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31.5 Setting up Correct Limits of Integration

To do this you must first decide on an order in which you wish to perform the integrals.

There are circumstances in which this does not matter much, and those in which the difference in the ease of doing the integral is very substantial. However, you can always decide to change the order of integration if you are dissatisfied with your choice.

So suppose we have an area integral and we wish to integrate over the variable x first and then y .

The first and most important step in deciding on limits of integration is to draw a picture of the region you wish to integrate over. This region will usually be bounded by a set of curves.

For certain choices of the variable y the limits of integration x will typically be the values of x that lie on two of these bounding curves for this y value. You integrate over y over those intervals of its values for which you get the same bounding curves in x . You then fix a y value in each of such intervals.

In most cases that you encounter, the values of x that you wish to integrate over will form an interval lying between two of these curves. You must determine which curves these are (occasionally they are the same curve) and then solve each curve equation for its x value with the y value assumed. These will be the limits for your x integration for this y value.

Under some circumstances the limits on x involve different curves for different y values. You must choose your x integration accordingly.

All this undoubtedly sounds vague and frighteningly difficult.

In fact, if you draw a picture, it is usually very easy to do. Without a picture it is indeed frighteningly difficult to get it right. And very easy to screw up.

We consider some examples:

Suppose first you wish to integrate over the triangle with base y = 0 and sides determined by the equations x = 0 and x + y = 1 .

There is an x interval to integrate over for all y values between 0 and 1.

If you fix y between 0 and 1, you will integrate over x between x = 0 and x + y = 1 , which means x = 1 − y .

The integral therefore becomes

∫ y = 0 y = 1 d y ( ∫ x = 0 x = 1 − y ( integrand ) d x )

Suppose you want to integrate over the triangle between y = x , y = 0 and x = 1 .

Again the values of y for which you have an x region are between y = 0 and y = 1 , and for any y value in that interval you want to integrate over x between x = y and x = 1 .

The limits of integration are then

∫ y = 0 y = 1 d y ( ∫ x = y x = 1 ( integrand ) d x )

Exercises:

31.1 Suppose you want to integrate over this same triangular region integrating y first and then x . What are the appropriate limits in that case?

31.2 Suppose instead we wish to integrate over x from y to 2, and then over y from y = 0 to 1. This defines a trapezoid. What are the limits of integration if we want to integrate over y first?

Tricky things can happen, here are some examples to look at:

1. The area bounded by x = 1 , x = 2 , y = x and y = 0 . This is simple if you integrate over y first. But if you integrate over x first you find the integral must be split into two parts.
Between y = 0 and y = 1 you must integrate x between x = 0 and x = 1 ; between y = 1 and y = 2 , you must integrate x between x = y and x = 2 .

2. The area bounded by x = π 6 , x = 5 π 6 , y = 0 and y = sin ⁡ x .
If you integrate over x first then for y between 0 and 1 2 you integrate from x = π 6 to 5 π 6 . ( 1 2 is the sine of x = π 6 and also x = 5 π 6 .)
For y above 1 2 , you must instead integrate between x = arcsin ⁡ y and x = π − arcsin ⁡ y , which represent two points on the bounding curve y = sin ⁡ x .

It should be clear from these examples that without an adequate picture it is almost impossible to get these things right.

Below is an applet for trying different limits in rectangular and polar coordinates. Try it on the examples above and on the following areas:

Exercises:

31.3 Consider the interior of a circle of radius A centered about the origin. What are the limits on x and y , or on r and θ ?

31.4 Find limits integrating on x first then y , and vice versa, for the area between y = x 3 , x = 1 , x = 2 and y = − x 3 .

31.5 Find limits of integration for the ellipse with boundary ( x A ) 2 + ( y B ) 2 = 1 , what are other potential variables?

Integrating over a volume can be done in exactly the same way.

You first select an order of integration, on the basis of your guess of what will be most convenient.

You then create a picture of the region of integration, and determine the intervals of integration starting from the last integration variable, working back to the first.

For the last one, you determine which of its values corresponds to a region that you wish to integrate over between each fixed set of bounding surfaces.

You then imagine that that variable, say y , is fixed in each such interval, and go on to the next variable (the next to last to be integrated over) and do the same thing in the surface obtained by fixing y .

When you are dealing with rectilinear coordinates this will be an area bounded by curves, and the rest of the procedure is exactly the same as before:

Find the appropriate interval bounds for z , say, and then imagine you have fixed a z value in each relevant interval then determine the limits on x .

Consider integration over a sphere of radius A , whose bounding surface obeys the equation x 2 + y 2 + z 2 = A 2 .

If we choose to integrate over x then z then y , we first notice that y can go from − A to A .

Given a value for y , z can vary from − A 2 − y 2 to A 2 − y 2 , and fixing z then x varies from − A 2 − y 2 − z 2 to A 2 − y 2 − z 2 .

Of course a sphere has much neater limits of integration in polar coordinates

0 < ρ < A , 0 < θ < 2 π , 0 < φ < π

The rectangular limits are tolerable though, if barely so.

Exercises:

31.6 Determine appropriate limits of integration if you wish to integrate some integrand over an ellipsoid with boundary ( x A ) 2 + ( y B ) 2 + ( z C ) 2 − 1 = 0 .

31.7 Determine appropriate limits in rectangular and cylindric coordinates for the region inside a cone of with boundary z = 1 and z = r 2 . (The origin is the bottom tip of this region.)

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