Double Integrals: Definition, Evaluation in Cartesian Coordinates & Area of Plane Regions

Explanation

Double integration extends single-variable calculus to functions of two independent variables over bounded planar domains. Evaluation is executed via iterated single integrals. The innermost integral treats the outer variable as a fixed parameter while integrating with respect to the inner variable. Careful determination of the bounding functions g₁(x) and g₂(x) is achieved by drawing test vertical or horizontal lines traversing the domain from entry boundary to exit boundary.

Key points

• Geometric Meaning: When $f(x,y)\ge 0$, ${\iint }_{R}f(x,y)dA$ represents the volume of the solid cylindrical column bounded above by surface $z=f(x,y)$ and below by plane region $R$.
• When $f(x,y)=1$, the double integral computes the exact Area of the planar region: $\text{Area}(R)={\iint }_{R}dxdy$.
• Type I (Vertical Slices): $R=\{(x,y)\mid a\le x\le b,g_1(x)\le y\le g_2(x)\}⟹{\iint }_{R}f(x,y)dA={\int }_a^b\left[{\int }_{g_1(x)}^{g_2(x)}f(x,y)dy\right]dx$.
• Type II (Horizontal Slices): $R=\{(x,y)\mid c\le y\le d,h_1(y)\le x\le h_2(y)\}⟹{\iint }_{R}f(x,y)dA={\int }_c^d\left[{\int }_{h_1(y)}^{h_2(y)}f(x,y)dx\right]dy$.
• Fubini's Theorem: If $f(x,y)$ is continuous on rectangular region $R=[a,b]\times [c,d]$, then ${\int }_a^b{\int }_c^{d}f(x,y)dydx={\int }_c^d{\int }_a^{b}f(x,y)dxdy$.

Example

Code examples

// Analytical Evaluation: Area Bounded by Parabola and Line
// Problem: Find the area enclosed by y = x^2 and y = 2x + 3
// Step 1: Find points of intersection:
//    x^2 = 2x + 3 => x^2 - 2x - 3 = 0 => (x - 3)(x + 1) = 0
//    Intersection points: x = -1 (y = 1) and x = 3 (y = 9)
// Step 2: For any x ∈ [-1, 3], vertical strip enters at y = x^2 (lower) and exits at y = 2x + 3 (upper)
// Step 3: Set up iterated double integral for Area:
//    Area = ∬_R dy dx = ∫_{-1}^3 [ ∫_{x^2}^{2x + 3} dy ] dx
//         = ∫_{-1}^3 [ y ]_{x^2}^{2x + 3} dx
//         = ∫_{-1}^3 (2x + 3 - x^2) dx
// Step 4: Evaluate antiderivative:
//    = [ x^2 + 3x - (x^3 / 3) ]_{-1}^3
//    At x = 3:  3^2 + 3(3) - (27/3) = 9 + 9 - 9 = 9
//    At x = -1: (-1)^2 + 3(-1) - (-1/3) = 1 - 3 + 1/3 = -5/3
//    Area = 9 - (-5/3) = 9 + 5/3 = 32 / 3 square units (≈ 10.67)

Exam questions

• [8 Marks] Evaluate ∬_R (x² + y²) dx dy over the region R bounded by y = x² and x = y².
• [7 Marks] Find the area of the region bounded by the parabola y² = 4ax and its latus rectum x = a using double integration.
• [7 Marks] Evaluate ∬_R x y (x + y) dx dy over the region bounded by y = x and y = x².