# Integration of Mutual Inductance

1. Dec 18, 2011

### jd_ee

Hello. I am working through the book "Analysis of Electric Machinery and Drive Systems" by Krause. On p.51, the mutual inductance between two windings (a and b) of a 3-phase machine (generator) is calculated. I am struggling to follow the integration performed on eqn 1.5-15 to arrive at eqn. 1.5-16

I have given the root of the problem: a double integral involving the product of trig. functions

1. The problem statement, all variables and given/known data

∫sin ρ ∫cos(ζ - 2∏/3)(A - Bcos(2(ζ - θ)) dζ dρ

first integral is done from ∏→2∏
second integral is done from ρ → ρ + ∏

result given in textbook is:

∏/2 * (A + Bcos(2(θ - ∏/3)))
2. Relevant equations

3. The attempt at a solution

I tried expanding out the cosine and sine terms but the problem quickly blows up... :(

EDIT: I did perform the integral using MATLAB Symbolic Toolbox. The result agrees with the given answer. I'm just at a loss of how to do this by hand. It seems very challenging so I'm wondering if I'm missing a trick.

MATLAB Code

%2011-12-18
%Derivation of main integral in eqn. 1.5-15 --> 1.5-16 on p.51
syms xi A B phi theta
first_int = int(cos(xi - 2*pi/3)*(A - B*cos(2*xi - 2*theta)), xi, phi, phi+pi);
second_int = int(first_int*sin(phi), phi, pi, 2*pi)

Last edited: Dec 18, 2011
2. Dec 18, 2011

### Dick

I'd try expanding cos(ζ - 2∏/3)(A - Bcos(2(ζ - θ))) to Acos(ζ - 2∏/3)-Bcos(ζ - 2∏/3)cos(2(ζ - θ)). The term with one cos should be no problem. For the second one use cos(X)cos(Y)=(1/2)(cos(X+Y)-cos(X-Y)).