Recipe

Perform a change of variables in a double integral using the Jacobian

Use when: R's boundary is awkward in x,y (a tilted parallelogram, a region between hyperbolas) but becomes simple under a given or chosen substitution x=x(u,v),\ y=y(u,v)

Goal: the integral rewritten over a simpler region S in u,v, with the Jacobian paying for the change

Taught in Change of variables in double integrals · a later section extends this recipe with more cases

1. 1

   Check that each pair of the region's boundary curves becomes a pair of constants under the substitution, so R maps to a rectangle S in u,v.

   Example

   Find \displaystyle\iint_R xy\,dA where R is bounded by x+2y=0, x+2y=3, x-y=0, and x-y=2. Let u=x+2y,\ v=x-y: each pair of boundary lines becomes one constant bound on u or v, so R maps to the rectangle S=\{0\le u\le3,\ 0\le v\le2\}.

   Why: a substitution is only worth using here if it turns the hard-to-integrate-over region into something with constant bounds; that's the entire reason to trade x,y for u,v in the first place.

2. 2

   Invert the substitution algebraically to get x=x(u,v) and y=y(u,v).

   Example

   From u=x+2y and v=x-y: subtracting gives u-v=3y, so y=\dfrac{u-v}{3}, and then x=v+y=\dfrac{u+2v}{3}.

   Why: both the Jacobian in the next step and the integrand itself need to be written as functions of u,v, which requires x and y each isolated in terms of them.

3. 3

   Form \dfrac{\partial(x,y)}{\partial(u,v)}=\det\begin{bmatrix}\partial x/\partial u & \partial x/\partial v\\ \partial y/\partial u & \partial y/\partial v\end{bmatrix}, then take \left|\dfrac{\partial(x,y)}{\partial(u,v)}\right|.

   Example

   \dfrac{\partial(x,y)}{\partial(u,v)}=\det\begin{bmatrix}1/3 & 2/3\\ 1/3 & -1/3\end{bmatrix}=\left(\tfrac13\right)\left(-\tfrac13\right)-\left(\tfrac23\right)\left(\tfrac13\right)=-\tfrac19-\tfrac29=-\tfrac13, so \left|\dfrac{\partial(x,y)}{\partial(u,v)}\right|=\dfrac13.

   • Not yet: the same idea in three variables, with a 3x3 Jacobian in place of this 2x2 one — covered in Change of variables in triple integrals

   Why: near any point the substitution is well approximated by this linear map, and a determinant is exactly the tool that measures how much a linear map scales area. That's precisely what the region's tiny pieces need correcting for.

4. 4

   Substitute x(u,v) and y(u,v) into the integrand, and integrate over the rectangle S with dA=\left|\dfrac{\partial(x,y)}{\partial(u,v)}\right|du\,dv.

   Example

   xy=\dfrac{u+2v}{3}\cdot\dfrac{u-v}{3}=\dfrac{u^2+uv-2v^2}{9}, so \displaystyle\iint_R xy\,dA=\iint_S \frac{u^2+uv-2v^2}{9}\cdot\frac13\,du\,dv = \frac1{27}\int_0^2\!\int_0^3 (u^2+uv-2v^2)\,du\,dv.

   • both the region and the integrand get simpler — the substitution is a clear win
   • the region becomes a rectangle but the integrand gets messier — it's usually still worth it, since constant bounds matter more to a hand computation than a slightly harder integrand

   Why: this is exactly the change of variables formula, and it's what makes the awkward original region disappear from the problem entirely once S takes its place.

5. 5

   Finish the double integral over S the same way any rectangular double integral is evaluated.

   Example

   \displaystyle\int_0^3 (u^2+uv-2v^2)\,du = \left[\frac{u^3}{3}+\frac{u^2v}{2}-2uv^2\right]_0^3 = 9+\frac{9v}{2}-6v^2. Then \displaystyle\int_0^2\left(9+\frac{9v}{2}-6v^2\right)dv = \left[9v+\frac{9v^2}{4}-2v^3\right]_0^2 = 18+9-16=11. So the answer is \dfrac{1}{27}\cdot11=\boxed{\dfrac{11}{27}}.

   Why: a rectangle's bounds are always constants on both variables, so this final integral is never harder than the ordinary iterated integrals from earlier in the chapter, whatever the original region looked like.