重积分(微积分)

由一元积分演化而来

  • 定积分
    • 二重积分:直角坐标、极坐标
    • 三重积分:柱面坐标、球坐标、直角坐标

1 二重积分的概念与性质

一、概念

引入:曲顶柱体的体积、平面薄片的质量

  • 定义1(分割求和取极限):∬Df(x,y) dσ=lim⁡λ→0∑i=1nf(xi,yi) Δσi\iint\limits_{D}f(x,y)\,d\sigma=\lim\limits_{\lambda\rightarrow 0}\sum\limits_{i=1}^n f(x_i,y_i)\,\Delta\sigma_iDf(x,y)dσ=λ0limi=1nf(xi,yi)Δσi(和式的极限),其中DDD积分区域x,yx,yx,y积分变量f(x,y)f(x,y)f(x,y)被积函数dσd\sigmadσ面积元素f(x,y) dσf(x,y)\,d\sigmaf(x,y)dσ被积表达式∑f(xi,yi) Δσi\sum f(x_i,y_i)\,\Delta\sigma_if(xi,yi)Δσi积分和λ=max⁡{di}且di=max⁡(x1,y1),(x2,y2)→Δτi{∣P1P2∣}\lambda=\max\{d_i\}且d_i=\max\limits_{(x_1,y_1),(x_2,y_2)\rightarrow\Delta\tau_i}\{|P_1P_2|\}λ=max{di}di=(x1,y1),(x2,y2)Δτimax{P1P2}.
    类比于定积分:∫abf(x) dx=lim⁡λ→0∑i=1nf(xi) Δxi\int_a^bf(x)\,dx=\lim\limits_{\lambda\rightarrow 0}\sum\limits_{i=1}^n f(x_i)\,\Delta x_iabf(x)dx=λ0limi=1nf(xi)Δxi.
  • 说明:
    • 连续必可积f(x,y)f(x,y)f(x,y)在闭区间连续,二重积分必存在
    • 面积元素:dσ=dxdyd\sigma=dxdydσ=dxdy
    • 几何意义f(x,y)f(x,y)f(x,y)DDD上的二重积分等于柱体体积的代数和(有正有负)

二、性质

  • 可加,可数乘
  • 单调,特别的∣∬Df(x,y) dσ∣≤∬D∣f(x,y)∣ dσ|\iint\limits_{D}f(x,y)\,d\sigma|\le\iint\limits_{D}|f(x,y)|\,d\sigmaDf(x,y)dσDf(x,y)dσ.
  • 估值不等式minσ≤∬Df(x,y) dσ≤MAXσmin\sigma \le\iint\limits_{D}f(x,y)\,d\sigma \le MAX\sigmaminσDf(x,y)dσMAXσ.
  • 中值定理:存在点(x0,y0)(x_0,y_0)(x0,y0),使得∬Df(x,y) dσ=f(x0,y0)σ\iint\limits_{D}f(x,y)\,d\sigma=f(x_0,y_0)\sigmaDf(x,y)dσ=f(x0,y0)σ.

2 二重积分的计算

一、直角坐标

  • X型V=V=V=∫ab[∫φ1(x)φ2(x)f(x,y) dy] dx=∫ab dx∫φ1(x)φ2(x)f(x,y) dy\int_a^b\left[\int_{\varphi_1(x)}^{\varphi_2(x)}f(x,y)\,dy\right]\,{\color{red}dx} = \int_a^b\,{\color{red}dx}\int_{\varphi_1(x)}^{\varphi_2(x)}f(x,y)\,dyab[φ1(x)φ2(x)f(x,y)dy]dx=abdxφ1(x)φ2(x)f(x,y)dy.
    Y型V=V=V=∫cd[∫ψ1(y)ψ2(y)f(x,y) dx] dy=∫cd dy∫ψ1(y)ψ2(y)f(x,y) dx\int_c^d\left[\int_{\psi_1(y)}^{\psi_2(y)}f(x,y)\,dx\right]\,{\color{red}dy} = \int_c^d\,{\color{red}dy}\int_{\psi_1(y)}^{\psi_2(y)}f(x,y)\,dxcd[ψ1(y)ψ2(y)f(x,y)dx]dy=cddyψ1(y)ψ2(y)f(x,y)dx.
  • 分割,选择次序
  • 做题时
    • 只要见到积分区域(有时需要拆分区域)具有对称性,就要想到考察被积函数的奇偶性
    • 考虑x、y能否轮换对称(关于y=x对称)
    • 直接积分图形复杂时,考虑可加(减)性

