Volume Under a Surface
A single integral finds area. A Double Integral finds volume under a surface \(f(x, y)\). For a rectangle \(R\), we integrate twice:
\[V = \iint_R f(x, y) dA = \int_c^d \int_a^b f(x, y) dx dy\]
According to Fubini's Theorem, you can integrate in either order (\(dx dy\) or \(dy dx\)) and get the same result.
Worked Examples
Example 1: Basic Volume
Evaluate \(\int_0^2 \int_0^1 (x + y) dx dy\).
- Inner: \(\int_0^1 (x+y) dx = [\frac{x^2}{2} + xy]_0^1 = 0.5 + y\).
- Outer: \(\int_0^2 (0.5 + y) dy = [0.5y + \frac{y^2}{2}]_0^2 = 1 + 2 = 3\).
- Result: 3.
The Bridge to Quantum Mechanics
Quantum Mechanics in 3D is built on multiple integrals. To find the probability of finding a particle in a 3D box, we must integrate the density \(|\psi|^2\) over all three dimensions: \(\iiint |\psi|^2 dx dy dz\). The logic of double integrals is the first step toward handling these higher-dimensional spaces. Fubini's theorem ensures that the physics doesn't depend on whether we measure the x-position or the y-position first.