# 二重积分的计算

1，如果区域为 $$D=\{(x,y)| a\le x\le b, g_1(x)\le y\le g_2(x)\}$$，

$\iint_Df(x,y)dA=\int_a^b\int_{g_1(x)}^{g_2(x)}f(x,y)dydx$

2，如果区域为 $$D=\{(x,y)| h_1(y)\le x\le , g_2(y), c\le y\le d\}$$，

$\iint_Df(x,y)dA=\int_c^d\int_{h_1(y)}^{h_2(y)}f(x,y)dxdy$

3，如果区域的上、下曲线不是同一个表达式，或者左、右曲线不是同一个表达式，则将区域分成几个部分，每一个部分的上、下曲线应该是一致的。例如 $$D=D_1+D_2$$

\begin{align*}\iint_Df(x,y)dA&=\iint_{D_1}f(x,y)dA+\iint_{D_2}f(x,y)dA\\ &=\int_a^c\int_{g_1(x)}^{g_2(x)}f(x,y)dydx+\int_c^b\int_{g_3(x)}^{g_4(x)}f(x,y)dydx\end{align*}

$\begin{cases}y=2x\\ y=x^2\end{cases}$ 交点为 $$(0,0)$$ 和 $$(2,4)$$。图形如下：

$D=\{(x,y)|0\le x\le 2, x^2\le y\le 2x\}$

\begin{align*}\iint_Df(x,y)dA&=\int_0^2\int_{x^2}^{2x}(x^2+y^2)dydx\\ &=\int_0^2(x^2y+\frac{1}{y^3})\Big|_{x^2}^{2x}\\ &=\int_0^2(2x^3+\frac{8}{3}x^3-x^4-\frac{1}{3}x^6)dx\\ &=\int_0^2(\frac{14}{3}x^3-x^4-\frac{1}{3}x^6)dx\\&=\frac{7}{6}x^4-\frac{1}{5}x^5-\frac{1}{21}x^7\Big|_0^2=\frac{216}{35}\end{align*}

\begin{align*}\iint_DxydA&=\int_{-2}^4\int_{y+1}^{\frac{y^2}{2}-3}xydxdy=\int_{-2}^4\frac{1}{2}x^2y\Big|_{y+1}^{\frac{y^2}{2}-3}dy\\ &=\int_{-2}^4\frac{y}{2}((y+1)^2-(\frac{y^2}{2}-3)^2)dy\\ &=\frac{1}{2}\int_{-2}^4y(4y^2-\frac{y^4}{4}+2y-8)dy\\ &=\frac{1}{2}(y^4-\frac{1}{24}y^6+\frac{2}{3}y^3-4y^2)\Big|_{-2}^4\\ &=36\end{align*}

$D=\{(x,y)|0\le x\le 1,\sqrt{x}\le y\le 1\}$

$D=\{(x,y)|0\le x\le y^2, 0\le y\le 1\}$

\begin{align*}\int_0^1\int_{\sqrt{x}}^1e^{y^3}dydx&=\int_0^1\int_0^{y^2}e^{y^2}dxdy=\int_0^1e^{y^3}x\Big|_0^{y^2}dy\\&=\int_0^2y^2e^{y^3}dy=\frac{1}{3}e^{y^3}\Big|_0^1=\frac{1}{3}(e-1)\end{align*}