Now let’s add limits of integration. If the limits of integration for u are a and b, then the limits of integration for x will be x(a) and x(b).

Z x(b) Z b dx Jacobians f(x) dx = f(x(u)) du. Math 131 Multivariate Calculus x(a) a du D Joyce, Spring 2014 We want to generalize this to multiple integrals. Jacobians for change of variables. Since dou- Before we do, however, let’s change the interval ble integrals are iterated integrals, we can use the of integration into a domain of integration since usual substitution method when we’re only work- when we generalize to two variables, we’ll be talking ∗ ing with one variable at a time. But there’s also about domains. Let D denote the interval [a, b], a way to substitute pairs of variables at the same which is a subset of R, and let D = [x(a), x(b)] time, called a change of variables. Some integrals be the image of that interval. Then the rule for can be evaluated most easily by change of variables. substitution becomes In particular, changing to polar coordinates is often Z Z dx helpful. f(x) dx = f(x(u)) du. ∗ du If the original variables are (x, y), and the new D D variables are (u, v), then there’s a function T : Actually, this isn’t always valid, since when we R2 → R2 that gives u and v in terms of x and y, change from intervals to domains, we lose the orien- that is, T(u, v) = (x(u, v), y(u, v). We’ll see that tation of the interval. Take, for instance, the inte- we need something called the Jacobian, denoted Z 1 ∂(x, y) gral −x dx, and apply the substitution x = −u, , to effect a change of variables in double 0 ∂(u, v) dx = −du. With limits on our intervals, we get integrals. First, we’ll review ordinary substitution for sin- Z 1 Z −1 −x dx = −u (−du), gle variables to see what we’re generalizing. Sec- 0 0 ond, we’ll look at a change of variables in the spe- 1 cial case where that change is effected by a linear which is correct, since both integrals equal − 2 . transformation T : R2 → R2. Finally, we’ll look at But, in terms of domains of integration, we have the general case where T doesn’t have to be linear. Z Z −x dx = −u (−du), Recall the change of variables for single in- [0,1] [−1,0] tegrals. Let’s start with an indefinite integral which is wrong, because the right integral means Z Z 0 f(x) dx −u (−du). The problem is that domains are u=−1 subsets without orientation. Thus, the correct rule and apply a substitution x = x(u). Note that we’re for substitution when using domains has absolute using x both as a variable and a function, but if values of the derivative: you prefer, you can use a different symbol for the Z Z dx function. After substitution, we get the integral f(x) dx = f(x(u)) du, D D∗ du Z Z dx f(x) dx = f(x(u)) du. du and that’s the form we’ll be generalizing.

1 Linear transformations. Consider a linear Change of variables for double integrals. We transformation T : R2 → R2. Such a linear trans- have to make two generalizations to make that last formation can be described by a 2 × 2 matrix A. equation into a rule for change of variables in dou- We identify ordered pairs as column vectors, so ble integrals. First, the integrand has to be changed x from the constant 1 to a general scalar-valued func- (x, y) ∈ R2 is identified with , and (u, v) ∈ R2 y tion f. Second, the transformation T has be gener- u alized from a linear transformation to a nonlinear is identified with . Then the equation v transformation T. (x, y) = T(u, v) = (au + bv, cu + dv), Let’s replace the constant integrand 1 by a func- tion f : R2 → R. We’ll simply get becomes the matrix equation au + bv a b u ZZ ZZ T(u, v) = = , cu + dv c d v f(x, y) dx dy = f(T(u, v))| det(A)| du dv. D D∗ so that x u Here, T(u, v) = (x(u, v), y(u, v)). Again, we’re = T(u, v) = A y v treating x and y as both variables and functions. The justification for this generalization is that  a b  where A is the matrix . the solids whose volumes the double integrals de- c d Note that the entries of the matrix A which de- scribe are being stretched/squeezed from the (u, v)- scribes the linear transformation t are actually par- plane to the (x, y)-plane, but their heights, which tial derivatives of T. are given by f, aren’t being changed at all. 2 2 ∂x ∂x Next, let T be any transformation R → R , not necessarily a linear transformation. (Although T is ∂u ∂v  A = ∂y ∂y  . a vector-valued function, and, in fact, it’s a vector ∂u ∂v field, we’ll call it a transformation because we’re A matrix A sends the unit square (the square treating it in a different way.) with two sides being the standard unit vectors i The matrix A of partial derivatives (which is a and j) to a parallelogram with two sides being the constant matrix when T is a linear transformation)  a   b  has a determinant which is called the Jacobian of columns of A, namely, and . The area c d T and denoted of this parallelogram is | det(A)|, the absolute value ∂x ∂x ∗ ∂(x, y) of the determinant of A. More generally, if D is DT(u, v) = = ∂u ∂v . 2 ∗ ∂y ∂y any region in R , and D = T(D ) is its image ∂(u, v) under this linear transformation, then the area of ∂u ∂v D is | det(A)| times the area of D∗. Although T is not a linear transformation, this Ja- Now, let’s look at double integrals. First consider cobian describes the stretching/squeezing at partic- the case when the integrand is the constant 1. Then ular points, and so the general change of variables RR D 1 dx dy equals the area of D. We can rewrite has the same equation. the final statement in the last paragraph ZZ ZZ ∂(x, y) Area(D) = | det(A)| Area(D∗), f(x, y) dx dy = f(T(u, v)) du dv, D D∗ ∂(u, v) in terms of integrals as ∂(x, y) ZZ ZZ where denotes the absolute value of the 1 dx dy = 1 | det(A)| du dv. ∂(u, v) D D∗ Jacobian.

2 Examples of change of variables in double integrals. Determine the value of

ZZ r x + y dA D x − 2y where D is the region in R2 enclosed by the lines y = x/2, y = 0, and x + y = 1.

Math 131 Home Page at http://math.clarku.edu/~djoyce/ma131/

3