eRepeatedIntegral

Example Change the order of integration．

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

Answer

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

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,

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

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,

$\displaystyle \iint_{\Omega} x^2 dxdy$	$\displaystyle =$	$\displaystyle \int_{y=0}^{3}(\int_{x=-1}^{1}x^2 dx)dy$
	$\displaystyle =$	$\displaystyle \int_{0}^{3}(\left[\frac{x^3}{3} \right]_{x=-1}^{1})dy = \int_{0}^{3}\frac{2}{3} dy$
	$\displaystyle =$	$\displaystyle [\frac{2}{3}y]_{0}^{3} = 2-0 = 2$

This technic is called the change of order of integration.

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