Double numeric integral of a function help please

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Hello everyone
I want to numerically integrate a function
1-x^2-y^2
Where x^2+y^2<=1 circular area
What I tired so far is First try:
quad2d(@(x,y) 1-x^2-y^2, -1,1,-1,sqrt(1-x^2))
Second try:
syms x y
int(int(1-x^2-y^2,y, -1,sqrt(1-x^2))x,-1,1)
How can I use sqrt(1-x^2) as the boundary of the integral for y?
I know this can be done with r and theta but I want to calculate the integral by cartesian coordinates.
Could someone offer a solution?

Accepted Answer

Roger Stafford
Roger Stafford on 26 Sep 2014
You need to make the integration limit sqrt(1-x^2) a function handle. See the documentation at
http://www.mathworks.com/help/matlab/ref/quad2d.html
In particular note where it says "q = quad2d(fun,a,b,c,d) approximates the integral of fun(x,y) over the planar region and fun is a function handle, c and d may each be a scalar or a function handle."
  1 Comment
Youssef  Khmou
Youssef Khmou on 26 Sep 2014
Edited: Youssef Khmou on 26 Sep 2014
it did not work yet, you have dblquad function?
f=@(x,y) 1-(x.^2)-(y.^2);
w=@(x) sqrt(1-x.^2);
q=dblquad(f,-1,1,-1,w);

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More Answers (1)

bogazici guvercin
bogazici guvercin on 26 Sep 2014
Thanks for the help
The function handle is not working with dblquad somehow but working with quad2d.
Thanks for the help!

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