Convert the triple integral into spherical coordinates.

- anonymous

Convert the triple integral into spherical coordinates.

- Stacey Warren - Expert brainly.com

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- schrodinger

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- anonymous

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

- anonymous

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

- anonymous

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.

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## More answers

- anonymous

|dw:1389713913315:dw|

- anonymous

Obviously, \(0\le r\le 4\) and \(0\le \theta \le 2\pi\)

- anonymous

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

- anonymous

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

- anonymous

No worries, lol.

- anonymous

|dw:1389714352578:dw|

- anonymous

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

- anonymous

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.

- anonymous

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

- anonymous

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.

- anonymous

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

- anonymous

Why does it look so complicated?! :S

- anonymous

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

- anonymous

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.

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

What is the exact question?

- anonymous

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.

- anonymous

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

- anonymous

alright

- anonymous

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

- anonymous

I figured it was the cone portion, yes.

- anonymous

(The cone part)

- anonymous

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

- anonymous

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.

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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.

- anonymous

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

- anonymous

I mean to write that integral in spherical coordinates.

- anonymous

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.

- anonymous

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

- anonymous

Yes.

- anonymous

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?

- anonymous

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

- anonymous

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

- anonymous

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.

- 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
\[ \int_{-a}^a f(x) dx = 0\]

- anonymous

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.

- anonymous

Consider the black triangle
|dw:1389718930585:dw|

- anonymous

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

- anonymous

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

- anonymous

And \(\phi\) in the first part
|dw:1389719149953:dw|

- anonymous

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

- anonymous

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.

- anonymous

@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!

- anonymous

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

- anonymous

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.

- anonymous

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

- experimentX

|dw:1389729705792:dw|

- 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.

- anonymous

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 ?

- 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|

- 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.

- 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.

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