[Analysis] Add notes for how to do multi-dim integrals

semester3/analysis-ii/cheat-sheet-jh/parts/vectors/integration/01_int_in_rn.tex

@@ -36,6 +36,7 @@ The integral of a continuous function $f: X \rightarrow \R$ with $X \subseteq \R
           \rmvspace
           In particular, if $Y$ empty (or size is ``negligible''), then $\int_{X} f(x) \dx x = \int_{X_1} f(x) \dx x + \int_{X_2} f(x) \dx x$
 \end{enumerate}
+%
 \setLabelNumber{all}{3}
 \shortdef For $m \leq n \in \N$, a \bi{parametrized $m$-set} in $\R^n$ is a continuous map $f: [a_1, b_1] \times \ldots \times [a_m, b_m] \rightarrow \R^n$,
 which is $C^1$ on $]a_1, b_1[ \times \ldots \times ]a_m, b_m[$.
@@ -43,6 +44,29 @@ $B \subseteq \R^n$ is \bi{negligible} if $\exists k \geq 0 \in \Z$ and parametri
 $X \subseteq f_1(x_1) \cup \ldots \cup f_k(X_k)$. A parametrized $1$-set in $\R^n$ is a parametrized curve.
 \shortex Any $\R \times \{ 0 \} \subseteq \R^2$ is negligible in $\R^2$, or more generally,
 if $H \subseteq \R^n$ is an affine subspsace of dimension $m < n$, then any subset of $\R^n$ that is contained in $H$ is negligible.
-Image of par. curve $\gamma: [a, b] \rightarrow \R^n$ is negligible, since $\gamma$ is a $1$-set in $\R^n$
-
+Image of par. curve $\gamma: [a, b] \rightarrow \R^n$ is negligible, since $\gamma$ is a $1$-set in $\R^n$\\
+%
 \shortproposition $X$ compact set, negligible. Then for any cont. function on $X$, $\displaystyle\int_{X} f(x) \dx x = 0$
+
+\mrmvspace
+\shade{gray}{Computing it}
+\bi{How to find the actual integrals from the intervals:} \hl{(Be careful with order of $x$ and $y$!)}
+\rmvspace
+\begin{itemize}[noitemsep]
+    \item Given an integral $\int_{D} f(x, y) \dx x \dx y$ for a set (or region) $X$ that is bounded by the coordinate axes and the line $x + y = 2$,\
+          the integral we can actually compute is $\int_{0}^{2} \int_{0}^{2} f(x, y) \dx y \dx x$.
+    \item Given an integral $\int_{X} g(x, y) \dx x \dx y$ with $X = [0, 1] \times [0, 2]$ and $g(x) = x^2 + y^2$. Then the integral should be obvious:
+          $\int_{0}^{1} \int_{0}^{2} g(x, y) \dx y \dx x$
+    \item \textit{Harder example} Given integral $\int_Y h(x, y) \dx x \dx y$ with $Y = \{ (x, y) \divides x \in [0, 1], y \leq 2x \land y \geq -2x \}$.
+          A good idea is to visualize the set: This one is a triangle and the integral is
+          $\int_{0}^{1} \int_{-2x}^{2x} h(x, y) \dx y \dx x$
+    \item \textit{Non-obvious example} For a set $U = \{ (x, y) : \sqrt{ x^2 + y^2 } \leq R \}$, we have the integral
+          $\displaystyle \int_{-R}^{R} \int_{-\sqrt{R^2 - x^2}}^{\sqrt{R^2 - x^2}} 1 \dx y \dx x$.
+          The new limits were attained by a simple inequality transformation, because in such equations, $y$ could be $0$ (and thus $|x|$ is limited by $R$)
+\end{itemize}
+
+\rmvspace
+\bi{How to compute the integral:} We compute each integral "inside out". For a definite integral, don't just find the anti-derivative, compute the actual integral!
+For an integral as seen in the harder example, we compute it as we normally would, simply using the $\pm 2x$ as the $a$ and $b$
+
+\rmvspace

semester3/analysis-ii/cheat-sheet-jh/parts/vectors/integration/03_change_of_variable_formula.tex

