Example Change the order of integration．

$$\iint_{\Omega}x^2 dxdy, \ \Omega: -1 \leq x \leq 1, \ 0 \leq y \leq 3$$

Answer

$$\Omega = \{(x,y) : -1 \leq x \leq 1, 0 \leq y \leq 3\}$$

First, we use v-simple to evaluate this integral. Consider the small rectangle $dy$ in the direction of $y$-axis. Accumulate $dy$ from the bottom to the top of the region. Then fill $\Omega$ using $dy$.　In other words, Integrate the function with respect to $y$. Then integrate with respect to $x$. Thus,

$$\iint_{\Omega} x^2 dxdy = \int_{x=-1}^{1}(\int_{y=0}^{3}x^2 dy)dx$$

$$= \int_{-1}^{1}(\left[x^2 y \right]_{y=0}^{3})dx = \int_{-1}^{1}3x^2 dx$$

$$= [x^3]_{-1}^{1} = 1 - (-1) = 2$$

Next, we use H-simple to evaluate this integral. Consider the small rectangle $dx$ in the direction of $x$-axis. Accumulate $dx$ from the left to the right of the region. Then fill $\Omega$ using $dx$.　In other words, Integrate the function with respect to $x$. Then integrate with respect to $y$. Thus,

$$\iint_{\Omega} x^2 dxdy = \int_{y=0}^{3}(\int_{x=-1}^{1}x^2 dx)dy$$

$$= \int_{0}^{3}(\left[\frac{x^3}{3} \right]_{x=-1}^{1})dy = \int_{0}^{3}\frac{2}{3} dy$$

$$= [\frac{2}{3}y]_{0}^{3} = 2-0 = 2$$

This technic is called the change of order of integration.