例1:求∬Dx2e−y2dxdy,其中D是以(0,0),(1,1),(0,1)为顶点的三角形\color{blue}求\iint\limits_{D}x^2e^{-y^2}dxdy,其中D是以(0,0),(1,1),(0,1)为顶点的三角形Dx2ey2dxdy,D(00),(1,1),(0,1)
解:
∵∫e−y2dy无法解出∴先积分x原式=∫01dy∫0yx2e−y2dx=∫01e−y2⋅y33dy =∫01e−y2⋅y26dy2=−16∫01ude−u=−13e {\color{red}\because \int e^{-y^2}dy无法解出\therefore 先积分x} \\ 原式 = \int_0^1 dy\int_0^y x^2e^{-y^2}dx = \int_0^1e^{-y^2}\cdot \dfrac{y^3}{3}dy\ {\color{red}=} \int_0^1e^{-y^2}\cdot \dfrac{y^2}{6}dy^2 = -\dfrac{1}{6}\int_0^1 u de^{-u} = -\dfrac{1}{3e} ey2dyx=01dy0yx2ey2dx=01ey23y3dy =01ey26y2dy2=6101udeu=3e1
例2:求∬Dy1+x2−y2dσ,其中D是由y=1,x=−1及y=x所围成的区域\color{blue}求\iint\limits_{D}y\sqrt{1+x^2-y^2}d\sigma,其中D是由y=1,x=-1及y=x所围成的区域Dy1+x2y2dσ,Dy=1,x=1y=x
解:
不妨以X型区域计算:原式=∫−11dx∫x1y1+x2−y2dy=...=−13∫−11[(1+x2−y2)32]x1dx=−13∫−11(∣x∣3−1)dx=−23∫01(x3−1)dx=12.不知x正负,平方开根号要绝对值,不然答案为奇函数错误 不妨以X型区域计算:原式 = \int_{-1}^1dx\int_x^1 y\sqrt{1+x^2-y^2}dy \\ = ...=-\dfrac{1}{3}\int_{-1}^1\left[(1+x^2-y^2)^{\dfrac{3}{2}}\right]_x^1dx = -\dfrac{1}{3}\int_{-1}^1({\color{red}|x|^3}-1 )dx = -\dfrac{2}{3}\int_0^1(x^3-1 )dx = \dfrac{1}{2}. \\ \color{red}不知x正负,平方开根号要绝对值,不然答案为奇函数错误 X=11dxx1y1+x2y2dy=...=3111(1+x2y2)23x1dx=3111(x31)dx=3201(x31)dx=21.x,
例3:求两个底圆半径都等于R的直交援助面积所围成的立体的体积\color{blue}求两个底圆半径都等于R的直交援助面积所围成的立体的体积R
解:
设两圆柱面方程为x2+y2=R2和x2+z2=R2V=8∫0Rdx∫0R2−x2R2−x2 dy=8∫0R(R2−x2)dx=163R3. 设两圆柱面方程为x^2+y^2=R^2和x^2+z^2=R^2 \\ V=8\int_0^Rdx\int_0^{\sqrt{R^2-x^2}} \sqrt{R^2-x^2}\,dy = 8\int_0^R(R^2-x^2)dx = \dfrac{16}{3}R^3. x2+y2=R2x2+z2=R2V=80Rdx0R2x2R2x2dy=80R(R2x2)dx=316R3.

