1) Introduction: Expanding to Two Dimensions 📚
After mastering single-variable integrals and partial derivatives, we now step into the realm of double integrals. These extend the concept of integration to two dimensions, allowing us to compute areas, volumes, and other quantities over regions in the plane. Double integrals are essential for applications in physics (e.g., mass distribution), probability, and engineering. This topic will guide you from the basics to advanced techniques, ensuring a complete understanding.
We’ll cover:
- Definition and Notation: Understanding double integrals and their setup.
- Computation Over Rectangles: Integrating over simple regions.
- Non-Rectangular Regions: Handling more complex areas.
- Applications: Calculating volume, area, and mass.
Quick Recap: A single integral \( \int_a^b f(x) \, dx \) gives area under a curve; a double integral extends this to a surface.
2) Definition and Notation 🎓
A double integral \( \iint_R f(x, y) \, dA \) computes the signed volume under the surface \( z = f(x, y) \) over a region \( R \) in the \( xy \)-plane. It’s the limit of Riemann sums over a grid.
Definition 20.1: Double Integral
For a function \( f(x, y) \) over a region \( R \), the double integral is \( \iint_R f(x, y) \, dA = \lim_{m, n \to \infty} \sum_{i=1}^m \sum_{j=1}^n f(x_{ij}, y_{ij}) \Delta A \), where \( \Delta A \) is the area of each subrectangle.
Example: \( \iint_R (x + y) \, dA \) over \( R = [0, 1] \times [0, 1] \).
**Iterated Integrals**: Computed as \( \int_a^b \int_c^d f(x, y) \, dy \, dx \) or \( \int_c^d \int_a^b f(x, y) \, dx \, dy \), depending on the order.
Example 1: Double Integral Over a Rectangle
Evaluate \( \iint_R 1 \, dA \) where \( R = [0, 2] \times [0, 3] \).
- This is the area: \( \int_0^2 \int_0^3 1 \, dy \, dx \).
- Inner integral: \( \int_0^3 1 \, dy = 3 \).
- Outer integral: \( \int_0^2 3 \, dx = 3 \cdot 2 = 6 \).
Answer: 6 (area of rectangle).
Example 2: Constant Function
Compute \( \iint_R 4 \, dA \) over \( R = [-1, 1] \times [0, 2] \).
- \( \int_{-1}^1 \int_0^2 4 \, dy \, dx \).
- Inner: \( \int_0^2 4 \, dy = 4 \cdot 2 = 8 \).
- Outer: \( \int_{-1}^1 8 \, dx = 8 \cdot 2 = 16 \).
Answer: 16.
3) Computation Over Rectangles 📐
For a rectangular region \( R = [a, b] \times [c, d] \), the double integral is an iterated integral. The order of integration (dx dy or dy dx) can often be interchanged if \( f \) is continuous.
Definition 20.2: Iterated Integral
\( \iint_R f(x, y) \, dA = \int_a^b \int_c^d f(x, y) \, dy \, dx = \int_c^d \int_a^b f(x, y) \, dx \, dy \), provided the integrals exist.
Example 3: Linear Function
Evaluate \( \iint_R (2x + y) \, dA \) over \( R = [0, 1] \times [0, 2] \).
- \( \int_0^1 \int_0^2 (2x + y) \, dy \, dx \).
- Inner: \( \int_0^2 (2x + y) \, dy = [2xy + \frac{y^2}{2}]_0^2 = 4x + 2 \).
- Outer: \( \int_0^1 (4x + 2) \, dx = [2x^2 + 2x]_0^1 = 2 + 2 = 4 \).
Answer: 4.
Example 4: Changing Order
Compute \( \iint_R xy \, dA \) over \( R = [0, 2] \times [1, 3] \) both ways.
- \( \int_0^2 \int_1^3 xy \, dy \, dx \): Inner \( \int_1^3 xy \, dy = x \cdot \frac{y^2}{2} \big|_1^3 = x \cdot \frac{9 - 1}{2} = 4x \), Outer \( \int_0^2 4x \, dx = 2x^2 \big|_0^2 = 8 \).
- \( \int_1^3 \int_0^2 xy \, dx \, dy \): Inner \( \int_0^2 xy \, dx = y \cdot \frac{x^2}{2} \big|_0^2 = y \cdot 2 = 2y \), Outer \( \int_1^3 2y \, dy = y^2 \big|_1^3 = 9 - 1 = 8 \).
Answer: 8 (same both ways).
4) Non-Rectangular Regions 🔍
Double integrals over non-rectangular regions require defining \( R \) by inequalities (e.g., \( x^2 + y^2 \leq 1 \)). Use limits that depend on the other variable.
Definition 20.3: Non-Rectangular Region
For \( R \) defined by \( a \leq x \leq b \), \( g_1(x) \leq y \leq g_2(x) \), the integral is \( \int_a^b \int_{g_1(x)}^{g_2(x)} f(x, y) \, dy \, dx \). Alternatively, switch orders if \( c \leq y \leq d \), \( h_1(y) \leq x \leq h_2(y) \).
Example 5: Triangular Region
Evaluate \( \iint_R x \, dA \) where \( R \) is \( 0 \leq x \leq 1 \), \( 0 \leq y \leq x \).
- \( \int_0^1 \int_0^x x \, dy \, dx \).
- Inner: \( \int_0^x x \, dy = x \cdot y \big|_0^x = x^2 \).
- Outer: \( \int_0^1 x^2 \, dx = \frac{x^3}{3} \big|_0^1 = \frac{1}{3} \).
