MHB Definite integral on elliptic integral where modulus is function of variable

Click For Summary
The discussion focuses on proving the integral equation involving a definite integral of a function that includes the complete elliptic integral of the first kind. The integral in question is expressed as $\int_{0}^{\frac{\pi }{2}} {\frac{\sin \theta}{\sqrt{Z^2+(R+h \tan \theta)^2}} K[k(\theta)]}d\theta=\frac{\pi }{2\sqrt{R^2 + (h+Z)^2}}$, where $k(\theta)$ is defined in terms of the parameters R, h, and Z. Participants share their attempts to solve the integral, noting that the complexity increases with various methods tried. One participant reformulates the identity to a new expression involving $\sqrt{\sin \theta \cos \theta}$ and suggests that multiplying both sides by $2\sqrt{Rh}$ simplifies the problem. The conversation emphasizes the challenges faced in deriving the proof and invites further input on potential strategies.
bshoor
Messages
5
Reaction score
1
How to prove:

$\int_{0}^{\frac{\pi }{2}} {\frac{\sin \theta}{\sqrt{Z^2+(R+h \tan \theta)^2}} K[k(\theta)]}=\frac{\pi }{2\sqrt{R^2 + (h+Z)^2}} $

where \[ k(\theta)=\sqrt\frac{4Rh \tan \theta}{Z^2+(R+h \tan \theta)^2}\]

and $ K[k(\theta)] $ is the complete elliptic integral of the first kind, defined by

\[ K[k(\theta)]= \int_0^{\frac{\pi }{2}}\frac{\,d\phi}{\sqrt{1-k^2(\theta)\sin^2 \phi}}\]

and h, R and Z $ \gt 0 $
 
Physics news on Phys.org
Can you show us what you have tried and where you are stuck? This will give our helpers a better idea how to provide help without perhaps offering suggestions that you may already be trying.
 
bshoor said:
How to prove:

$\int_{0}^{\frac{\pi }{2}} {\frac{\sin \theta}{\sqrt{Z^2+(R+h \tan \theta)^2}} K[k(\theta)]}=\frac{\pi }{2\sqrt{R^2 + (h+Z)^2}} $

where \[ k(\theta)=\sqrt\frac{4Rh \tan \theta}{Z^2+(R+h \tan \theta)^2}\]

and $ K[k(\theta)] $ is the complete elliptic integral of the first kind, defined by

\[ K[k(\theta)]= \int_0^{\frac{\pi }{2}}\frac{\,d\phi}{\sqrt{1-k^2(\theta)\sin^2 \phi}}\]

and h, R and Z $ \gt 0 $
Please make correction of the post:
$\int_{0}^{\frac{\pi }{2}} {\frac{\sin \theta}{\sqrt{Z^2+(R+h \tan \theta)^2}} K[k(\theta)]}d\theta=\frac{\pi }{2\sqrt{R^2 + (h+Z)^2}} $
 
MarkFL said:
Can you show us what you have tried and where you are stuck? This will give our helpers a better idea how to provide help without perhaps offering suggestions that you may already be trying.

I have tried in different ways. But the integral becomes more and more complicated. Anyway the identity can be modified to :

$\int_{0}^{\frac{\pi }{2}}{\sqrt {\sin \theta cos\theta}k(\theta) K[k(\theta)]}d\theta=\pi \sqrt{ \frac{Rh}{R^2 + (h+Z)^2}} $

by multiplying both side by $ 2 \sqrt{Rh} $
 
Thread 'Problem with calculating projections of curl using rotation of contour'
Hello! I tried to calculate projections of curl using rotation of coordinate system but I encountered with following problem. Given: ##rot_xA=\frac{\partial A_z}{\partial y}-\frac{\partial A_y}{\partial z}=0## ##rot_yA=\frac{\partial A_x}{\partial z}-\frac{\partial A_z}{\partial x}=1## ##rot_zA=\frac{\partial A_y}{\partial x}-\frac{\partial A_x}{\partial y}=0## I rotated ##yz##-plane of this coordinate system by an angle ##45## degrees about ##x##-axis and used rotation matrix to...

Similar threads

Replies
2
Views
3K
  • · Replies 8 ·
Replies
8
Views
3K
Replies
4
Views
4K
  • · Replies 29 ·
Replies
29
Views
4K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 16 ·
Replies
16
Views
2K
  • · Replies 20 ·
Replies
20
Views
4K
  • · Replies 5 ·
Replies
5
Views
2K