symbolic integrating only one variable of three

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.

Answer 1

[Walter Roberson]

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.

[KOZI]

Sir i have a new clue. I must numericaly integrate for all values of x and z without using symbolic. how i do that?

[Walter Roberson]

f = matlabFunction(F1+F2, 'vars', [x, z, th1]);
g = @(X,Z) arrayfun( @(x, z) integral(@(th1) f(x,z,th1), a1, a2), X, Z);

h = integral2(g, xmin, xmax, zmin, zmax);

[KOZI]

No,i think you misunderstood me. O will integrade var th1 numericaly for different values of x,z. So i suppose i must use msgrid for x,z What is the second intgral?

[Walter Roberson]

f = matlabFunction(F1+F2, 'vars', [x, z, th1]);
g = @(x,z) integral(@(th1) f(x,z,th1) , a1,a2);

[X, Y] = meshgrid(x_values, z_values);

h = arrayfun(g, X, Y);
surf(x_values, z_values, h, 'edgecolor', 'none');

you might need to use

surf(x_values, z_values, h.', 'edgecolor', 'none');

[Walter Roberson]

Or is the point that you cannot even use symbolic toolbox to form F1+F2 ?

[KOZI]

I think i will canot thank you enough.thanks a lot.o will try it when i got home. I hope it works bacause i have to give the homework tommorow morning

[KOZI]

r=@(x,z)(x.^2  + z.^2).^0.5;
th=@(x,z)acos(z/r(x,z));
d=2;
a=1;
d1=(a.^2)/d;
L=1.5;
R1=@(x,z,th1)(r(x,z).^2 + d^2 - 2*r(x,z)*d*(cos(th(x,z))*cos(th1) + sin(th(x,z))*sin(th1))).^0.5;
R2=@(x,z,th1)(r(x,z).^2 + d^2 - 2*r(x,z)*d1*(cos(th(x,z))*cos(th1) + sin(th(x,z))*sin(th1))).^0.5;
F1=@(x,z,th1)d/R1(x,z,th1);
F2=@(x,z,th1)-(a/R2(x,z,th1));
a1=-L/d;
a2=L/d;
f = matlabFunction(F1(x,z,th1)+F2(x,z,th1), 'vars', [x, z, th1]);
g = @(x,z) integral(@(th1) f(x,z,th1) , a1,a2);
x = -2:0.25:2;
z = -2:0.25:2;
[X, Y] = meshgrid(x, z);
h = arrayfun(g, X, Y);
surf(x, z, h, 'edgecolor', 'none');

error

Error using symengine
Operands are invalid.

Error in sym/privBinaryOp (line 973)
            Csym = mupadmex(op,args{1}.s, args{2}.s, varargin{:});

Error in  /  (line 353)
        X = privBinaryOp(A, B, 'symobj::mrdivide');

Error in Untitled>@(x,z,th1)d/R1(x,z,th1)

Error in Untitled (line 14)
f = matlabFunction(F1(x,z,th1)+F2(x,z,th1), 'vars', [x, z, th1]);

[Walter Roberson]

I will look at this later; I have an appointment to go to now.

[Walter Roberson]

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.

Answer 2

[madhan ravi]

just change your last Line to this:

g =  int( f(x,z,th1) ,th1, a1,a2)

[KOZI]

I tried it but it runs for ever

[madhan ravi]

as sir Walter mentioned matlabs symbolic engines not able to calculate such integral

[KOZI]

The problem is ths i am sure it can be solved because our teacher told us. Maybe i am trying to do it in wrong way. So i just want to integrate for var th1 from a1 to a2 and have a new functiom that will have only x and z as variables

[Walter Roberson]

Your g is suitable for numeric calculations not symbolic, as your g uses integral() which is for numeric calculations. The Answer that I posted shows how to generate the correct function for numeric calculations.

If you need the symbolic solution in terms of x and z then MATLAB is not able to calculate it. It is a long symbolic formula with a number of arctan and cos in it (I think it was).

[KOZI]

Sorry for the many questions but i dont know anything about matlab, i just need it for one homework.So there is not a way to get a symbolic solution?if instead of using x,z use syms r, th so that i will not have acos? Will it be possible to get solution in terms of r,th? Maybe i made a terible mistake not to inform you that when i solve the integrall then i want to make diff(f,r) and then for r=1 and th from 0 to 2pi plot it. Thanks again

[KOZI]

I just cant understand where is the problem, bacause in the interval practicaly x,z are constants only th1 should be the thing

[Torsten]

What makes you think that a function of the form

f(x) = 1/sqrt(a^2+b*cos(x)+c*sin(x))

has an analytical antiderivative ?

[KOZI]

Nothing except tha my proffesor told me it has.

[Torsten]

Maybe we can find an easier solution if you tell us what the underlying problem is.

[KOZI]

g that comes from f represents the electrical potential using the theory of superfluity. Thats the reason of integration. So i have to find surface density and then plot it for r=a and th from 0 to 2pi. But to find that i also need eltectric field to do that. Where in my code electric field=- diff(g,r). So wihout calculating g there is not much i can do. And that is only the first question. After that i have many things to do, but i also need g.

[Walter Roberson]

-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]

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.