Asked by Heather Holzer
on 23 Aug 2019

- Q=150;
- k=1.49;
- n=0.02;
- So=0.0048;
- b=3;
- g=32.2;
- x1=0;
- y1=3; %x2=200;
- h=20;
- x2=200;
- syms y;
- f_xy=(So-((Q*n)/(k*b*(y^(5/3))))^2)/(1-((Q^2)/((b^2)*g*y^3)));
- func=int(f_xy,y,[x1 x2]);
- vfunc=vpa(func);
- y=y1+vfunc;
- ys=string(y);
- yf=str2double(ys);

this is what shows up in the varaibles for ys= "int((10000/(22201*y^(10/3)) - 3/625)/(12500/(161*y^3) - 1), y, 0, 200) + 3"

Also the above equation is not how it should look with variables plugged in.

Answer by Torsten
on 23 Aug 2019

Accepted Answer

Q=150;

k=1.49;

n=0.02;

So=0.0048;

b=3;

g=32.2;

x1=0;

y1=3; %x2=200;

h=20;

x2=200;

f_xy=@(y)(So-((Q*n)./(k*b*(y.^(5/3)))).^2)./(1-((Q^2)./((b^2)*g*y.^3)));

vfunc=integral(f_xy,x1,x2);

y=y1+vfunc

Heather Holzer
on 23 Aug 2019

is this close?

sym y

x2=[20,20,200];

f_xy=(So-((Q*n)./(k*b*(y.^(5/3)))).^2)./(1-((Q^2)./((b^2)*g*y.^3)));

vfunc=int(f_xy,y,[x1 x2]);

y=y1+vfunc;

ys=table(y);

Torsten
on 23 Aug 2019

Q=150;

k=1.49;

n=0.02;

So=0.0048;

b=3;

g=32.2;

x1=0;

y1=3; %x2=200;

h=20;

x2=[20,40,60];

f_xy=@(y)(So-((Q*n)./(k*b*(y.^(5/3)))).^2)./(1-((Q^2)./((b^2)*g*y.^3)));

vfunc = zeros(numel(x2,1));

for i=1:numel(x2)

vfunc(i) = integral(f_xy,x1,x2(i));

end

y = y1+vfunc

But - as already said - you should check whether the integral exists at all.

Heather Holzer
on 23 Aug 2019

I will. Thank you so so very much for your excellence!

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