# 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 2018 年 3 月 1 日
コメント済み: Steven Lord 2018 年 3 月 1 日
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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### 採用された回答

Steven Lord 2018 年 3 月 1 日
"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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Steven Lord 2018 年 3 月 1 日
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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