vault backup: 2026-10-05 12:37:10

College/AU 26/MATH 224 (Multivar)/Class 10-5.md

@@ -9,22 +9,25 @@ $$\int_{a}^b \int_{c}^d f(x, y) dy dx$$
 
 This can be extended to more general domains that are not rectangles such as $[a,b] \times [c,d]$
 
-Example:
+#### Example:
 finding the volume under the function $f(x,y) = xy$ over the domain bound by 
 $$\begin{gathered} 1<x<2 \\[1.5ex] x < y < x^2 \end{gathered}$$
 We can think of finding the volume of the solid by using slices parallel to the $YZ$ plane. With the domain provided this area can be found with 
 $$ Area(x^*)=\int_{x}^{x^2}x^*ydy$$
 Since $x$ is bound by $1<x<2$ the double integral to find the volume can be written as 
-$$\begin{gather} \int_{1}^2 \left[ \int_{x}^{x^2}xy \  dy \right] dx \\[1.5ex] \int_{1}^2 \left[ \ \left.\frac{xy^2}{2} \right\vert_{x}^{x^2} \  \right] dx \\[1.5ex]  \int_{1}^2 \left[ \frac{x}{2}((x^2)^2-(x)^2)  \right] dx \\[1.5ex]  \int_{1}^2  \frac{x^5}{2}-\frac{x^3}{2}  dx \\[1.5ex]   \left.\frac{x^6}{12}-\frac{x^4}{8} \right\vert_{1}^2 \\[1.5ex] \left( \frac{16}{3} - 2 \right) - \left( \frac{1}{12} - \frac{1}{8} \right)  \end{gather}$$
+$$\begin{gather} \int_{1}^2 \left[ \int_{x}^{x^2}xy \  dy \right] dx \\[1.5ex] \int_{1}^2 \left[ \ \left.\frac{xy^2}{2} \right\vert_{x}^{x^2} \  \right] dx \\[1.5ex]  \int_{1}^2 \left[ \frac{x}{2}((x^2)^2-(x)^2)  \right] dx \\[1.5ex]  \int_{1}^2  \frac{x^5}{2}-\frac{x^3}{2}  dx \\[1.5ex]   \left.\frac{x^6}{12}-\frac{x^4}{8} \right\vert_{1}^2 \\[1.5ex] \left( \frac{16}{3} - 2 \right) - \left( \frac{1}{12} - \frac{1}{8} \right) \\[1.5ex] \frac{27}{8}  \end{gather}$$
-27/8
 
-Theorem:
+> [!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
 $$\int \int_{D} f(x,y)$$
 can be rewritten as
-$$\int_{a}^b \int_{g_{1}(x)}^{g_{2}(2)} f(x,y) \ dydx$$
+$$\int_{a}^b \int_{g_{1}(x)}^{g_{2}(x)} f(x,y) \ dydx$$
 This can also be done if $D$ is bound by functions on the x-axis instead of the y-axis
 
-
+### Property:
-If $D=D_{1} D_{2}$ and $D_{1}$ and $D_{2}$ do not intersect except at their boundaries then 
+If $D=D_{1} \cup D_{2}$ and $D_{1}$ and $D_{2}$ do not intersect except at their boundaries then 
-$$\int \int_{D} f(x,y)dA = \int \int_{D_{1}} f(x,y) \ dA + \int \int_{D_{2}} f(x,y)$$
+$$\int \int_{D} f(x,y) \ dA = \int \int_{D_{1}} f(x,y) \ dA + \int \int_{D_{2}} f(x,y) \ dA$$
+This can be used to split up integrals similar to area in 2D