# integration of a function of two variables in which the limits of integral on one are functions of the other

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Ranjan Sonalkar il 1 Mar 2018
Commentato: Steven Lord il 1 Mar 2018
I want to integrate the function f(x,theta)={1-exp[-alpha/sin(theta)]} over x and theta. x has definite limits, but the limits of the integral over theta, are functions of x. So, I don't think I can use intergral2. Any suggestions?
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### Risposta accettata

Steven Lord il 1 Mar 2018
"q = integral2(fun,xmin,xmax,ymin,ymax) approximates the integral of the function z = fun(x,y) over the planar region xmin ≤ x ≤ xmax and ymin(x) ≤ y ≤ ymax(x)."
The limits on y can be functions of x. The "Evaluate Double Integral in Polar Coordinates" example on that page includes an example where the upper limit on r is a function of theta.
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Ranjan Sonalkar il 1 Mar 2018
Still having trouble, however. The output of the inner integral over theta is divided by the difference of the limits, (theta2-theta1), where theta1 and theta2 are the functions of the outer variable, x. So, I have defined theta1 and theta2 as functions of x, just as rmax in the example. But when I divide the integrand by the difference [./(theta2-theta1)] I am getting an error - "Undefined operator './' for input arguments of type 'function_handle'."
Steven Lord il 1 Mar 2018
You can't perform arithmetic on function handles. You can perform arithmetic on the values you receive from evaluating function handles.
fh = @sin;
x = fh(pi/4).^2 % works
y = fh.^2 % does not work
Evaluate your theta2 and theta1 functions for the values of theta that integral2 passed into your integrand function.

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