@@ -1,9 +1,35 @@
 \newsectionNoPB
 \subsection{Change of Variable Formula}
-\compacttheorem{Change of variable formula} $\overline{X}, \overline{Y} \subseteq \R^n$ compact, $\varphi : \overline{X} \rightarrow \overline{Y}$ continuous.
-For the open sets $X, Y$, negligible sets $B, C$ and restriction of $\varphi : X \rightarrow Y$ to open set $X$ is a $C^1$ bijection,
+\compacttheorem{Change of variable formula} $\overline{X}, \overline{Y} \subseteq \R^n$ compact, $\phi : \overline{X} \rightarrow \overline{Y}$ continuous.
+For the open sets $X, Y$, negligible sets $B, C$ and restriction of $\phi : X \rightarrow Y$ to open set $X$ is a $C^1$ bijection,
 we can write $\overline{X} = X \cup B$ and $\overline{Y} = Y \cup C$.
-The Jacobian $J_\varphi(x)$ is invertible at all $x \in X$.
-For any cont. func. $f$ on $\overline{Y}$ we have $\displaystyle \int_{\overline{X}} f(\varphi(x)) |\det(J_\varphi(x))| \dx x = \int_{\overline{Y}} f(y) \dx y$
-% TODO: Add notes from TA's notes for how to apply it
-\rmvspace
+The Jacobian $J_\phi(x)$ is invertible at all $x \in X$.
+For any continuous function $f$ on $\overline{Y}$ we have $\displaystyle \int_{\overline{Y}} f(y) \dx y = \int_{\overline{X}} f(\phi(x)) |\det(J_\phi(x))| \dx x$
+
+\hrmvspace
+\shade{orange}{Computing the determinant} Let $A = \begin{bmatrix}a & b\\c & d\end{bmatrix}$, then $\det(A) = ad - bc$.
+For 3D: $B = \begin{bmatrix}
+        a_1 & b_1 & c_1 \\
+        a_2 & b_2 & c_2 \\
+        a_3 & b_3 & c_3 \\
+    \end{bmatrix}$\\
+and $\det(B) = a_1 \cdot b_2 \cdot c_3 + b_1 \cdot c_2 \cdot a_3 + c_1 \cdot a_2 \cdot b_3 - c_1 \cdot b_2 \cdot a_3 - b_1 \cdot a_2 \cdot c_3 - a_1 \cdot c_2 \cdot b_3$
+
+\shade{gray}{How to use it} We could use it for example to switch from Cartesian to polar coordinates (then $x = r \cdot \cos(\varphi)$ and $y = r \cdot \sin(\varphi)$).
+
+\bi{Finding $\phi$:} Given integral $\displaystyle \int_{B} (1 - x^2 - y^2)^{\frac{n - 2}{2}} \dx x \dx y$ with $B = \{ (x, y) : x^2 + y^2 \leq 1 \}$.
+Here, it should immediately ring a bell that this can be rewritten using polar coordinates with $x^2 + y^2$ simplifying to $r^2$.
+Thus, $\phi(r, \phi) = (r \cos(\varphi), r \sin(\varphi))$. The boundaries then have to be determined from the reference boundaries using the inverse function of $\phi$
+
+\bi{Computing the integral:} When applying the formula, we replace all variables with their counterparts in $\phi$ (see above how to),
+we change the integration boundaries to fit our new variables and finally multiply everything by the Jacobian of $\phi$
+
+\bi{Example:} Using the integral from above, we get:
+\begin{align*}
+    \int_{0}^{1} \int_{0}^{2\pi} (1 - (r \cos(\varphi))^2 - (r \sin(\varphi))^2)^{\frac{n - 2}{2}} \cdot r \dx \varphi \dx r
+    = \int_{0}^{1} \int_{0}^{2\pi} (1 - r^2)^{\frac{n - 2}{2}} \cdot r \dx \varphi \dx r
+\end{align*}
+
+\bi{Likeliest case:} Changing into polar coordinates, then we replace $x = r \cos(\varphi)$ and $y = r \sin(\varphi)$ and replace$\dx x \dx y = r \dx r \dx \varphi$
+
+\bi{Connection to Analysis I:} This is just the generalization of the substitution rule for integrals

semester3/analysis-ii/cheat-sheet-jh/parts/vectors/integration/04_green_formula.tex

@@ -1,4 +1,4 @@
-\newsectionNoPB
+\newsection
 \subsection{The Green Formula}
 \compactdef{Simple parametrized curve} $\gamma : [a, b] \rightarrow \R^2$ is a closed parametrized curve s.t.
 $\gamma(t) \neq \gamma(s)$ (if $s \neq t$ and $\{ s, t \} = \{ a, b \}$), s.t. $\gamma'(t) \neq 0$ for $a < t < b$.
@@ -20,3 +20,4 @@ $\gamma_i$ as above, then
 \begin{align*}
     \text{Vol}(X) = \sum_{i = 1}^{k} \int_{\gamma_i} x \dx \vec{s} = \sum_{i = 1}^{k} \int_{a_i}^{b_i} \gamma_{i, 1}(t) \gamma_{i, 2}'(t) \dx t
 \end{align*}
+% TODO: Notes from TA starting W13 P5