Integration of Multivariable Functions Over a Domain

Consider the function: \[ f(x) = x^2 \] The process of integration for this function starts in the form of adding the areas of many rectangles via a Riemann sum. The heights of these rectangles are determined by the value of the function at the relevant x value, and the width is decided by the amount of partitions over the bounds of integration. The process of integration sums up the area of infinitely thin rectangles over the relevant bounds. In this case there is a function of one variable over a domain in \( \mathbb{R} \). Now consider the function: \[ f(x,y) = x^2 + y^2 \] Now we have a function of 2 variables. An integral for this function can now be integrated over \(\mathbb{R}^2\), which can be viewed as an area rather than a line in \( \mathbb{R}\). The other major difference for integration over a two dimensional domain is having two variables to integrate. This means that the integral is taking with respect to an area or \( dA \). For this integral, the Riemann sum consists of rectangular prisms as opposed to rectangles. Their heights are still defined by the function, and their widths and lengths are defined by the number of partitions over the domain. For an integration with respect to \( dA \), the rectangular prisms have two dimensions that are infinitely small, and the sum of the volume of these skinny prisms is equivalent to the integral of the function with respect to the domain.

Approximation for a Two Variable Function

As an example, consider the function: \[ f(x,y) = xy + x + y \] this function is interesting as it has multiplication and addition of two variables. This function will be interesting to integrate via approximation and normal integration. The easiest domain to consider is a rectangle created by the x and y bounds of the integral. A rectangle bound is defined when all four bounds are constants. To create a an approximation with 25 rectangular prisms, the function will be integrated on the on the domain: \[{(x,y) | x \in [-2.5,2.5], y \in [-2.5,2.5]}- \] this domain is a square of side length 5 and area 25 centered at the origin. If the square is divided into smaller squares of area 1, there are now 25 regions that are defined to approximate the value of the integral. In this approximation, the center point of each smaller square will be used to define the height of the approximating prism.

How to Approximate

Let's start by looking at the bottom left most square. This would be the square whose center point is at (-2,-2). The length and width of this square also define the base of the rectangular prism that approximates this section of the region. The only thing left to find is the height. The height would be the z value obtained from the function f(x,y), where x and y are the center point of the relevant square. In this case, the height of the prism is: \[ f(-2,-2) = (-2)(-2) + (-2) + (-2) = 4 - 2 - 2 = 0 \] This means that the height and volume of this first region is 0. The following table shows the z value/height of all prisms:

Finding the volume for the prisms is quite easy since the base has an area of 1, so the height and volume of any given prism are equal. An interesting aspect for this function and domain is the fact that some of hte heights are negative, implying that the volume is also negative. While the volume itself is not negative, it does fall below the z axis making the value negative as a result. This is the same as integration of a 1 variable function, when the y value at a given x value falls below the x axis or more accurately y = 0. The same occurs for the two variable function when the value of z falls below the plane xy = 0. In both cases, the area/volume is technically negative and subtracts from the total. Taking all of this into consideration, the total volume of the 25 prisms comes out to a whopping total of 0 units. This indicates the approximation has equal volumes below and above the xy plane.

Carrying out the Integration

Now the accuracy of the approximation can be assessed by carrying out an integral with respect to the area of the domain. The integral is of the form: \[ \iint_{\mathbb{R}^2} f(x,y) \, dA \] There are 3 components to this integral: the domain of integration, the 2 variable function, and the respective domain. The function is already defined above. The domain of integration has been defined as the square of side length 5 centered at the origin. This means that with respect to x, the bounds are -2.5 and 2.5. The bounds with respect to y are also -2.5 and 2.5 since the region is a square. Lastly, there is the definition of dA as the product of dx and dy. Putting that all together the integral looks like: \[ \int_{-2.5}^{2.5} \int_{-2.5}^{2.5} xy + x + y \,dy \,dx \] When integrating with respect to two variables, the inside variable, in this case dy, is integrated with respect to first. This means that x is treated like a constant until the second integral. \[ \int_{-2.5}^{2.5} \int_{-2.5}^{2.5} xy + x + y \,dy \,dx = \int_{-2.5}^{2.5} (.5xy^2 + xy + .5y^2) \vert_{-2.5}^{2.5} \, dx = \int_{-2.5}^{2.5} 5x \, dx = 2.5x^2 \vert_{-2.5}^{2.5} = 0\] So, the approximation was perfect as the volume over the domain is actually 0.

Ignoring Sign for a Bit

It seems a little unsatisfactory to say that this solid created by the taking the volume under the curve has a volume of zero. Let's say that we want to find the volume of the solid, but the volume can never be negative. For the approximation aspect of this, it is as simple as taking the absolute value of all the values from the table and adding them together. This comes out to a total of 54. To find the total volume of the integrated object while disregarding sign, the domains would have to be changed a bit. Whenever the z value is negative for the domains of x and y, that domain would have to integrate the negation of f(x,y). This means that there would be 2 integrals with two different areas of integration. However, since the integral over the entire domain was 0, it is known that these two integrals will be equal and opposite. Instead of carrying out both integrals, the value of one can be found and doubled to get the same answer. Since this is the case, let's take the integral over the domain in which z is positive which is: \[ {(x,y) | x,y \in [0,2.5]} \] The integral looks like: \[ \int_{0}^{2.5} \int_{0}^{2.5} xy + x + y \,dy \,dx = \int_{0}^{2.5} 5.625x + 3.125 \, dx = 2.8125x^2 + 3.125x \vert_{0}^{2.5} = 25.39\] Doubling this answer yields a value of 50.78 units cubed which varies from the approximation by about 6.34%. The approximation does well on two fronts. The first being that it was fairly close in estiamting the volume of the solid and was totally correct in saying that the positive and negative volumes cancelled. The created 3-D model for this function over the domain has a square base with side length 5. The difference between the highest and lowest point of the model is 12 units. This model would have to be scaled down quite a bit to fit in a 1.5 cubic inch space. If going by the integrated volume, this model would have to be scaled down by a factor of about 33 times. The following is approximated model: