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symbolic integrating only one variable of three

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KOZI
KOZI 2018-11-3
编辑： Walter Roberson 2018-11-6
Hellow every one. I have one function with three variables. Variables typicaly are x,y,z where y=0 but in function i am using spherical coordinates which i have define in the begining of my code. I have one extra variable th1 which i want to integrate and the new function will have only x,z(r,th in spherical ). My problem is i cannot do that because when i run this:
syms x z th1;
r=(x^2 + z^2)^0.5;
th=acos(z/r);
d=2;
a=1;
d1=(a^2)/d;
L=1.5;
R1=(r^2 + d^2 - 2*r*d*(cos(th)*cos(th1) + sin(th)*sin(th1)))^0.5;
R2=(r^2 + d1^2 - 2*r*d1*(cos(th)*cos(th1) + sin(th)*sin(th1)))^0.5;
F1=d/R1;
F2=-(a/R2);
a1=-L/d;
a2=L/d;
f=@(x,z,th1)(F1+F2)
g = @(x,z) integral(@(th1) f(x,z,th1) , a1,a2)
I get :
f =
function_handle with value:
@(x,z,th1)(F1+F2)
I just want the symbolic expression of the above integration. I am stuck a lot of hours looking in already answered post but nothing so far. thanks in advance.

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采纳的回答

Walter Roberson
Walter Roberson 2018-11-3
f = matlabFunction(F1+F2, 'vars', [x, z, th1]);
It turns out there is a closed form solution for g, but it is long and a bit complicated, and MATLAB cannot find it.
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Walter Roberson
Walter Roberson 2018-11-5
I will look at this later; I have an appointment to go to now.
Walter Roberson
Walter Roberson 2018-11-5
I ran that code preceded by
syms x z th1
and it worked. It did give two messages about
Warning: Infinite or Not-a-Number value encountered.
One of those messages is for (0,0). r(0,0) = 0, and so acos(z/r(x,z)) is acos(z/0) which is 0/0 which is nan. This is the case for all th1 for (0,0)
The other of the messages is for (0,2). F1(0,2,th1) = 2^(1/2)/(2*(1 - cos(th1))^(1/2)) and when th1 is sufficiently close to 0, 1-cos(th1) becomes 0, leading to a division by 0, which leads to an infinite result. This the case for two of the th1 values that are generated internally, both near +/- 7E-9, but it is enough to "poison" the entire integral, so g(0,2) is infinite.

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更多回答（1 个）

madhan ravi
madhan ravi 2018-11-3
just change your last Line to this:
g = int( f(x,z,th1) ,th1, a1,a2)
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Walter Roberson
Walter Roberson 2018-11-5
-2*(-a^2+(b^2+c^2)^(1/2))*(-(sin(x-arctan(-b,c))*(b^2+c^2)^(1/2)+a^2)/(-a^2 +(b^2+c^2)^(1/2)))^(1/2)*(-(sin(x-arctan(-b,c))-1)*(b^2+c^2)^(1/2)/(a^2+(b^ 2+c^2)^(1/2)))^(1/2)*((sin(x-arctan(-b,c))+1)*(b^2+c^2)^(1/2)/(-a^2+(b^2+c^ 2)^(1/2)))^(1/2)*EllipticF((-(sin(x-arctan(-b,c))*(b^2+c^2)^(1/2)+a^2)/(-a^ 2+(b^2+c^2)^(1/2)))^(1/2),(-(-a^2+(b^2+c^2)^(1/2))/(a^2+(b^2+c^2)^(1/2)))^( 1/2))/(b^2+c^2)^(1/2)/cos(x-arctan(-b,c))/((a^2*(b^2+c^2)^(1/2)+b^2*sin(x- arctan(-b,c))+c^2*sin(x-arctan(-b,c)))/(b^2+c^2)^(1/2))^(1/2)
Is the antiderivative of that function, Torsten
Torsten
Torsten 2018-11-6
编辑：Walter Roberson 2018-11-6
My suggestion is to differentiate f with respect to r, set r=a and use "integral" to do the integration.
Interchanging integration and differentiation is justified by "Leibniz integral rule":
https://en.wikipedia.org/wiki/Leibniz_integral_rule
Best wishes
Torsten.

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