例4:更换积分次序,I=∫04 dy∫−4−y4y−y2f(x,y) dx\color{blue}更换积分次序,I=\int_0^4\,dy\int_{-\sqrt{4-y}}^{\sqrt{4y-y^2}}f(x,y)\,dxI=04dy4y4yy2f(x,y)dx.
解:
I=∫−20 dx∫04−x2f(x,y) dy+∫02 dx∫2−4−x22+4−x2f(x,y) dy. I=\int_{-2}^0\,dx\int_0^{4-x^2}f(x,y)\,dy+\int_0^2\,dx\int_{2{\color{red}-}\sqrt{4-x^2}}^{2{\color{red}+}\sqrt{4-x^2}}f(x,y)\,dy. I=20dx04x2f(x,y)dy+02dx24x22+4x2f(x,y)dy.
在这里插入图片描述

例5:计算z=xy,x+y+z=1,z=0所围闭合区域的体积\color{blue}计算z=xy,x+y+z=1,z=0所围闭合区域的体积z=xy,x+y+z=1,z=0.
解:法1
分为两部分,一部分z=xy为顶,另一部分z=1−x−y为顶∵xy=1−x−y  ⟺  y=1−x1+x∴V=∫01 dx∫01−x1+xxy dy+∫01 dx∫1−x1+x1−x1−x−y dy. 分为两部分,一部分z=xy为顶,另一部分z=1-x-y为顶 \\ \because xy=1-x-y \iff y=\dfrac{1-x}{1+x} \\ \therefore V=\int_0^1\,dx\int_0^{\frac{1-x}{1+x}}xy\,dy+\int_0^1\,dx\int_{\frac{1-x}{1+x}}^{1-x}1-x-y\,dy. z=xyz=1xyxy=1xyy=1+x1xV=01dx01+x1xxydy+01dx1+x1x1x1xydy.
在这里插入图片描述

​ 法2:参考例9,截面法
例6:平面区域D由曲线{x=t−sin⁡t,y=1−cos⁡t(0≤t≤2π)与x轴围成,计算∬D(x+2y) dxdy\color{blue}平面区域D由曲线\begin{cases}x=t-\sin t, \\ y=1-\cos t\end{cases}(0\le t\le 2\pi)与x轴围成,计算\iint\limits_D (x+2y)\,dxdyD线{x=tsint,y=1cost(0t2π)xD(x+2y)dxdy.
将参数方程选做换元积分的变量替换设D:0≤x≤2π,0≤y≤g(x),则∬D(x+2y) dxdy=∫02π dx∫0g(x)(x+2y) dy=∫02π[xg(x)+g2(x)] dx=∫02π[(t−sin⁡t)(1−cos⁡t)+(1−cos⁡t)2](1−cos⁡t) dt= ... =3π2+5π. \color{red}将参数方程选做换元积分的变量替换 \\ 设D:0\le x\le 2\pi,0\le y\le g(x),则 \\ \iint\limits_D (x+2y)\,dxdy=\int_0^{2\pi}\,dx\int_0^{g(x)}(x+2y)\,dy=\int_0^{2\pi}[xg(x)+g^2(x)]\,dx \\ =\int_0^{2\pi}[(t-\sin t)(1-\cos t)+(1-\cos t)^2](1-\cos t)\,dt =\,...\,=3\pi^2+5\pi. D:0x2π,0yg(x),D(x+2y)dxdy=02πdx0g(x)(x+2y)dy=02π[xg(x)+g2(x)]dx=02π[(tsint)(1cost)+(1cost)2](1cost)dt=...=3π2+5π.

