Double Integrals over General Regions
Textbook Reference
| Primary source | OpenStax Calculus Volume 3, Section 5.2: “Double Integrals over General Regions” |
| Direct link | https://openstax.org/books/calculus-volume-3/pages/5-2-double-integrals-over-general-regions |
| Textbook used in class | Stewart, Calculus, Section 15.3: “Double Integrals over General Regions” (Examples 3, 4, 5) |
Quick Reference
Reversing order: If an integral $\int_a^b\int_{g_1(x)}^{g_2(x)} f\,dy\,dx$ is hard, redraw the region and rewrite as $\int_c^d\int_{h_1(y)}^{h_2(y)} f\,dx\,dy$.
Area: $\text{Area}(D) = \iint_D 1\,dA$.
Splitting regions: $\iint_D f\,dA = \iint_{D_1}f\,dA + \iint_{D_2}f\,dA$ when $D = D_1 \cup D_2$ and $D_1, D_2$ overlap only on a curve.
Motivation
Sometimes an iterated integral in one order cannot be evaluated because the antiderivative does not exist in closed form. Reversing the order of integration (switching from Type I to Type II or vice versa) can unlock a tractable antiderivative. This is the most powerful technique for dealing with iterated integrals that look impossible.
Key Concepts
1. Reversing the Order of Integration
To reverse the order:
- Sketch the region of integration from the original limits.
- Describe the region the other way (if originally Type I, redescribe as Type II).
- Write the new iterated integral.
The value of the integral does not change -- only the description of the region changes.
2. Using the Area Formula
Since $\iint_D 1\,dA = \text{Area}(D)$, you can verify your region by computing its area two ways: geometrically and as an integral.
Worked Example
Evaluate $\int_0^1\int_x^1 \sin(y^2)\,dy\,dx$ by reversing the order. (Stewart 15.3, Example 4.)
The inner integral $\int \sin(y^2)\,dy$ has no closed form. Reverse the order.
Identify the region from the original limits: $0 \leq x \leq 1$, $x \leq y \leq 1$. This is the triangle with $0 \leq x \leq y$, $0 \leq y \leq 1$.
Redescribe as Type II: for each $y \in [0,1]$, $x$ runs from $0$ to $y$.
New iterated integral: $$\int_0^1\int_0^y \sin(y^2)\,dx\,dy.$$
Evaluate: $$= \int_0^1 \left[x\sin(y^2)\right]_{x=0}^{x=y}\,dy = \int_0^1 y\sin(y^2)\,dy.$$
Let $u = y^2$, $du = 2y\,dy$: $$= \int_0^1 \frac{1}{2}\sin(u)\,du = \frac{1}{2}[-\cos u]_0^1 = \frac{1}{2}(1-\cos 1).$$
The answer is $\dfrac{1-\cos 1}{2} \approx 0.230$.
reversing the order changes the value of the integral. Fubini’s theorem guarantees that the two orders give the same value (for continuous $f$). Reversing the order changes ONLY the limits and the order of $dx$ and $dy$ -- NOT the integrand itself (except that limits that were functions of $x$ now become functions of $y$). The value is unchanged.
Leveled Practice
Problem 1. Reverse the order of integration and evaluate: $\int_0^4\int_{\sqrt{x}}^{2} e^{y^3}\,dy\,dx$.
Show answer
Region: $0 \leq x \leq 4$, $\sqrt{x} \leq y \leq 2$. Since $y \geq \sqrt{x}$, we have $x \leq y^2$.
Type II: $0 \leq y \leq 2$, $0 \leq x \leq y^2$.
$$\int_0^2\int_0^{y^2} e^{y^3}\,dx\,dy = \int_0^2 y^2 e^{y^3}\,dy.$$
Let $u = y^3$: $= \tfrac{1}{3}[e^{y^3}]_0^2 = \tfrac{1}{3}(e^8 - 1)$.