How to do type II double integration on matlab

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Ibraheem
Ibraheem 2022-10-28
评论： Ibraheem 2022-10-28
I have no problem using type I (dydx) and when I use it i get the correct answer but if i try replicating the same method using type 2(dxdy) i get a diffrent answer which is wrong can someone explain what i am getting wrong.
Type 1
ymax = @(x) cos(x);
xmax = pi/2;
% ymax = upperboud of y I had to make a function for it since we can't put
% y below in integral2
rrr = integral2(fun,0,xmax,0,ymax);
%%% the volume is 0.4674 using type I which is correct
Type 2
% finding the volume using type 2
fun = @(x,y) (x.^2);
xmax = @(y) acos(y);
% xmax = upperbound of x, we had to make a function for it since we can't put
% x below in integral 2
rr = integral2(fun,0,xmax,0,1)
i got a wrong answer.
I would be very grateful if someone can help me thanks.

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采纳的回答

David Hill
David Hill 2022-10-28
Ran in:
Use symbolic integration
syms x y
f=x^2;
int(int(f,x,0,acos(y)),y,0,1)
ans = 
double(ans)
ans = 0.4674

更多回答（1 个）

John D'Errico
John D'Errico 2022-10-28
编辑：John D'Errico 2022-10-28
Ran in:
First, type I and II are not any common names for integrals that I know of. It might be what your teacher calls them. Maybe the new math, something that constantly seems to change. Regardless, it appears the difference lies merely in which order you perform the integration. So first, what volume are you computing an integral for?
The integration kernel is x^2, so you want to compute an integral of x^2 over a region in the (x,y) plane.
x varies from 0 to pi/2, y varies from 0 to cos(x).
First, what should we get analytically?
syms x y
symbintegral = int(int(x.^2,y,[0,cos(x)]),x,[0,pi/2])
symbintegral = 
vpa(symbintegral)
ans = 
0.46740110027233965470862274996904
So you are correct there. What is the domain of integration? DRAW A PICTURE IF NECESSARY! I should have done that first. My bad.
X = linspace(0,pi/2);
Ymax = cos(X);
domainpoly = polyshape([X,pi/2,0],[Ymax,0,0])
Warning: Polyshape has duplicate vertices, intersections, or other inconsistencies that may produce inaccurate or unexpected results. Input data has been modified to create a well-defined polyshape.
domainpoly =
polyshape with properties: Vertices: [101×2 double] NumRegions: 1 NumHoles: 0
plot(domainpoly)
So a quasi-pseudo-vaguely-triangular area, where one side of the triangle (the hypotenuse) is a cosine function. First, we can integrate with respect to x on the inner integral, then on y.
fun1 = @(x,y) x.^2;
xmax = pi/2;
ymax = @(x) cos(x);
integral2(fun1,0,xmax,0,ymax)
ans = 0.4674
That works perfectly. Now, you want to swap the order of integration. REMEMBER that integral2 does the inner integral on the FIRST parameter of the function you give it! So you need a different function, with the arguments swapped!
fun2 = @(y,x) x.^2;
ymax = 1;
xmax = @(y) acos(y);
integral2(fun2,0,ymax,0,xmax)
ans = 0.4674
Easy peasy.

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