二、极坐标

  • 由于Δσi≈ρiΔρiΔθi\Delta\sigma_i\approx \rho_i \Delta\rho_i \Delta\theta_iΔσiρiΔρiΔθi,元素面积可以表示为dσ=ρdρdθd\sigma = \rho d\rho d\thetadσ=ρdρdθ.
    因此lim⁡λ→0∑i=1nf(xi,yi) Δσi=lim⁡λ→0∑i=1nf(ρicos⁡θi,ρisin⁡θi) ρiΔρiΔθi\lim\limits_{\lambda\rightarrow 0}\sum\limits_{i=1}^n f(x_i,y_i)\,\Delta\sigma_i = \lim\limits_{\lambda\rightarrow 0}\sum\limits_{i=1}^n f(\rho_i\cos\theta_i,\rho_i\sin\theta_i)\,\rho_i\Delta\rho_i\Delta\theta_iλ0limi=1nf(xi,yi)Δσi=λ0limi=1nf(ρicosθi,ρisinθi)ρiΔρiΔθi
    即==∬Df(x,y) dσ=∬Df(ρcos⁡θ,ρsin⁡θ) ρ dρ dθ\iint\limits_{D}f(x,y)\,d\sigma = \iint\limits_{D}f(\rho\cos\theta,\rho\sin\theta)\,\rho\,d\rho\,d\thetaDf(x,y)dσ=Df(ρcosθ,ρsinθ)ρdρdθ==.
  • 其中按顺序:∬Df(ρcos⁡θ,ρsin⁡θ) ρ dρ dθ=∫αβ dθ∫φ1(θ)φ2(θ)f(ρcos⁡θ,ρsin⁡θ) ρ dρ\iint\limits_{D}f(\rho\cos\theta,\rho\sin\theta)\,\rho\,d\rho\,d\theta = \int_{\alpha}^{\beta}\,d\theta \int_{\varphi_1(\theta)}^{\varphi_2(\theta)}f(\rho\cos\theta,\rho\sin\theta)\,\rho\,d\rhoDf(ρcosθ,ρsinθ)ρdρdθ=αβdθφ1(θ)φ2(θ)f(ρcosθ,ρsinθ)ρdρ.

例7:转换成直角坐标系I=∫0π2 dθ∫acos⁡θ2acos⁡θ4a2−ρ2 ρ dρ\color{blue}转换成直角坐标系I= \int_0^{\frac{\pi}{2}}\,d\theta \int_{a\cos\theta}^{2a\cos\theta}\sqrt{4a^2-\rho ^2}\,\rho\,d\rhoI=02πdθacosθ2acosθ4a2ρ2ρdρ.
解:
难点:画图I=I=∫0a dx∫ax−x22ax−x24a2−x2−y2 dy+∫02a dx∫02ax−x24a2−x2−y2 dy. 难点:画图 I=I=\int_0^a\,dx\int_{\sqrt{ax-x^2}}^{\sqrt{2ax-x^2}}\sqrt{4a^2-x ^2-y^2}\,dy+\int_0^{2a}\,dx\int_0^{\sqrt{2ax-x^2}}\sqrt{4a^2-x ^2-y^2}\,dy. I=I=0adxaxx22axx24a2x2y2dy+02adx02axx24a2x2y2dy.
在这里插入图片描述

