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# Double iterated integrals to calculate areas of regions in the plane

Here’s how to calculate areas using double iterated integrals. We will see how to calculate areas depending on the order of integration chosen, with exercises solved step by step.

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Let’s go!

## Calculates areas with double iterated integrals with the order of integration dy.dx

Let’s consider a flat region R bounded by:  Which is represented graphically: This region R is composed of infinite rectangles dx, represented as a vertical rectangle: This vertical rectangle dx has two very important characteristics:

1. Moves horizontally between the limits of x a and b.
2. Depending on its position, its height varies adapting to functions g2(x) and g1(x), always remaining within both functions, so its height is limited above by the function above g2(x) and below by the function below g1(x).

Therefore, with this rectangle we can deduce the integration limits for the variable x, which are the two limits between which the rectangle can be moved horizontally, therefore, that is why the area of that region is given by the defined integral: On the other hand, we can rewrite the integrating g2(x)-g1(x) as a new defined integral, but integrated for the variable «y».

Let’s see how.

The limits of this integral are determined by the height of the rectangle dx, i.e. between the functions g2(x) and g1(x) and being «y» the integration variable, the differential in this case is dy: We solve this integral using Barrow’s rule and it fits:  Therefore, by replacing in the integral defined above the integrand g2(x)-g1(x) with the integral with respect to «y», the area of region R can be expressed as the iterated integral: The vertical rectangle implies the order of integration dy.dx.

## Calculates areas with double iterated integrals with the order of integration dx.dy

Similarly, we can calculate the area of a region by changing the order of integration to dy.dx.

In this case, region R is delimited by:  Note that now, the functions among which the variable x is limited, depends on «y».

And as before, this region R is composed of infinite rectangles, but this time the rectangles are horizontal and correspond to dy. Therefore, dy is represented as a horizontal rectangle: This horizontal rectangle dy has two very important characteristics:

1. Moves vertically between the limits of «y» c and d.
2. Depending on its position, its length varies, adapting to functions h2(y) and h1(y), always remaining within both functions, so its length is limited above by the function h2(y) and below by the function h1(y).

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Therefore, with this rectangle we can deduce the integration limits for the variable and, which are the two limits between which the rectangle can be moved vertically.

The integration limits for variable x are determined by the length of the rectangle, that is, by both functions.

In this case, the area of region R can be expressed as the iterated integral: The horizontal rectangle, implies the order of integration dx.dy..

## Double iterated integrals for area calculation

In summary, if we want to calculate the value of the area of a region in the plane by means of an iterated integral, this will be given by:

1- If R is defined by:  where g1 and g2 are continuous in [a,b], then the area of R will be: 2- If R is defined by:  where h1 and h2 are continuous in [c,d], then the area of R will be: How do we know which iterated integral to use for the calculation of areas?

When the functions between which the area is defined are given as a function of x, the rectangle representing the area will be vertical and therefore the order of integration will be dy.dx.

On the other hand, if the functions are given in function of «y», the representative rectangle will be horizontal and the order of integration will be dy.dx.

However, we can change the order of integration by changing the boundaries of the region.

For each specific problem, one of the two orders will make the calculations simpler. The order of integration chosen affects the difficulty of the calculations, but not the result. In each case, the most convenient order of integration is chosen.

Let’s go see him with some determined exercises.

## Resolved exercises of calculation of areas with double iterated integrals

### Exercise resolved 1

Calculate the area delimited by these two functions using a double iterated integral:  As the functions are defined in function of x, we are going to use the integration order dy.dx First of all, we are going to calculate the cut-off points of these two functions, since the cut-off points will be the integration limits of the variable x, that is, the limits between which the vertical rectangle dx moves horizontally.

To find the cut-off points, we match the two functions: We are left with a second-degree equation, so we pass all terms to one member and equal zero: We simplify terms: And we solve the equation. The solutions are: So the limits for dx are -2 and 1. We always put in the lower limit the lowest number and in the upper limit the highest limit.

The limits for the variable «y» are defined by the height of the rectangle, which moves between the two functions. The function at the top is the upper limit and the function at the bottom is the lower limit.

To know this we can draw both functions: We see that it is above the function: So it will be this one that determines the upper limit and the other function that determines the lower limit.

We can also substitute the same value of x for each function and whose value of the function is greater, will be the function that is above. We are going to substitute for example by x=0 in both functions: We see again that the 4-x² function results in a value greater than x+2, so it is shown that 4-x² is the upper limit.

Therefore, once we know the limits, we can write to the double iterated integral: What’s left: Now we first integrate the integral that remains inside, that is, the one that depends on dy, by means of Barrow’s rule: We replace the result of the integral with the upper limit and subtract the result of replacing the result of the integral with the lower limit: We operate within parentheses:  We are left with an integral that depends on the variable x. We integrate by applying Barrow’s rule: We subtract by substituting the upper limit minus the lower limit: We operate and come up with the solution:   Don’t forget that the solution must be in square units, as we are calculating areas.

### Exercise resolved 2

Draw the region whose area is given by the integral: Represent that same area by changing the order of integration to dy.dx or verify that the two integrals give the same result.

We see that the order of integration is dx.dy and that the region is defined by the limits:  This means that in this case dy is represented by a horizontal rectangle, that the outer limits of the double integral, 0 and 2, are the limits between which dy can move vertically and the inner limits and² and 4, are the limits between which the length of the dy rectangle is adapted.

In other words, the area is bounded on the left by the function y² and on the right by the line x=4. It is also dimensioned lower by the x axis (y=0) and higher by the value 4 (y=4): Let’s calculate the value of this integral: We integrate and apply the Barrow rule for the internal integral, integrated with respect to x:  Now we do the same for the integral that we have left, integrated with respect to and:   Now let’s change the order of integration to dy.dx.

To do this, we first define the function in terms of x, clearing the y: Now, in the previous region, we place the vertical triangle dx: We see that the triangle can be moved horizontally between the values of x 0 and 4: Which are the external summarization limits for variable x.

We also see that the height of the dx rectangle moves between the x-axis (y=0) and the function (now defined as a function of x): Which are the internal limits of integration for the variable «and», then the area of the region is given by the integral: We are going to demonstrate that the result is the same as with the previous integral.

We integrate the inner integral and apply the Barrow rule:  We do the same with the integral we have left:    And we see that indeed the result is the same.

## Calculation of areas with more than one double iterated integral

Sometimes it is not possible to calculate the area of a region with a single integral. In these cases, the region can be divided into sub-regions, whose areas can be calculated using double iterated integrals.

The total area will be the sum of the areas.

Let’s see an example:

Calculate the area of the region R included between the parabola y=4x-x², the x-axis and above the straight line y=-3x+6: First, we divide the area of two parts, at the point where the line intersects the x-axis: In both regions it is convenient to use the vertical rectangle dx and therefore the order of integration will be dy.dx.

The region on the left is defined between points 1 and 2 (the cut-off points of the line with the parabola and the x-axis respectively) which is where the rectangle can move horizontally: and correspond to the outer limits.

The inner limits to integrate with respect to and, are determined by the height of the rectangle which is bounded by the straight line below and by the parabola above: In the region on the right, the vertical triangle dx can move between points 2 and 4 (cut-off point of the line with the x-axis and of the parabola with the x-axis): The height of the rectangle is determined by the x-axis (y=o) and the parabola. For which the inner limits to integrate with respect to «and» are: Therefore, the area will be determined by the sum of the two integrals: That we solve them and that’s it:  