CODEWITHSUNDEEP
javanumerical-methodsnumerical
stiv
stiv

Reputation: 33

How to solve 2D integral?

I've been trying to implement trapezoid rule for double integral. I've tried many approaches, but I can't get it to work right.

static double f(double x) {
    return Math.exp(- x * x / 2);
}

// trapezoid rule
static double trapezoid(double a, double b, int N) {
    double h = (b - a) / N;
    double sum = 0.5 *  h * (f(a) + f(b));
    for (int k = 1; k < N; k++)
        sum = sum + h * f(a + h*k);
    return sum;
}

I understand the method for a single variable integral, but I don't know how to do it for a 2D integral, say: x + (y*y). Could someone please explain it briefly?

Upvotes: 3

Views: 2068

Answers (3)

Steve21
Steve21

Reputation: 11

Consider using the class jhplot.F2D from the DataMelt Java program. You can integrate and visualize 2D functions doing something like: 

f1=F2D("x*y",-1,1,-1,1) # define in a range
print f1.integral()

Upvotes: 0

null0pointer
null0pointer

Reputation: 1543

If you're intent on using the trapezoid rule then you would do it like so:

// The example function you provided.
public double f(double x, double y) {
    return x + y * y;
}

/**
 * Finds the volume under the surface described by the function f(x, y) for a <= x <= b, c <= y <= d.
 * Using xSegs number of segments across the x axis and ySegs number of segments across the y axis. 
 * @param a The lower bound of x.
 * @param b The upper bound of x.
 * @param c The lower bound of y.
 * @param d The upper bound of y.
 * @param xSegs The number of segments in the x axis.
 * @param ySegs The number of segments in the y axis.
 * @return The volume under f(x, y).
 */
public double trapezoidRule(double a, double b, double c, double d, int xSegs, int ySegs) {
    double xSegSize = (b - a) / xSegs; // length of an x segment.
    double ySegSize = (d - c) / ySegs; // length of a y segment.
    double volume = 0; // volume under the surface.

    for (int i = 0; i < xSegs; i++) {
        for (int j = 0; j < ySegs; j++) {
            double height = f(a + (xSegSize * i), c + (ySegSize * j));
            height += f(a + (xSegSize * (i + 1)), c + (ySegSize * j));
            height += f(a + (xSegSize * (i + 1)), c + (ySegSize * (j + 1)));
            height += f(a + (xSegSize * i), c + (ySegSize * (j + 1)));
            height /= 4;

            // height is the average value of the corners of the current segment.
            // We can use the average value since a box of this height has the same volume as the original segment shape.

            // Add the volume of the box to the volume.
            volume += xSegSize * ySegSize * height;
        }
    }

    return volume;
}

Hope this helps. Feel free to ask any questions you may have about my code (warning: The code is untested).

Upvotes: 3

bdecaf
bdecaf

Reputation: 4732

Many ways to do it.

If you already know it for 1d you could make it like this:

  1. write a function g(x) that calculates the 1d integral over f(x,y) for a fixed x
  2. then integrate the 1d integral over g(x)
  3. Success :)

That way you can basically have as many dimensions as you like. Though it scales poorly. For larger problems it might be neccesary to use monte carlo integration.

Upvotes: 0

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