高等数学 A2 周维祺
设积分区域可以写作
{a≤x≤bφ1(x)≤y≤φ2(x)\begin{cases}a\le x\le b \\ \varphi_1(x)\le y\le \varphi_2(x)\end{cases} {a≤x≤bφ1(x)≤y≤φ2(x)
其中φ1(x),φ2(x)\varphi_1(x),\varphi_2(x)φ1(x),φ2(x)连续
设有DDD上的连续函数z=f(x,y)z=f(x,y)z=f(x,y)
用平面x=x0x=x_0x=x0截z=f(x,y)z=f(x,y)z=f(x,y)与DDD围成的柱体，得一截面
此截面的顶即z=f(x0,y)z=f(x_0,y)z=f(x0,y)，φ1(x0)≤y≤φ2(x0)\varphi_1(x_0)\le y\le \varphi_2(x_0)φ1(x0)≤y≤φ2(x0)
其面积即∫φ1(x0)φ2(x0)f(x0,y)dy\int_{\varphi_1(x_0)}^{\varphi_2(x_0)} f(x_0,y)dy∫φ1(x0)φ2(x0)f(x0,y)dy
再对x0x_0x0由aaa到bbb积分，即得
∬Df(x,y)dxdy=∫ab(∫φ1(x)φ2(x)f(x,y)dy)dx\iint_D f(x,y)dxdy=\int_a^b(\int_{\varphi_1(x)}^{\varphi_2(x)} f(x,y)dy)dx ∬Df(x,y)dxdy=∫ab(∫φ1(x)φ2(x)f(x,y)dy)dx
同理，若积分区域可以写作 c≤y≤dc\le y\le d\quadc≤y≤d, ψ1(y)≤x≤ψ2(y)\psi_1(y)\le x\le \psi_2(y)ψ1(y)≤x≤ψ2(y) 其中ψ1(y),ψ2(y)\psi_1(y),\psi_2(y)ψ1(y),ψ2(y)连续，则
∬Df(x,y)dxdy=∫cd(∫ψ1(y)ψ2(y)f(x,y)dx)dy\iint_D f(x,y)dxdy=\int_c^d(\int_{\psi_1(y)}^{\psi_2(y)} f(x,y)dx)dy ∬Df(x,y)dxdy=∫cd(∫ψ1(y)ψ2(y)f(x,y)dx)dy
设DDD是由直线x=2,y=1,y=x所围成的区域，计算z=xyz=xyz=xy在DDD上的积分
D:1≤x≤2,1≤y≤xD:1\le x\le 2, 1\le y \le xD:1≤x≤2,1≤y≤x
∬Dxydxdy=∫12(∫1xxydy)dx=∫12x⋅12(x2−1)dx\iint_D xydxdy=\int_{1}^2(\int_1^x xydy)dx=\int_1^2x\cdot\frac{1}{2}(x^2-1)dx∬Dxydxdy=∫12(∫1xxydy)dx=∫12x⋅21(x2−1)dx
∫12x⋅12(x2−1)dx=[x48−x24]12=98\int_1^2x\cdot\frac{1}{2}(x^2-1)dx=\left[\frac{x^4}{8}-\frac{x^2}{4}\right]_1^2=\frac{9}{8}∫12x⋅21(x2−1)dx=[8x4−4x2]12=89
设D1D_1D1是区域:{(x,y):x2+y2≤2,∣y∣≥1}\{(x,y): x^2+y^2\le 2, |y|\ge 1\}{(x,y):x2+y2≤2,∣y∣≥1}
设D2D_2D2是抛物线x=y2x=y^2x=y2以及直线y=x−2y=x-2y=x−2所围成的区域
分别求f(x,y)=xyf(x,y)=xyf(x,y)=xy在D1D_1D1和D2D_2D2上的积分
(1−11−11−11⋱⋱)\begin{pmatrix} 1 & -1 & & & \\ & 1 & -1 & & \\ & & 1 & -1 & \\ & & & 1 & \ddots \\ & & & & \ddots \\ \end{pmatrix}⎝⎛1−11−11−11⋱⋱⎠⎞
D=[−1,1]×[−1,1]D=[-1,1]\times [-1,1]D=[−1,1]×[−1,1]，f(x,y)=x/yf(x,y)=x/yf(x,y)=x/y
计算
∫01x2−y2(x2+y2)2dx,以及∫01x2−y2(x2+y2)2dy\int_0^1\frac{x^2-y^2}{(x^2+y^2)^2}dx, \quad以及\quad \int_0^1\frac{x^2-y^2}{(x^2+y^2)^2}dy ∫01(x2+y2)2x2−y2dx,以及∫01(x2+y2)2x2−y2dy
