INSTRUCTOR: Let D be the region in xyz space, defined by 1 is less than or equal to x is less than or equal to 2. 0 is less than or equal to xy is less than or equal to 2. And 0 is less than or equal to z is less than or equal to 1. Evaluate the triple integral over the region D with respect to z then y then x of the function x squared y plus 3 xyz by using the transformation u equals x, v equals xy, and w equals 3z.

Let's first write our transformations. This will be T of u comma v comma w. And that will be equal to parentheses u comma v over u comma one third w close parentheses. And that is us rearranging those values that we were given, x equals u, y equals v over x which would be v over u, and z equals w over 3 because w equals 3z, just rearranging that.

Now we can write our Jacobian matrix, J of u comma v comma w. That'll be equal to the determinant of a 3 by 3 matrix. And our first row, a value of 1 for the first entry, second entry is negative v over u squared. And our third entry is 0.

Our second row, first entry of 0, second entry of 1 divided by u, and a third entry of 0, and our third row, the first entry of 0, a second entry of 0, and a third of 1/3. And when we find the determinant of that matrix, using probably cofactor expansion or something like that, we get 1 over the denominator 3u, end fraction. Now, along with this, we want to define what our region is so we can set up our integral.

We can define the region D as the set of ordered triples u comma v comma w such that 1 is less than or equal to-- So let me set that up differently. So the ordered pairs u comma v comma w such that 0 is less than or equal to w is less than or equal to 3. 0 is less than or equal to v is less than or equal to 2. And 1 is less than or equal to u is less than or equal to 2.

Now, that will set up our integral as being the triple integral. First outermost integral is from 0 to 3 with respect to w of the integral from 0 to 2 with respect to v of the integral from 1 to 2 with respect to u of the function 1 over 3u, end fraction, open parentheses, uv plus vw, close parentheses. Now take a moment here and convince yourself that x squared y becomes u times v and that 3 xyz becomes vw.

Look at the relationships we have. u equals x, v equals xy, w equals 3z. And I hope you can convince yourself of that fairly quickly. All right, we'll take our innermost integral. This will become integral from 0 to 3 with respect to w of the integral from 0 to 2 with respect to v of the function 1/3 times v plus 1/3 times vw times natural log of u.

Actually, that's going to be a natural log of 2. That function is 1/3 v plus 1/3 natural log of 2 vw. OK, yeah, that's correct. All right, now that will be equal to the integral from 0 to 3 with respect to w of 2/3 plus 2/3 ln of 2 times w. And evaluating this one final time with respect to w, this becomes 2 plus natural log of 8 or 2 plus 3 natural log of 2.