General Change of Variables
Textbook Reference
| Primary source | OpenStax Calculus Volume 3, Section 5.7: “Change of Variables in Multiple Integrals” |
| Direct link | https://openstax.org/books/calculus-volume-3/pages/5-7-change-of-variables-in-multiple-integrals |
| Textbook used in class | Stewart, Calculus, Section 15.10: “Change of Variables in Multiple Integrals” (Examples 3, 4) |
Quick Reference
Change of variables theorem: $$\iint_R f(x,y)\,dA = \iint_S f(g(u,v), h(u,v))\left|\frac{\partial(x,y)}{\partial(u,v)}\right|\,du\,dv.$$
where $x = g(u,v)$, $y = h(u,v)$ is a one-to-one transformation mapping $S$ (in $uv$-space) to $R$ (in $xy$-space).
Motivation
Polar coordinates are a special case of this general theorem. The theorem allows ANY invertible change of variables to simplify a double integral -- either by transforming a complicated region into a rectangle, or by simplifying the integrand (or both).
The strategy: look at the region $R$ and the integrand; find a transformation that maps a simpler region $S$ to $R$, or that simplifies the integrand. Common choices include linear transformations (mapping parallelograms to rectangles) and elliptical substitutions (mapping ellipses to circles).
Key Concepts
1. Elliptical Substitution
For an ellipse $x^2/a^2 + y^2/b^2 = 1$, the substitution $x = au$, $y = bv$ maps the ellipse to the unit circle $u^2+v^2 = 1$. The Jacobian is $ab$ (the area of the ellipse is $\pi ab$ times the area of the unit circle $\pi$).
2. Parallelogram Substitution
If the region $R$ is a parallelogram bounded by $y = mx+k_1$, $y = mx+k_2$, $y = nx+l_1$, $y = nx+l_2$, set $u = y-mx$, $v = y-nx$ to transform $R$ to a rectangle.
Worked Example
Evaluate $\iint_R x\,dA$ where $R$ is the region bounded by $x-y=0$, $x-y=1$, $x+y=0$, $x+y=3$, using $u = x-y$, $v = x+y$. (Stewart 15.10, Example 3.)
Find the Jacobian. We need $\partial(x,y)/\partial(u,v)$.
From $u = x-y$ and $v = x+y$: $x = (u+v)/2$, $y = (v-u)/2$.
$$J = \frac{\partial(x,y)}{\partial(u,v)} = \begin{vmatrix}1/2 & 1/2 \\ -1/2 & 1/2\end{vmatrix} = \frac{1}{4}+\frac{1}{4} = \frac{1}{2}.$$
New region $S$ in $(u,v)$-space: $0 \leq u \leq 1$, $0 \leq v \leq 3$ (a rectangle).
Integrand: $x = (u+v)/2$.
$$\iint_R x\,dA = \int_0^3\int_0^1 \frac{u+v}{2}\cdot\frac{1}{2}\,du\,dv = \frac{1}{4}\int_0^3\int_0^1(u+v)\,du\,dv.$$
$$= \frac{1}{4}\int_0^3\left[\frac{u^2}{2}+vu\right]_0^1 dv = \frac{1}{4}\int_0^3\left(\frac{1}{2}+v\right)dv = \frac{1}{4}\left[\frac{v}{2}+\frac{v^2}{2}\right]_0^3 = \frac{1}{4}\cdot\frac{3+9}{2} = \frac{12}{8} = \frac{3}{2}.$$
the Jacobian in the formula is $\partial(u,v)/\partial(x,y)$ (the inverse Jacobian). The formula uses $|\partial(x,y)/\partial(u,v)|$ -- the Jacobian of the FORWARD transformation from $(u,v)$ to $(x,y)$. If you find the Jacobian of the inverse (from $(x,y)$ to $(u,v)$), you must take its reciprocal. Computing the forward Jacobian directly (from $x = g(u,v)$, $y = h(u,v)$) avoids this inversion.
Leveled Practice
Problem 1. Use the substitution $x = 2r\cos\theta$, $y = 3r\sin\theta$ to evaluate $\iint_D 1\,dA$ where $D$ is the ellipse $x^2/4 + y^2/9 \leq 1$.
Show answer
The substitution maps the unit disk $r \leq 1$ to the ellipse.
$J = \partial(x,y)/\partial(r,\theta)$: $x_r = 2\cos\theta$, $x_\theta = -2r\sin\theta$, $y_r = 3\sin\theta$, $y_\theta = 3r\cos\theta$.
$J = 6r\cos^2\theta + 6r\sin^2\theta = 6r$.
$\iint_D dA = \int_0^{2\pi}\int_0^1 6r\,dr\,d\theta = 6\pi$.
This confirms the area of the ellipse is $\pi ab = \pi(2)(3) = 6\pi$.