Objective: Recognize when a function of two variables is integrable over a rectangular region
Double integral of a function f(x,y) over a rectangular region R
Let's start by considering the space above a rectangular region R. Let z = f(x,y) ≥ 0 be a function of two variables defined over R as follows:
The graph of f is a surface above the xy plane where z = f(x,y) is the height of the surface at the point (x,y). Let's consider the volume of the solid between the surface S and the rectangular region R. The base of the solid is the rectangle R. Our goal is to find the volume V of the solid S.
Let's divide the interval [a b] in m sub-intervals and the interval [c d] in n sub-intervals. This allows to divide the rectangular region R into small rectangles Rij with area ΔA and sides Δx and Δy. Therefore we have:
Let's consider a thin rectangular box above Rij with height f(xij, yij).
Considering all the thin rectangular boxes for all the subrectangles, we obtain an approximate volume of the solid S as follows:
This sum is known as the double Rieman sum and can be used to approximate the volume of the solid. The double sum means that, for each subrectangle, we evaluate the function at the chosen point, multiply by the area and add all the results.
As seen in the case of a single variable function, we obtain a better approximation when m and n become larger.
Here we are ready to define the double integral of a function f(x,y) over a rectangular region R in the xy plane.
Definition
The double integral of the function f (x,y) over the rectangular region R in the xy plane is given by:
Example
Since the number of subintervals respectively on the x-axis and the y-axis is known i.e m = n =2, we don't use the limit notation to calculate the volume. Therefore:
Then V = [3(1)² - 1] * 1 + [3(2)² - 1] * 1 + [3(1)² - 2] * 1 + [3(2)² - 2] * 1
V = (3/4 - 1/2) + (27/4 - 1/2) + (3/4 - 3/2) + (27/4 - 3/2)
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