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
- \int1^2 (1/2) du/u = (1/2) ln(u) |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 = g1(x) and y = g2(x).
- The integral of a function f over a type 1 region D is:
\int \intD f(x, y) dA = \inta^b \int{g1(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 = h1(y) and x = h2(y).
- The integral of a function f over a type 2 region D is:
\int \intD f(x, y) dA = \intc^d \int{h1(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{-1}^1 \int{2x^2}^{x^2+1} (x + 2y) dy dx - Inner integral:
\int{2x^2}^{x^2+1} (x + 2y) dy = xy + y^2 |{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{-a}^a g(x) dx = 2 \int0^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{-1}^1 (-3x^4 - x^3 + 2x^2 + x + 1) dx = 2 \int0^1 (-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{-2}^4 \int{(y^2/2)-3}^{y+1} xy dx dy - \int{-2}^4 \int{(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 \int0^1 \int{\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:
\int0^1 \int0^{y^2} \sqrt{y^3 + 1} dx dy - Evaluate the inner integral:
\int0^{y^2} \sqrt{y^3 + 1} dx = \sqrt{y^3 + 1} * x |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:
\int0^1 y^2 \sqrt{y^3 + 1} dy = 1/3 \int1^2 \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 \intR f(x, y) dA = \intα^β \int_a^b f(r cos(θ), r sin(θ)) r dr dθ
- 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
- $\int0^{2\pi} \int0^2 2 r cos(θ) - rsin(θ) r dr dθ$