iterated integrals?

Question

how do I know which is the correct order in ANY iterated integral: dxdy or dydx?

also, how do I do this one:
SSR x(x-1)e^(xy)dA where R is a triangular region bounded by x=0, y=0, x+y=2

Answer 1

The right order is the order the symbols dx and dy are presented. If dx comes fists, then you first integrate with respect to x and, then, with respect to y.

The integral above is not difficult to do, but requires a lot of work. I'll sketch a solution. First of all, we have to determine the integration limits, So, we have to find the x intercept of the line x + y = 2. We readily see it is (2 0). So, x varies from o to 2 and y varies from 0 to 2 -x, that is, the integration limit of y depends on x.

Therefore, we gotta compute Int(0 to 2) (Int (0 to 2-x) x(x-1)e^(xy)dy ) dx. First, let's evaluate the inner integral in y. We get x(x-1) Int (0 to 2-x) e^(xy) dy = x(x-1) [e^(xy)/x] (y=0 to y=2 -x) = (x-1)[e^(x(2-x)) - 1]. Now we have to integrat this function of x from 0 to 2. It's not difficult to find it's antiderivative, but it's kinda tiring. I`m sure you can do it. If you have any question, email me.

Answer 2

The order of integration depends on how you set up the limits. If both sets of limits are simple constants, you can integrate in either order. However, if the limits on y contain a fuinction of x you must integrate on y first. If you try to do it the other way there will be xs laying around in your final answer. Of course the inverse is true if the x limit is a function of y.

In your specific example, you can define the limits either way. Looking at the limit of x+y = 2, it is easy to show that the lower x limit of zero means that y cannot exceed 2. Similarly x cannot exceed 2. You can set the limits up as:

x ranges from 0 to 2
y ranges from 0 to 2-x
Then integrate on y first since the y upper limit is a function of x

Or:

y ranges from 0 to 2
x ranges from 0 to 2-y
Then integrate on x first since the x upper limit is a function of y

Although you can integrate in either order. Integrating on y first is much easier because there is only a single y in the integrand so it doesn't complicate things anywhere near as much:

INT(x*(x-1)*e^(x*y)) dy = (x-1)*e^(x*y)

For limits 0 to 2 - x: (x-1)*(e^(x*(2-x)) - 1) = (x-1)*(e^(2*x - x^2) - 1)

You need to integrate this on x. This looks like a pain but it isn't so bad. Since this is homework, there must be a trick. Check out a form of e^u * du. u = 2*x - x^2, du = 2 - 2x = -2*(x-1).

What a coincidence, -2*(x-1)*e^(2*x - x^2) is of the form e^u * du. Homeworkis so much easier than real life. So integrating on x gives:

-1/2 * e^(2*x - x^2) - (x^2)/2 + x

Put in the limits of 0 to 2:

( -1/2 * e^(2*2 - 2^2) - (2^2)/2 + 2) - ( -1/2 * e^(2*0 - 0^2) - (0^2)/2 + 0) = (-1/2 * 1 - 2 + 2) - (-1/2 * 1 - 0 + 0) = 0

That's right, all that work for an area of zero. There is as much above the xy plane as below.