Double Integrals for Rectangles

Double Integrals for Rectangles:

The center point method:

Presume that we need to find the integral of a function, f(x, y) on a rectangle:

R = {(x, y) :a ≤ x ≤ b, c ≤ y ≤ d}.

In calculus you erudite to do this by an iterated integral:

413_iterated integral.jpg

You also must have learned that the integral is the limit of the Riemann sums of the function as the size of the sub-rectangles goes to zero. This signifies that the Riemann sums are approximations of the integral which is exactly what we need for numerical methods.

For a rectangle R we start by subdividing into smaller sub-rectangles {Rij} in a systematic way. We will divide [a, b] into m subintervals as well as [c, d] into n subintervals. Afterwards Rij will be the “intersection” of the i-th subinterval in [a, b] with the j-th subinterval of [c, d]. In this way the complete rectangle is subdivided into mn sub-rectangles numbered as in given figure.

A Riemann sum utilizing this subdivision would have the form:

116_riemann sum.jpg

Subdivision of the rectangle R = [a, b] × [c, d] into sub-rectangles Rij

Where Aij= ΔxiΔyj is the area of Rij, and x*ij is a point in Rij. The theory of integrals notify us that if f is continuous then this calculation will converge to the same number no matter how we choose x*ij. For illustration we could choose x*ij to be the point in the lower left corner of Rij and the sum would still converge as the size of the sub-rectangles goes to zero. Nevertheless in practice we wish to choose x*ij in such a way to make S as accurate as possible even when the sub-rectangles are not very small. The noticeable choice for the best point in Rij would be the center point. The center point is most probable of all points to have a value of f close to the average value of f. If we denote the center points by cij then the sum becomes:

2102_center point.jpg

Note that if the subdivision is consistently spaced then Δx ≡ (b − a)/m and Δy ≡ (d − c)/n and therefore in that case:

2041_evenly spaced subdividon.jpg

The four corners method:

An extra good idea would be to take the value of f not only at one point however as the average of the values at several points. An clear choice would be to evaluate f at all four corners of each Rijthen average those. If we note that the lower left corner is (xi, yj), the upper left is (xi, yj+1) the lower right is (xi+1, yi) as well as the upper right is (xi+1, yi+1) then the corresponding sum will be:

142_four corners method.jpg

Which we will call the four-corners method. If the sub-rectangles are consistently spaced then we can simplify this expression. Notice that f(xi, yj) gets counted multiple times depending on where (xi, yj) is located. For illustration if (xi, yj) is in the interior of R then it is the corner of 4 sub-rectangles.

Consequently the sum becomes:

2493_sum of four corner method.jpg

Where A = ΔxΔy is the area of the sub-rectangles. We are able to think of this as a weighted average of the values of f at the grid points (xi, yj). The weights utilized are represented in the matrix:

1785_area of subrectangles.jpg

We could execute the four-corner method by forming a matrix (fij) of f values at the grid points then doing entry-wise multiplication of the matrix with the weight matrix. Then the integral would be acquired by summing all the entries of the resulting matrix and multiplying that by A/4. The formula would be:

137_four corner formula.jpg

Notice that the four-corner method coincides among applying the trapezoid rule in each direction.

Therefore it is in fact a double trapezoid rule.

The double Simpson method:

The subsequent improvement one might make would be to take an average of the center point sum Cmnand the four corners sum Fmn. Nevertheless a more standard way to acquire a more accurate method is the Simpson double integral. It is acquire by applying Simpson’s rule for single integrals to the iterated double integral. The resulting method needs that both m and n be even numbers and the grid are evenly spaced. If this is the case we summarize the values f(xi, yj) with weights represented in the matrix:

1926_double simpson formula.jpg

The amount of the weighted values is multiplied by A/9 and the formula is:

78_double simpson formula.jpg

Mat lab has a built in command for double integrals on rectangles dblquad(f,a,b,c,d). Here is an instance

> f = inline(’sin(x.*y)./sqrt(x+y)’,’x’,’y’)
> I = dblquad(f,0.5,1,0.5,2)

Below is a Mat lab function which will produce the matrix of weights needed for Simpson’s rule for double integrals. It utilize the function mysimpweights.

function W = mydblsimpweights(m,n)
% Produces the m by n matrix of weights for Simpson’s rule
% for double integrals
% Inputs: m -- number of intervals in the row direction.
% must be even.
% n -- number of intervals in the column direction.
% must be even.
% Output: W -- a (m+1)x(n+1) matrix of the weights
if rem(m,2)~=0 | rem(n,2)~=0
error(’m and n must be even’)
end
u = mysimpweights(m);
v = mysimpweights(n);
W = u*v’

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