例8:闭区域D:x2+y2≤y(1),x≥0(2),f(x,y)为D上连续函数,且f(x,y)=1−x2−y2−8π∬Df(u,v) dudv(∗),求f(x,y)\color{blue}闭区域D:x^2+y^2\le y(1),x\ge 0(2),f(x,y)为D上连续函数,且f(x,y)=\sqrt{1-x^2-y^2}-\dfrac{8}{\pi}\iint\limits_ D f(u,v)\,dudv(*),求f(x,y)D:x2+y2y(1),x0(2),f(x,y)Df(x,y)=1x2y2π8Df(u,v)dudv(),f(x,y).
解:
设A=∬Df(u,v) dudv,又∬D dxdy=π8,则在(∗)两边求二重积分得到:A=∬D1−x2−y2 dxdy−∬D8πA dxdy.∴A=12∬D1−x2−y2 dxdy=12∫0π2 dθ∫0sin⁡θρ1−ρ2 dρ=12∫0π2[−13(1−ρ2)32]0sin⁡θ dθ=π12−19.∴f(x,y)=1−x2−y2−8π(π12−19). 设A=\iint\limits_ D f(u,v)\,dudv,又\iint\limits_ D \,dxdy=\dfrac{\pi}{8}, \\则{\color{red}在(*)两边求二重积分}得到:A=\iint\limits_ D\sqrt{1-x^2-y^2}\,dxdy-\iint\limits_ D\dfrac{8}{\pi}A\,dxdy. \\ \therefore A=\dfrac{1}{2}\iint\limits_ D\sqrt{1-x^2-y^2}\,dxdy = \dfrac{1}{2}\int_0^{\pi\over2}\,d\theta\int_0^{\sin\theta}\rho\sqrt{1-\rho^2}\,d\rho = \dfrac{1}{2}\int_0^{\pi\over2}\left[-\dfrac{1}{3}(1-\rho^2)^{3\over2}\right]_0^{\sin\theta} \,d\theta = \dfrac{\pi}{12}-\dfrac{1}{9}. \\ \therefore f(x,y) = \sqrt{1-x^2-y^2}-\dfrac{8}{\pi}(\dfrac{\pi}{12}-\dfrac{1}{9}). A=Df(u,v)dudv,Ddxdy=8π,()A=D1x2y2dxdyDπ8Adxdy.A=21D1x2y2dxdy=2102πdθ0sinθρ1ρ2dρ=2102π[31(1ρ2)23]0sinθdθ=12π91.f(x,y)=1x2y2π8(12π91).

三、坐标变换

3 三重积分

一、概念

定义1:∭Ωf(x,y,z) dσ=lim⁡λ→0∑i=1nf(xi,yi,z1) Δσi\iiint\limits_{\Omega}f(x,y,z)\,d\sigma=\lim\limits_{\lambda\rightarrow 0}\sum\limits_{i=1}^n f(x_i,y_i,z_1)\,\Delta\sigma_iΩf(x,y,z)dσ=λ0limi=1nf(xi,yi,z1)Δσi.
三重积分也记作:∭Ωf(x,y,z) dσ=∭Ωf(x,y,z) dx dy dz\iiint\limits_{\Omega}f(x,y,z)\,d\sigma=\iiint\limits_{\Omega}f(x,y,z)\,dx\,dy\,dzΩf(x,y,z)dσ=Ωf(x,y,z)dxdydz.

二、计算

  • 先一后二(投影法”压薄“)
    ∭Ωf(x,y,z) dx dy dz=∬DF(x,y) dx dy=∬D[∫z1(x,y)z2(x,y)f(x,y,z) dz] dx dy \iiint\limits_{\Omega}f(x,y,z)\,dx\,dy\,dz = \iint\limits_{D}F(x,y)\,dx\,dy = \iint\limits_{D} \left[\int_{z_1(x,y)}^{z_2(x,y)}f(x,y,z)\,dz \right]\,dx\,dy Ωf(x,y,z)dxdydz=DF(x,y)dxdy=D[z1(x,y)z2(x,y)f(x,y,z)dz]dxdy

