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 87e6348..6ea3264 100644
--- a/College/AU 26/MATH 224 (Multivar)/Class 10-5.md	
+++ b/College/AU 26/MATH 224 (Multivar)/Class 10-5.md	
@@ -16,4 +16,15 @@ We can think of finding the volume of the solid by using slices parallel to the
 $$ Area(x^*)=\int_{x}^{x^2}x^*ydy$$
 Since $x$ is bound by $1<x<2$ the double integral to find the volume can be written as 
 $$\begin{gather} \int_{1}^2 \left[ \int_{x}^{x^2}xy \  dy \right] dx \\[1.5ex] \int_{1}^2 \left[ \ \left.\frac{xy^2}{2} \right\vert_{x}^{x^2} \  \right] dx \\[1.5ex]  \int_{1}^2 \left[ \frac{x}{2}((x^2)^2-(x)^2)  \right] dx \\[1.5ex]  \int_{1}^2  \frac{x^5}{2}-\frac{x^3}{2}  dx \\[1.5ex]   \left.\frac{x^6}{12}-\frac{x^4}{8} \right\vert_{1}^2 \\[1.5ex] \left( \frac{16}{3} - 2 \right) - \left( \frac{1}{12} - \frac{1}{8} \right)  \end{gather}$$
-27/8
\ No newline at end of file
+27/8
+
+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
+$$\int_{a}^b \int_{g_{1}(x)}^{g_{2}(2)} f(x,y) \ dydx$$
+This can also be done if $D$ is bound by functions on the x-axis instead of the y-axis
+
+
+If $D=D_{1} D_{2}$ and $D_{1}$ and $D_{2}$ do not intersect except at their boundaries then 
+$$\int \int_{D} f(x,y)dA = \int \int_{D_{1}} f(x,y) \ dA + \int \int_{D_{2}} f(x,y)$$
\ No newline at end of file