 months ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.1
And \(\phi\) in the first part dw:1389719149953:dw
 10 months ago

Concentrationalizing Group TitleBest ResponseYou've already chosen the best response.0
Yes, arctan(1/2) because of the sides of the black triangle you drew.
 10 months ago

Concentrationalizing Group TitleBest ResponseYou've already chosen the best response.0
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.
 10 months ago

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

RolyPoly Group TitleBest ResponseYou've already chosen the best response.1
Amendment to the fifth line. The fourth and the fifth line should be \[\int\limits_{2}^{2}\int\limits_{\sqrt{4x^{2}}}^{\sqrt{4x^{2}}}\int\limits_{x^{2}+y^{2}}^{4}xdzdydx\]\[=\int\limits_{0}^{2 \pi}\int\limits_{0}^{\arctan(1/2)}\int\limits_{0}^{4 \sec \phi} \rho^{3} \sin^{2} \phi \cos \theta d \rho d \phi d \theta + \int\limits_{0}^{2 \pi}\int\limits_{\arctan(1/2)}^{\frac{ \pi }{ 2 }}\int\limits_{0}^{\cot \phi \csc \phi} \rho^{3} \sin^{2} \phi \cos \theta d \rho d \phi d \theta\]
 10 months ago

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.0
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.
 10 months ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.1
I also have to go now. Will check this post again as soon as possible.
 10 months ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
dw:1389729705792:dw
 10 months ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
dw:1389729892139:dwdw: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.
 10 months ago

Concentrationalizing Group TitleBest ResponseYou've already chosen the best response.0
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 ?
 10 months ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
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:dwdw:1389785535697:dw
 10 months ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
note that in first region r goes from to to .. that straight line, on the second region ... r goes from 0 to that parabola.
 10 months ago

Mendicant_Bias Group TitleBest ResponseYou've already chosen the best response.0
What we're graphing is a paraboloid with a circular crosssection, right? The thing that initially confused me, looking at the first few posts, is that a cone is depicted.
 25 days ago
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