  • 先二后一(截面法”集中“)
    ∭Ωf(x,y,z) dx dy dz=∫C1C2F(x) dz=∫C1C2[∬Dzf(x,y,z) dx dy] dz \iiint\limits_{\Omega}f(x,y,z)\,dx\,dy\,dz = \int_{C_1}^{C_2} F(x)\,dz = \int_{C_1}^{C_2} \left[\iint\limits_{D_z}f(x,y,z)\,dx\,dy\right] \,dz Ωf(x,y,z)dxdydz=C1C2F(x)dz=C1C2Dzf(x,y,z)dxdydz
    例9:计算I=∭Ωe∣z∣ dxdydz,其中Ω为x2+y2+z2≤1所围成的立体\color{blue}计算I=\iiint\limits_{\Omega}e^{|z|}\,dxdydz,其中\Omega为x^2+y^2+z^2\le 1所围成的立体I=ΩezdxdydzΩx2+y2+z21.
    解:
    ∵e∣z∣为偶函数,采用先二后一的思想,I=2∫01e∣z∣ dz∬Dxy dxdy. 其中∬Dxy dxdy=π(r(z))2,r(z)=1−z2.∴I=2∫01ez(1−z2) dz=2π[−(z−1)2ez]01=2π. \because e^{|z|}为偶函数,采用先二后一的思想,\\ I=2\int_0^1e^{|z|}\,dz \iint\limits_{D_{xy}}\,dxdy.\ 其中\iint\limits_{D_{xy}}\,dxdy=\pi (r(z))^2,r(z)=\sqrt{1-z^2}. \\ \therefore I = 2\int_0^1 e^z(1-z^2)\,dz = 2\pi\left[-(z-1)^2e^z\right]_0^1 = 2\pi. ezI=201ezdzDxydxdy. Dxydxdy=π(r(z))2,r(z)=1z2.I=201ez(1z2)dz=2π[(z1)2ez]01=2π.
    例10:计算I=∭Ωx dxdydz,其中Ω由z=xy,x+y+z=1及z=0所围成的立体\color{blue}计算I=\iiint\limits_{\Omega}x\,dxdydz,其中\Omega由z=xy,x+y+z=1及z=0所围成的立体I=ΩxdxdydzΩz=xy,x+y+z=1z=0.
    解:
    发现当x为一定值时,平行于Oyz面的切面切立体的截面为一过原点的三角形,则原式表示成I=∫01x dx∬DyzS(x) dydz.对于S(x),当x=k时,则z=ky,z=1−k−y,z=0,sin⁡θ=k1+k2有S=12(1−k)(1+k21−k1+k)sin⁡θ=k(1−k)22(1+k).∴I=12∫01x(1−x)21+x dx+1=12∫12(t−2)2(t−1)2t dt=12∫01t3−6t2+13t−12+4t dt=2ln2−118. 发现当x为一定值时,平行于O_{yz}面的切面切立体的截面为一{\color{red}过原点的三角形},则原式表示成I=\int_0^1 x\,dx \iint\limits_{D_{yz}}S(x)\,dydz. \\ 对于S(x),当x=k时,则z=ky,z=1-k-y,z=0,\sin\theta=\dfrac{k}{\sqrt{1+k^2}} \\ 有S=\dfrac{1}{2}(1-k)(\sqrt{1+k^2}\dfrac{1-k}{1+k})\sin\theta = \dfrac{k(1-k)^2}{2(1+k)}. \\ \therefore I = \dfrac{1}{2}\int_0^1 \dfrac{x(1-x)^2}{1+x}\,dx+1 = \dfrac{1}{2}\int_1^2 \dfrac{(t-2)^2(t-1)^2}{t}\,dt = \dfrac{1}{2}\int_0^1 t^3-6t^2+13t-12+\dfrac{4}{t}\,dt = 2ln2-\dfrac{11}{8}. x,Oyz,I=01xdxDyzS(x)dydz.S(x),x=kz=ky,z=1ky,z=0sinθ=1+k2kS=21(1k)(1+k21+k1k)sinθ=2(1+k)k(1k)2.I=21011+xx(1x)2dx+1=2112t(t2)2(t1)2dt=2101t36t2+13t12+t4dt=2ln2811.
    在这里插入图片描述

