Double Integrals

Polar Rectangles and Integration

• Polar rectangle enables finding equivalent limits of integration in terms of $R$ and $θ$.

Factoring Integrals

• If the integrand can be factored as a function of $x$ times a function of $y$, the integral can be separated.
• Example:
  $\int \int x/(x^2+1) * y^2 dA = \int y^2 dy * \int x/(x^2+1) dx$

U-Substitution Example

• $u = x^2 + 1$
• $du = 2x dx$
• $du/2 = x dx$
• If $x = 1$, then $u = 2$
• If $x = 0$, then $u = 1$
• $\int<em>1^2 (1/2) du/u = (1/2) ln(u) |</em>1^2 = (1/2) ln(2)$

Combining Results

• If $\int \int x/(x^2+1) * y^2 dA = \int y^2 dy * \int x/(x^2+1) dx = 18 * (1/2)ln(2) = 9ln(2)$

Generalized Rectangles and Strong Fubini Theorem

• Generalized rectangles allow iterated integrals on non-rectangular regions.
• Sometimes called the strong version of the Fubini Theorem.

Type 1 Regions

• Bounded by two vertical lines $x = a$ and $x = b$.
• $y$ varies between two functions of $x$, $y = g<em>1(x)$ and $y = g</em>2(x)$.
• The integral of a function $f$ over a type 1 region $D$ is:
  $\int \int<em>D f(x, y) dA = \int</em>a^b \int<em>{g</em>1(x)}^{g_2(x)} f(x, y) dy dx$

Type 2 Regions

• Bounded by two horizontal lines $y = c$ and $y = d$.
• $x$ varies between two functions of $y$, $x = h<em>1(y)$ and $x = h</em>2(y)$.
• The integral of a function $f$ over a type 2 region $D$ is:
  $\int \int<em>D f(x, y) dA = \int</em>c^d \int<em>{h</em>1(y)}^{h_2(y)} f(x, y) dx dy$

Example: Evaluating Integral on Generalized Rectangle

• Evaluate $\int \int_D (x + 2y) dA$ where $D$ is bounded by $y = 2x^2$ and $y = 1 + x^2$.
• First step: sketch the region.
• Find the intersection points of the two curves:
  2x^2 = x^2 + 1 => x^2 = 1 => x = \pm 1
• Set up the double integral:
  $\int<em>{-1}^1 \int</em>{2x^2}^{x^2+1} (x + 2y) dy dx$
• Inner integral:
  $\int<em>{2x^2}^{x^2+1} (x + 2y) dy = xy + y^2 |</em>{2x^2}^{x^2+1} = x(x^2+1) + (x^2+1)^2 - [x(2x^2) + (2x^2)^2]$

Simplifying the Integrand

• Simplify the expression:
  $x^3 + x + x^4 + 2x^2 + 1 - 2x^3 - 4x^4 = -3x^4 - x^3 + 2x^2 + x + 1$

Using Symmetry to Simplify Integration

• Fact 1: If $g(x)$ is an even function, then $\int<em>{-a}^a g(x) dx = 2 \int</em>0^a g(x) dx$.
• Fact 2: If $g(x)$ is an odd function, then $\int_{-a}^a g(x) dx = 0$.
• Apply these facts to the integral:
  $\int<em>{-1}^1 (-3x^4 - x^3 + 2x^2 + x + 1) dx = 2 \int</em>0^1 (-3x^4 + 2x^2 + 1) dx$
• Integrate:
  $2 [-3x^5/5 + 2x^3/3 + x]_0^1 = 2 [-3/5 + 2/3 + 1] = 2 [-9/15 + 10/15 + 15/15] = 2 [16/15] = 32/15$

Example 2: Generalized Rectangles

• Evaluate $\int \int_D xy dA$ where $D$ is bounded by $y = x - 1$ and $y^2 = 2x + 6$.
• Sketch the region.
• Solve for the intersection points:
  (x-1)^2 = 2x + 6 => x^2 - 2x + 1 = 2x + 6 => x^2 - 4x - 5 = 0 => (x-5)(x+1) = 0
• $x = 5$ or $x = -1$
• Corresponding $y$ values: If $x = 5$, $y = 4$. If $x = -1$, $y = -2$.
• Express $x$ as a function of $y$: $x = y + 1$ and $x = (y^2/2) - 3$.
• Set up the integral:
  $\int<em>{-2}^4 \int</em>{(y^2/2)-3}^{y+1} xy dx dy$
• $\int<em>{-2}^4 \int</em>{(y^2/2)-3}^{y+1} xy dx dy = 1/2 \int_{-2}^4 y((y+1)^2 - ((y^2/2)-3)^2)dy$

Example: Interchanging Order of Integration

• Evaluate $\int<em>0^1 \int</em>{\sqrt{x}}^1 \sqrt{y^3 + 1} dy dx$
• Sketch the region. The region is bounded by $x = 0$, $x = 1$, $y = \sqrt{x}$, and $y = 1$.
• Change to type II. The region is bounded by $y = 0$, $y = 1$, $x = 0$, and $x = y^2$.
• The integral becomes:
  $\int<em>0^1 \int</em>0^{y^2} \sqrt{y^3 + 1} dx dy$
• Evaluate the inner integral:
  $\int<em>0^{y^2} \sqrt{y^3 + 1} dx = \sqrt{y^3 + 1} * x |</em>0^{y^2} = y^2 \sqrt{y^3 + 1}$
• Evaluate the outer integral using u-substitution:
  • $u = y^3 + 1$
  • $du = 3y^2 dy$
  • The integral becomes:
    $\int<em>0^1 y^2 \sqrt{y^3 + 1} dy = 1/3 \int</em>1^2 \sqrt{u} du = [2/9 * u^(3/2)]_1^2 = 2/9 * (2^(3/2) - 1)$

Polar Coordinates

• If f is continuous on polar rectangle R defined by: $R = {(r, θ) | a ≤ r ≤ b, α ≤ θ ≤ β}$
• Then
  $\int \int<em>R f(x, y) dA = \int</em>α^β \int_a^b f(r cos(θ), r sin(θ)) r dr dθ$

Transformation Equations

• x = r cos θ
• y = r sin θ
• dA = r dr dθ

Example - disk to polar cordinates

• Evaluate Integral by making a change to polar coordinates: $\int \int_D 2x-y dA$, where D is a disk radius 2
• $\int0^{2\pi} \int0^2 2 r cos(θ) - rsin(θ) r dr dθ$