Double Integral = \int \int_R f(x, y) d d = \lim_{m, n \rightarrow \infty} \sum_{i = 1}^m \sum_{j = 1}^n f(x_i, y_j) \Delta Area_{ij}
For Rectangle: \begin{align*} \int \int_R f(x, y) dA &= \int_c^d (\int_a^b f(x, y) dx) dy\\ &= \int_a^b (\int_c^d f(x, y) dy) dx \end{align*}
Average Value:
In 2D:
In 3D:
2D Example: area between y = x^2 and y = \sqrt{x} \begin{align*} R &= \{(x, y) | 0 \leq x \leq 1, x^2 \leq y \leq \sqrt[3]{x}\}\\ &= \{(x, y) | 0 \leq y \leq 1, x^2 \leq y^3 \leq \sqrt{y}\}\\ \end{align*}
\begin{align*} \int \int_R 1 dA &= \int_0^1 \int_{x^2}^{\sqrt[3]{x}} 1 dy dx\\ &= \int_0^1 (y |_{x^2}^{\sqrt[3]{x}}) dx\\ &= \int_0^1 \sqrt[3](x) - x^2 dx\\ &= \frac{5}{12}\\ \end{align*}
3D Switch Order Example: signed volume of 3xy under region between y = x^3, y = x^2+1, x = 0, x = 1 \begin{align*} D &= \{(x, y) | 0 \leq x \leq 1, x^3 \leq y \leq x^3 + 1\}\\ &= D_1 \cup D_2\\ &= \{(x, y) | 0 \leq y \leq 1, 0 \leq x \leq \sqrt[3]{y}\}\\ \cup &\{(x, y) | 1 \leq y \leq 2, \sqrt[3]{y-1} \leq x \leq 1\}\\ V &= \int_0^1 \int_{x^3}^{x^3 + 1} 3xy dy dx\\ &= \int_0^1 \int_0^{\sqrt[3]{y}} 3xy dx dy + \int_1^2 \int_{\sqrt[3]{y-1}}^1 3xy dx dy\\ \end{align*}
Polar Rectangle: R = \{(r, \theta) | a \leq r \leq b, \alpha \leq \theta \leq \beta\} (a sector)
If z = f(r, \theta) continuous on R, then \int \int_R f(r, \theta) dA = \lim_{m, n \rightarrow \infty} \sum_{i = 1}^m \sum_{j = 1}^n f(r_{ij} \theta_{ij}) \Delta A_{ij} where \Delta A_{ij} = r_{ij} \Delta r_i \Delta \theta_j
Note R_{ij} = \{(r, \theta) | r_i \leq r \leq r_{i+1}, \theta_j \leq \theta \leq \theta_{j+1}\}
Therefore \begin{align*} \Delta A_{ij} &= \frac{\Delta \theta_i}{2 \pi} \times \pi r_{i+1}^2 - \frac{\Delta \theta_i}{2 \pi} \times \pi r_i^2\\ &= \frac{r_i + r_{i+1}}{2} \times \Delta r_i \Delta \theta_i\\ \end{align*} where we define r_{ij} = \frac{r_i + r_{i+1}}{2}
Probability Density Function (pdf):
greater to 0 at each point: (\forall t \in \mathbb{R})(f(t) \geq 0)
sum to 1: \int_R f(t) dt = 1
P(a \leq x \leq b) = \int_a^b f(t) dt
Joint Density Function: for continuous random variable x and y
(\forall (x, y) \in \mathbb{R}^2)(f(x, y) \geq 0)
\int \int_{\mathbb{R}^2} f(x, y) dA = 1
(\forall D \subseteq \mathbb{R}^2)(P((x, y) \in D) = \int \int_D f(x, y) dA)
X~Exp(n): \frac{1}{n}e^{-X/n}
Expected Value: E(x) = \int_R x f(x) dx if x is a continuous random variable with pdf f(x)
The volume of standard n-complex is \frac{1}{n!}
u-substitution: dx = \frac{dx^3}{2x^2} integrate by part: \int f(x)g(x) dx = f(x) \int g(x) dx - \int (\int g(x) dx) \frac{d f(x)}{dx} dx
// TODO: review integrate by part
Planar Transformation: a function T : \mathbb{R}^2 \rightarrow \mathbb{R}^2 (from (u, v) coordinates to (x, y) coordinates)
T should be injective (also locally surjective, with inverse), because we don't want region to overlap and cancel each other out or counted twice
T should have continuous first order partial derivatives for its component functions
then T is a C^1 (continuous, first derivative continuous) transformation
Jacobion: if T(u, v) = \langle{x(u, v), y(u, v)}\rangle, then \frac{\partial (x, y)}{\partial (u, v)} = \det \begin{vmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v}\\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v}\\ \end{vmatrix}
T(x, y, z) = \langle{x(u, v, w), y(u, v, w)}\rangle, then \frac{\partial (x, y, z)}{\partial (u, v, w)} = \det \begin{vmatrix} \frac{\partial x}{\partial u} & \frac{\partial y}{\partial u} & \frac{\partial z}{\partial u}\\ \frac{\partial x}{\partial v} & \frac{\partial y}{\partial v} & \frac{\partial z}{\partial v}\\ \frac{\partial x}{\partial w} & \frac{\partial y}{\partial w} & \frac{\partial z}{\partial w}\\ \end{vmatrix}
Then \begin{align*} &\int \int_R f(x, y) dA\\ = &\int \int_S f(x, (u, v), y(u, v)) \left\|\frac{\partial (x, y)}{\partial (u, v)}\right\| du dv\\ \end{align*}
Special Case for Cylindrical Coordinates:
Special Case for Spherical Coordinates:
Jacobian Matrix: Best linear map approximate f at (x_0, y_0). Even though the transformation is not linear in global sense, it can be linear in local sense. (See Video)
We need the absolute value because we don't care about orientation of the region when integrate
Multiplying the Jacobian has the effect to warp rectangular region into sector-like region
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