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Integrate[Sqrt[(roh^2 - r^2)/(r^2 - rb^2)], {r, rb, roh}, 
 Assumptions -> r > rb \[And] roh > rb \[And] roh > 1]

Outputs to:

 ConditionalExpression[(
         rb roh (-Sqrt[(1/rb)] Sqrt[-rb] EllipticE[rb^2/roh^2] + 
            I EllipticE[ArcSin[roh/rb], rb^2/roh^2]))/Abs[rb], rb + roh >= 0]

EDIT1:

Integrate[Sqrt[(roh^2 - r^2)/(r^2 - rb^2)], {r, rb, roh}, 
 Assumptions -> rb > 0 && roh > rb]
Integrate[Sqrt[(m^2 - r^2)/(r^2 - m^2 + 1)], {r, Sqrt[m^2 - 1], m}, 
 Assumptions -> m > 1]
Clear["Global`*"];
Plot[-I m (EllipticE[1 - 1/m^2] - 
    EllipticE[ArcSin[m/Sqrt[-1 + m^2]], 1 - 1/m^2]), {m, 1 + 10^-6, 
  4}, GridLines -> Automatic]

Additionally to simplify using a single parameter $m>1$ defined through $(roh = m, rb= \sqrt{m^2-1})$ one obtains plot:

enter image description here

Thanks for understanding/comment/intuition /about how a plot with $\sqrt{-1}$ coefficient can work, when arguments of $(K,E)$ type elliptic integrals appear to be real.

WA plot

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2
  • $\begingroup$ Looking at result, without assumptions shows a term $\sqrt\frac{-1}{\rho}$. But then you say in the assumptions that $\rho>1$, therefore the complex number shows up. To get rid of the complex number, then remove the assumption that causes it. $\endgroup$
    – Nasser
    Sep 4, 2021 at 1:04
  • $\begingroup$ If an expression with complex components reduce to real than it must be the imaginary portions cancel. $\endgroup$
    – josh
    Dec 11, 2021 at 16:43

1 Answer 1

3
$\begingroup$
Integrate[Sqrt[(roh^2 - r^2)/(r^2 - rb^2)], {r, rb, roh}, Assumptions -> rb > 0 && roh > rb ]

(* roh (EllipticK[1 - rb^2/roh^2] - EllipticE[1 - rb^2/roh^2]) *)
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