#uw/notes #uw/class/math224 
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Recall that:
$$\int \int_{R} f(x, y) dA$$
is the signed area under the function $f(x,y)$ over the domain $R$ where 
$$R=[a,b] \times [c,d]$$
and it can be written as 
$$\int_{a}^b \int_{c}^d f(x, y) dy dx$$

This can be extended to more general domains that are not rectangles such as $[a,b] \times [c,d]$

#### Example:
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 
$$\begin{gather} \int_{1}^2 \left[ \int_{x}^{x^2}xy \  dy \right] dx \\[1.5ex] \int_{1}^2 \left[ \ \left.\frac{xy^2}{2} \right\vert_{x}^{x^2} \  \right] dx \\[1.5ex]  \int_{1}^2 \left[ \frac{x}{2}((x^2)^2-(x)^2)  \right] dx \\[1.5ex]  \int_{1}^2  \frac{x^5}{2}-\frac{x^3}{2}  dx \\[1.5ex]   \left.\frac{x^6}{12}-\frac{x^4}{8} \right\vert_{1}^2 \\[1.5ex] \left( \frac{16}{3} - 2 \right) - \left( \frac{1}{12} - \frac{1}{8} \right) \\[1.5ex] \frac{27}{8}  \end{gather}$$

### 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
$$\int_{a}^b \int_{g_{1}(x)}^{g_{2}(x)} f(x,y) \ dydx$$
This can also be done if $D$ is bound by functions on the x-axis instead of the y-axis


> [!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