(提示：令x=ytantx=y\tan tx=ytant)
∫01(∫01x2−y2(x2+y2)2dx)dy=−π4≠π4=∫01(∫01x2−y2(x2+y2)2dy)dx\int_0^1(\int_0^1\frac{x^2-y^2}{(x^2+y^2)^2}dx)dy=-\frac{\pi}{4}\neq\frac{\pi}{4}=\int_0^1(\int_0^1\frac{x^2-y^2}{(x^2+y^2)^2}dy)dx ∫01(∫01(x2+y2)2x2−y2dx)dy=−4π=4π=∫01(∫01(x2+y2)2x2−y2dy)dx
反常积分：积分区域或被积函数无界
设x=g(u,v),y=h(u,v)x=g(u,v),y=h(u,v)x=g(u,v),y=h(u,v)的偏导都存在且连续
称矩阵J=(gugvhuhv)J=\begin{pmatrix}g_u & g_v \\ h_u & h_v \end{pmatrix}J=(guhugvhv)为映射(u,v)↦(x,y)(u,v)\mapsto(x,y)(u,v)↦(x,y)的雅可比矩阵
称∣J∣=det(gugvhuhv)=hvgu−hugv|J|=\det\begin{pmatrix}g_u & g_v \\ h_u & h_v \end{pmatrix}=h_vg_u-h_ug_v∣J∣=det(guhugvhv)=hvgu−hugv为该映射的雅可比式
计算{x=ρcosθy=ρsinθ\begin{cases}x=\rho\cos\theta \\ y=\rho\sin\theta \end{cases}{x=ρcosθy=ρsinθ的雅可比式
设z=f(x,y)z=f(x,y)z=f(x,y)是有界闭区域DDD上的连续函数
x=g(u,v),y=h(u,v)x=g(u,v),y=h(u,v)x=g(u,v),y=h(u,v)的偏导都存在且连续
设(u,v)↦(x,y)(u,v)\mapsto(x,y)(u,v)↦(x,y)的映射是单射
设∣J∣≠0|J|\neq0∣J∣=0
∬Df(x,y)dxdy=∬Df(g(u,v),h(u,v))∣J∣dudv\iint_D f(x,y)dxdy=\iint_D f\left(g(u,v),h(u,v)\right)|J|dudv ∬Df(x,y)dxdy=∬Df(g(u,v),h(u,v))∣J∣dudv
∬Df(x,y)dxdy=∬Df(ρcosθ,ρsinθ)ρ dρdθ\iint_D f(x,y)dxdy=\iint_D f(\rho\cos\theta,\rho\sin\theta)\rho \;d\rho d\theta ∬Df(x,y)dxdy=∬Df(ρcosθ,ρsinθ)ρdρdθ
计算f(x,y)=e−x2−y2f(x,y)=e^{-x^2-y^2}f(x,y)=e−x2−y2在D={(x,y):x2+y2≤r2}D=\{(x,y):x^2+y^2\le r^2\}D={(x,y):x2+y2≤r2}上的积分
∬Df(x,y)dxdy=∬De−ρ2ρ dρdθ=∫02π(∫0re−ρ2ρ dρ)dθ\iint_D f(x,y)dxdy=\iint_D e^{-\rho^2}\rho \;d\rho d\theta=\int_0^{2\pi}(\int_0^re^{-\rho^2}\rho \;d\rho)d\theta ∬Df(x,y)dxdy=∬De−ρ2ρdρdθ=∫02π(∫0re−ρ2ρdρ)dθ
∫02π(∫0re−ρ2ρ dρ)dθ=−π[e−ρ2]0r=π(1−e−r2)→π,若r→∞\int_0^{2\pi}(\int_0^re^{-\rho^2}\rho \;d\rho)d\theta=-\pi[e^{-\rho^2}]_0^r=\pi(1-e^{-r^2})\to\pi, 若r\to\infty ∫02π(∫0re−ρ2ρdρ)dθ=−π[e−ρ2]0r=π(1−e−r2)→π,若r→∞
计算 f(x,y)=ln(1+x2+y2)f(x,y)=\ln(1+x^2+y^2)f(x,y)=ln(1+x2+y2)在 D={(x,y):x2+y2≤1,x,y≥0}D=\{(x,y):x^2+y^2\le 1, x,y\ge 0\}D={(x,y):x2+y2≤1,x,y≥0}上的积分