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David1132

Joined: 12 Dec 2005
Posts: 51

Posted: Tue Jul 11, 2006 7:06 pm    Post subject: Advice on messy integral?

I'm struggeling with a weird triple integral:

z / (1 + x^2 + y^2) dxdydz

over the volume defined as x^2 + y^2 + z^2 <= 1 and z >= sqrt(x^2 + y^2)
(Which is the volume that is inside the unit sphere, and also inside the
upside-down flipped cone with the pointy end at origo and it's base at
z=1/sqrt(2) )

I've tried to approach this in several ways, but every time i end up with
_really_ messy solutions, like integrating complex ln functions to hell when
I try cylindrical coordinates and even worse situations with spherical
coordinates.

How would I do to solve this problem in the best, nicest and simplest way?
I really hope anyone can help... I would be very grateful for advice.

/ David
Lynn Kurtz
science forum Guru

Joined: 02 May 2005
Posts: 603

Posted: Tue Jul 11, 2006 8:36 pm    Post subject: Re: Advice on messy integral?

On Tue, 11 Jul 2006 21:06:20 +0200, "David" <dani4965@student.uu.se>
wrote:

 Quote: I'm struggeling with a weird triple integral: z / (1 + x^2 + y^2) dxdydz over the volume defined as x^2 + y^2 + z^2 <= 1 and z >= sqrt(x^2 + y^2) (Which is the volume that is inside the unit sphere, and also inside the upside-down flipped cone with the pointy end at origo and it's base at z=1/sqrt(2) ) I've tried to approach this in several ways, but every time i end up with _really_ messy solutions, like integrating complex ln functions to hell when I try cylindrical coordinates and even worse situations with spherical coordinates. How would I do to solve this problem in the best, nicest and simplest way? I really hope anyone can help... I would be very grateful for advice. / David

The two surfaces intersect where their z values are equal:

2(x^2 + y^2) = 1, or a circle of radius 1/sqrt(2)

Just use cylindrical coordinates:

x^2 + y^2 = r^2, dx dy dz = r dr dtheta dz

where r and theta traverse the circle. You do get a ln in the answer
but there's nothing difficult about it.

--Lynn
Lynn Kurtz
science forum Guru

Joined: 02 May 2005
Posts: 603

Posted: Tue Jul 11, 2006 8:44 pm    Post subject: Re: Advice on messy integral?

On Tue, 11 Jul 2006 20:36:32 GMT, "[Mr.] Lynn Kurtz"
<kurtzDELETE-THIS@asu.edu> wrote:

 Quote: The two surfaces intersect where their z values are equal: 2(x^2 + y^2) = 1, or a circle of radius 1/sqrt(2) Just use cylindrical coordinates: x^2 + y^2 = r^2, dx dy dz = r dr dtheta dz where r and theta traverse the circle. You do get a ln in the answer but there's nothing difficult about it.

Woops, that answer may have been a bit hasty and optimistic....

--Lynn
matt271829-news@yahoo.co.
science forum Guru

Joined: 11 Sep 2005
Posts: 846

Posted: Tue Jul 11, 2006 8:47 pm    Post subject: Re: Advice on messy integral?

David wrote:
 Quote: I'm struggeling with a weird triple integral: z / (1 + x^2 + y^2) dxdydz over the volume defined as x^2 + y^2 + z^2 <= 1 and z >= sqrt(x^2 + y^2) (Which is the volume that is inside the unit sphere, and also inside the upside-down flipped cone with the pointy end at origo and it's base at z=1/sqrt(2) ) I've tried to approach this in several ways, but every time i end up with _really_ messy solutions, like integrating complex ln functions to hell when I try cylindrical coordinates and even worse situations with spherical coordinates. How would I do to solve this problem in the best, nicest and simplest way? I really hope anyone can help... I would be very grateful for advice. / David

As far as I can see, the whole of the cone you describe is inside the
unit sphere, so I don't see why the unit sphere is relevant. There
isn't a typo somewhere is there? Or maybe I'm misunderstanding
something...
David1132

Joined: 12 Dec 2005
Posts: 51

Posted: Tue Jul 11, 2006 9:00 pm    Post subject: Re: Advice on messy integral?

Yes, the cone is inside the unit sphere, so the volume can be said to
consist of the cone plus the upper slice of the sphere; e.g. the cone has a
convex base (or is it concave? being from sweden, my english isn't too
good...)

<matt271829-news@yahoo.co.uk> skrev i meddelandet
 Quote: David wrote: I'm struggeling with a weird triple integral: z / (1 + x^2 + y^2) dxdydz over the volume defined as x^2 + y^2 + z^2 <= 1 and z >= sqrt(x^2 + y^2) (Which is the volume that is inside the unit sphere, and also inside the upside-down flipped cone with the pointy end at origo and it's base at z=1/sqrt(2) ) I've tried to approach this in several ways, but every time i end up with _really_ messy solutions, like integrating complex ln functions to hell when I try cylindrical coordinates and even worse situations with spherical coordinates. How would I do to solve this problem in the best, nicest and simplest way? I really hope anyone can help... I would be very grateful for advice. / David As far as I can see, the whole of the cone you describe is inside the unit sphere, so I don't see why the unit sphere is relevant. There isn't a typo somewhere is there? Or maybe I'm misunderstanding something...
Lynn Kurtz
science forum Guru

Joined: 02 May 2005
Posts: 603

Posted: Tue Jul 11, 2006 9:08 pm    Post subject: Re: Advice on messy integral?

