From ce223daf750188624a76166a60c3dc4fe164c971 Mon Sep 17 00:00:00 2001
From: ben-jaynes <1btjaynes@gmail.com>
Date: Mon, 5 Oct 2026 15:06:34 -0700
Subject: [PATCH] vault backup: 2026-10-05 15:06:34

---
 College/AU 26/MATH 224 (Multivar)/Class 10-5.md | 12 +++++-------
 1 file changed, 5 insertions(+), 7 deletions(-)

diff --git a/College/AU 26/MATH 224 (Multivar)/Class 10-5.md b/College/AU 26/MATH 224 (Multivar)/Class 10-5.md
index e3250e0..70db885 100644
--- a/College/AU 26/MATH 224 (Multivar)/Class 10-5.md	
+++ b/College/AU 26/MATH 224 (Multivar)/Class 10-5.md	
@@ -17,9 +17,6 @@ $$ Area(x^*)=\int_{x}^{x^2}x^*ydy$$
 Since $x$ is bound by $1<x<2$ the double integral to find the volume can be written as 
 $$\begin{gather} \int_{1}^2 \left[ \int_{x}^{x^2}xy \  dy \right] dx \\[1.5ex] \int_{1}^2 \left[ \ \left.\frac{xy^2}{2} \right\vert_{x}^{x^2} \  \right] dx \\[1.5ex]  \int_{1}^2 \left[ \frac{x}{2}((x^2)^2-(x)^2)  \right] dx \\[1.5ex]  \int_{1}^2  \frac{x^5}{2}-\frac{x^3}{2}  dx \\[1.5ex]   \left.\frac{x^6}{12}-\frac{x^4}{8} \right\vert_{1}^2 \\[1.5ex] \left( \frac{16}{3} - 2 \right) - \left( \frac{1}{12} - \frac{1}{8} \right) \\[1.5ex] \frac{27}{8}  \end{gather}$$
 
-> [!NOTE] Theorem
-> Contents
-
 ### Theorem:
 If the region on the $XY$ plane for the region being integrated over of $f(x,y)$  ($D$) is continuous and bound by $x=a$ and $x=b$  in the x-axis and two functions $g_{1}(x)$ and $g_{2}(x)$ on the y-axis then the integral
 $$\int \int_{D} f(x,y)$$
@@ -27,7 +24,8 @@ can be rewritten as
 $$\int_{a}^b \int_{g_{1}(x)}^{g_{2}(x)} f(x,y) \ dydx$$
 This can also be done if $D$ is bound by functions on the x-axis instead of the y-axis
 
-### Property:
-If $D=D_{1} \cup D_{2}$ and $D_{1}$ and $D_{2}$ do not intersect except at their boundaries then 
-$$\int \int_{D} f(x,y) \ dA = \int \int_{D_{1}} f(x,y) \ dA + \int \int_{D_{2}} f(x,y) \ dA$$
-This can be used to split up integrals similar to area in 2D
+
+> [!note] Property
+> If $D=D_{1} \cup D_{2}$ and $D_{1}$ and $D_{2}$ do not intersect except at their boundaries then 
+> $$\int \int_{D} f(x,y) \ dA = \int \int_{D_{1}} f(x,y) \ dA + \int \int_{D_{2}} f(x,y) \ dA$$
+> This can be used to split up integrals similar to area in 2D