INSTRUCTOR: Using the substitutions x equals v and y equals the square root of u plus v, end square root, evaluate the interval the double integral over the region R of the function y sin open parentheses y squared minus x, close parentheses, where R is the region bounded by the lines, or the curves, y equals the square root of x, x equals 2, and y equals 0. So first, let's go and just write what that transformation is. So the transformation, T of u comma v, is equal to parentheses v comma the square root of u plus v, end square root and then close parentheses. That's based on our x equals and y equals relationships we have in this question. That'll be important in a minute.

Now, we have these boundaries of y equals x squared, x equals 2, and y equals 0. And actually, there's one more boundary. And that is that these intersect when x equals 0. So thinking about the points of that region, x equals 0 is a boundary.

So I'm just going to write all of these, x equals 0, x equals 2, y equals square root of x, and y equals 0. Let's convert these to be in terms of our variables v and u. Now, since x is equal to v, that means that x equals 0 means v equals 0. And similarly, x equals 2 means that v equals 2.

However, when we look at y equals the square root of x, well, since the equation we have to relate this is the square root of u plus v, that means that the square root of u plus v as a quantity equals the square root of v because that's x is equal to v. That actually necessitates that u is equal to 0.

So that restriction converts into u being 0. And then y equals 0. For y equals 0, we would say the square root of u plus v, end square root, equals 0, which gives us a relationship that u is equal to negative v.

Now, I've done this on purpose so that I have two of my statements-- Well, two of them automatically came as v equals boundaries. But I wanted that last one in terms of u so that I would have two boundaries for v and two boundaries for u. So we can describe our region R, describe our region R as the set of all ordered pairs u comma v such that v is bounded between 0 and 2, that is, 0 is less than or equal to v is less than or equal to 2. And negative v is less than or equal to u is less than or equal to 0.

Now, the order here I'm also thinking in is important based on what we know about this. All right, now that we have the transformation, we need to find the Jacobian of that. And in fact, we specifically need the absolute value of the Jacobian. So let's go ahead and find that.

So the Jacobian matrix of u comma v is equal to the determinant of a first row of 0. And our second element in that first row is 1/2, open parentheses u plus v, close parentheses, to the negative 1/2. That is the derivative of y with respect to u. And then our second row is 1. And our second element in that second row is 1/2, open parentheses u plus v, close parentheses, to the negative 1/2.

Now, when we take the determinant of that matrix, finding our diagonal products there, this is going to be equal to negative 1 over, this is our whole denominator, 2 square root of u plus v, end square root. Now we are going to want the absolute value of this, which not to be confused with the determinant. The absolute value of that will become, this will be equal to 1 over denominator of 2 square root of u plus v, end square root.

All right, now, let's revisit the integral that we actually have. Our integral over the region R, in fact, let me write this is as double integral over the region R, of the function y sin, open parentheses, y squared minus x, close parentheses. That's dA, will be equal to the integral from 0 to 2 with respect to v of the integral from negative v to 0 with respect to u of our new function, which will be 1 over a denominator of 2 square root of u plus v, end square root, end denominator, multiplied by square root of u plus v, end square root, times sin have u.

We include the absolute value of our Jacobian in this product. And that's where that's coming from. But we converted, so we said y was equal to the square root of u plus v, end square root, made that substitution. And then, because, let's see, because u would be equal to y squared minus v, that would, again, v and x being the same, y squared minus x becomes u. All right, now, let's actually find the antiderivative here because the simplifies nicely.

This becomes the integral from 0 to 2 dv with respect to v of the integral from negative v to 0 with respect to u of 1/2 times sin u. Find the antiderivative and then evaluate that. This becomes the integral from 0 to 2 dv, or with respect to v, of the function negative 1/2 plus 1/2 half times cosine v.

Now in reality, it's actually a cosine of negative v. But because cosine is an even function, yes, symmetric about the y-axis, because that's the case, we can actually trade that out for just a cosine of v. And then we can evaluate this from the 0 to 2, after we find the antiderivative. And that will be 1/2 sin of 2, close parentheses, minus 1.