Answer: \( \frac{1}{3} \).
Example 6: Circular Region
Compute \( \iint_R 1 \, dA \) where \( R \) is \( x^2 + y^2 \leq 1 \).
- Use \( x \) from \(-1\) to \(1\), \( y \) from \(-\sqrt{1 - x^2}\) to \(\sqrt{1 - x^2}\).
- \( \int_{-1}^1 \int_{-\sqrt{1 - x^2}}^{\sqrt{1 - x^2}} 1 \, dy \, dx \).
- Inner: \( \int_{-\sqrt{1 - x^2}}^{\sqrt{1 - x^2}} 1 \, dy = 2\sqrt{1 - x^2} \).
- Outer: \( \int_{-1}^1 2\sqrt{1 - x^2} \, dx \) (area of circle, \( \pi r^2 = \pi \cdot 1^2 = \pi \)).
Answer: \( \pi \).
5) Applications of Double Integrals 🔍
Double integrals calculate area, volume under a surface, and properties like mass or center of mass. For a density \( \rho(x, y) \), mass is \( \iint_R \rho(x, y) \, dA \).
Definition 20.4: Applications
- Area: \( \iint_R 1 \, dA \). - Volume: \( \iint_R f(x, y) \, dA \) if \( f \geq 0 \). - Mass: \( \iint_R \rho(x, y) \, dA \) with density \( \rho \).
Example 7: Volume Under a Surface
Find the volume under \( z = 4 - x^2 - y^2 \) over \( R = [0, 1] \times [0, 1] \).
- \( \iint_R (4 - x^2 - y^2) \, dA \).
- \( \int_0^1 \int_0^1 (4 - x^2 - y^2) \, dy \, dx \).
- Inner: \( \int_0^1 (4 - x^2 - y^2) \, dy = [4y - x^2 y - \frac{y^3}{3}]_0^1 = 4 - x^2 - \frac{1}{3} \).
- Outer: \( \int_0^1 (4 - x^2 - \frac{1}{3}) \, dx = [4x - \frac{x^3}{3} - \frac{x}{3}]_0^1 = 4 - \frac{1}{3} - \frac{1}{3} = \frac{11}{3} \).
Answer: \( \frac{11}{3} \).
Example 8: Mass with Density
Find the mass of a plate \( R = [0, 1] \times [0, 1] \) with density \( \rho(x, y) = x + y \).
- \( \iint_R (x + y) \, dA \).
- \( \int_0^1 \int_0^1 (x + y) \, dy \, dx \).
- Inner: \( \int_0^1 (x + y) \, dy = [xy + \frac{y^2}{2}]_0^1 = x + \frac{1}{2} \).
- Outer: \( \int_0^1 (x + \frac{1}{2}) \, dx = [\frac{x^2}{2} + \frac{x}{2}]_0^1 = \frac{1}{2} + \frac{1}{2} = 1 \).
Answer: Mass = 1 unit.
6) Practice Questions 🎯
Fundamental Practice Questions 🌱
Instructions: Evaluate the double integral over the given region. 📚
\( \iint_R 1 \, dA \) where \( R = [0, 2] \times [0, 3] \)
\( \iint_R x \, dA \) where \( R = [0, 1] \times [0, 2] \)
\( \iint_R (x + y) \, dA \) where \( R = [0, 1] \times [0, 1] \)
\( \iint_R y^2 \, dA \) where \( R = [-1, 1] \times [0, 2] \)
\( \iint_R \cos(x) \, dA \) where \( R = [0, \pi] \times [0, 1] \)
\( \iint_R x^2 + y^2 \, dA \) where \( R = [0, 1] \times [0, 1] \)
\( \iint_R e^x \, dA \) where \( R = [0, 1] \times [0, 2] \)
\( \iint_R xy \, dA \) where \( R = [0, 2] \times [0, 1] \)
\( \iint_R \sin(x + y) \, dA \) where \( R = [0, \pi/2] \times [0, \pi/2] \)
\( \iint_R 1 \, dA \) where \( R = [-2, 2] \times [0, 1] \)
\( \iint_R x^2 y \, dA \) where \( R = [0, 2] \times [0, 3] \)
Challenging Practice Questions 🌟
Instructions: Solve these advanced double integrals or applications. 🧠
Evaluate \( \iint_R (x^2 + y^2) \, dA \) where \( R \) is \( 0 \leq x \leq 1 \), \( 0 \leq y \leq \sqrt{1 - x^2} \).
Compute the volume under \( z = 6 - 2x - 3y \) over \( R = [0, 1] \times [0, 2] \).
Find the mass of a plate \( R = [0, 2] \times [0, 1] \) with density \( \rho(x, y) = x^2 + y \).
Determine \( \iint_R e^{x+y} \, dA \) where \( R \) is \( 0 \leq x \leq 1 \), \( x \leq y \leq 1 \).
Evaluate the area of the region \( R \) defined by \( x^2 + y^2 \leq 4 \) using a double integral.
7) Summary & Cheat Sheet 📋
7.1) Double Integral
\( \iint_R f(x, y) \, dA \) computes volume or properties over \( R \).
7.2) Iterated Integral
\( \int_a^b \int_c^d f(x, y) \, dy \, dx \) or reverse order for rectangles.
7.3) Applications
Area, volume, mass via \( \iint_R f(x, y) \, dA \).
You’ve mastered double integrals! Next, we’ll tackle triple integrals. 🎉