On Tue, 11 Jul 2006 20:44:56 GMT, "[Mr.] Lynn Kurtz"
<kurtzDELETE-THIS@asu.edu> wrote:

 Quote: Woops, that answer may have been a bit hasty and optimistic.... --Lynn

Here's your integral (Maple format) in spherical coordinates, but I'm
guessing you knew that:

2*Pi*int(rho^3*cos(p)*sin(p)/(1 + rho^2*sin(p)^2),[rho,p] =
Region(0..Pi/4,0..1));

Maple gives an explicit answer about 3 lines of output full of ln's
and sin(1) and cos(1) terms. For what it's worth, here it is but I
have no idea whether it will be readable even with monospaced font:

-1/32*Pi*(-Pi^2-64*ln(2)-Pi^2*cos(1)^2*ln(Pi^2-Pi^2*cos(1)^2+16)+Pi^2*
cos(1)^2*ln(1-cos(1))+Pi^2*cos(1)^2*ln(cos(1)+1)+Pi^2*ln(Pi^2-Pi^2*cos
(1)^2+16)-Pi^2*ln(1-cos(1))-Pi^2*ln(cos(1)+1)-4*Pi^2*ln(2)+16*ln(Pi^2-
Pi^2*cos(1)^2+16)+Pi^2*cos(1)^2+2*ln(sin(1))*Pi^2+4*Pi^2*cos(1)^2*ln(2

Not particularly simple, eh?

--Lynn
science forum beginner

Joined: 07 May 2006
Posts: 38

Posted: Tue Jul 11, 2006 9:23 pm    Post subject: Re: Advice on messy integral?

On Tue, 11 Jul 2006 21:08:01 GMT, "[Mr.] Lynn Kurtz"
<kurtzDELETE-THIS@asu.edu> wrote:

 Quote: On Tue, 11 Jul 2006 20:44:56 GMT, "[Mr.] Lynn Kurtz" kurtzDELETE-THIS@asu.edu> wrote: Woops, that answer may have been a bit hasty and optimistic.... --Lynn Here's your integral (Maple format) in spherical coordinates, but I'm guessing you knew that: 2*Pi*int(rho^3*cos(p)*sin(p)/(1 + rho^2*sin(p)^2),[rho,p] >Region(0..Pi/4,0..1)); Maple gives an explicit answer about 3 lines of output full of ln's and sin(1) and cos(1) terms. For what it's worth, here it is but I have no idea whether it will be readable even with monospaced font: -1/32*Pi*(-Pi^2-64*ln(2)-Pi^2*cos(1)^2*ln(Pi^2-Pi^2*cos(1)^2+16)+Pi^2* cos(1)^2*ln(1-cos(1))+Pi^2*cos(1)^2*ln(cos(1)+1)+Pi^2*ln(Pi^2-Pi^2*cos (1)^2+16)-Pi^2*ln(1-cos(1))-Pi^2*ln(cos(1)+1)-4*Pi^2*ln(2)+16*ln(Pi^2- Pi^2*cos(1)^2+16)+Pi^2*cos(1)^2+2*ln(sin(1))*Pi^2+4*Pi^2*cos(1)^2*ln(2 Not particularly simple, eh? --Lynn

Appears the OP multiposted this question to sci.math where its been
answered a couple of times now.
matt271829-news@yahoo.co.
science forum Guru

Joined: 11 Sep 2005
Posts: 846

Posted: Tue Jul 11, 2006 9:42 pm    Post subject: Re: Advice on messy integral?

David wrote:
 Quote: Yes, the cone is inside the unit sphere, so the volume can be said to consist of the cone plus the upper slice of the sphere; e.g. the cone has a convex base (or is it concave? being from sweden, my english isn't too good...)

Convex.

Right, I get it now. The confusion stemmed from the wording "the volume
that is inside the unit sphere, and also inside the upside-down flipped
cone", specifically the word "also" which to me implies the
intersection of the two figures. But no matter!
matt271829-news@yahoo.co.
science forum Guru

Joined: 11 Sep 2005
Posts: 846

Posted: Tue Jul 11, 2006 11:05 pm    Post subject: Re: Advice on messy integral?

 Quote: David wrote: Yes, the cone is inside the unit sphere, so the volume can be said to consist of the cone plus the upper slice of the sphere; e.g. the cone has a convex base (or is it concave? being from sweden, my english isn't too good...) Convex. Right, I get it now. The confusion stemmed from the wording "the volume that is inside the unit sphere, and also inside the upside-down flipped cone", specifically the word "also" which to me implies the intersection of the two figures. But no matter!

Well, I see it's been answered at sci.math, but now that I've done it
anyway I may as well tell you that using cylindrical coordinates I get
the same answer of 3/2*pi*Log(3/2) - pi/2.

Interestingly, the value of the integral over the cone (with plane
base) looks to be exactly the same as the value over the "cap" (the
region between the surface of the sphere and the plane base of the
cone), both being 3/4*pi*Log(3/2) - pi/4.

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