  • 利用柱面坐标(case:被积函数具有f(x2+y2)f(\sqrt{x^2+y^2})f(x2+y2)的形式,积分区域是直柱体,且投影面用极坐标表示方便):
    {x=ρcos⁡θ.y=ρsin⁡θ.z=z. \begin{cases}x=\rho\cos\theta. \\ y=\rho\sin\theta. \\z=z. \end{cases} x=ρcosθ.y=ρsinθ.z=z.
    ∭Ωf(x,y,z) dx dy dz=∭Ωf(ρcos⁡θ,ρsin⁡θ,z)ρ dρdθdz=∫αβ dθ∫φ1φ2 ρ dρ∫z1z2f(ρcos⁡θ,ρsin⁡θ,z) dz\iiint\limits_{\Omega}f(x,y,z)\,dx\,dy\,dz = \iiint\limits_{\Omega}f(\rho\cos\theta, \rho\sin\theta, z)\rho\,d\rho d\theta dz = \int_{\alpha}^{\beta}\,d\theta \int_{\varphi_1}^{\varphi_2}\,{\color{red}\rho}\,d\rho\int_{z_1}^{z_2}f(\rho\cos\theta,\rho\sin\theta,z)\,dzΩf(x,y,z)dxdydz=Ωf(ρcosθ,ρsinθ,z)ρdρdθdz=αβdθφ1φ2ρdρz1z2f(ρcosθ,ρsinθ,z)dz

    例11:将I=∭Ωf(x,y,z) dxdydz表示成柱面坐标系的累次积分,其中Ω是由2z=x2+y2,z=2,z=1所围成的立体\color{blue}将I=\iiint\limits_{\Omega}f(x,y,z)\,dxdydz表示成柱面坐标系的累次积分,其中\Omega是由2z=x^2+y^2,z=2,z=1所围成的立体I=Ωf(x,y,z)dxdydzΩ2z=x2+y2,z=2,z=1.
    解:
    I=∫02π dθ∫02ρ dρ∫12f(ρcos⁡θ,ρsin⁡θ,z) dz+∫02π dθ∫22ρ dρ∫ρ222f(ρcos⁡θ,ρsin⁡θ,z) dz. I=\int_0^{2\pi}\,d\theta \int_0^{\sqrt{2}}\rho\,d\rho \int_1^2f(\rho\cos\theta,\rho\sin\theta,z)\,dz+\int_0^{2\pi}\,d\theta \int_{\sqrt{2}}^2\rho\,d\rho \int_{\rho^2\over2}^2f(\rho\cos\theta,\rho\sin\theta,z)\,dz. I=02πdθ02ρdρ12f(ρcosθ,ρsinθ,z)dz+02πdθ22ρdρ2ρ22f(ρcosθ,ρsinθ,z)dz.
    在这里插入图片描述

  • 利用球面坐标(case:被积函数具有f(x2+y2+z3)f(\sqrt{x^2+y^2+z^3})f(x2+y2+z3)的形式,积分区域是球面或锥面):
    {x=rsin⁡φcos⁡θ.y=rsin⁡φsin⁡θ.z=rcos⁡φ. \begin{cases}x=r\sin\varphi\cos\theta. \\ y=r\sin\varphi\sin\theta. \\z=r\cos\varphi. \end{cases} x=rsinφcosθ.y=rsinφsinθ.z=rcosφ.
    ∭Ωf(x,y,z) dx dy dz=∭Ωf(rsin⁡φcos⁡θ,rsin⁡φsin⁡θ,rcos⁡φ)r2sin⁡φ dr dφ dθ\iiint\limits_{\Omega}f(x,y,z)\,dx\,dy\,dz = \iiint\limits_{\Omega}f(r\sin\varphi\cos\theta, r\sin\varphi\sin\theta, r\cos\varphi){\color{red}r^2\sin\varphi}\,dr\,d\varphi\,d\thetaΩf(x,y,z)dxdydz=Ωf(rsinφcosθ,rsinφsinθ,rcosφ)r2sinφdrdφdθ.

    例12:将I=∭Ωf(x,y,z) dxdydz表示成球面坐标系的累次积分,其中Ω是由2z=x2+y2,z=2,z=1所围成的立体\color{blue}将I=\iiint\limits_{\Omega}f(x,y,z)\,dxdydz表示成球面坐标系的累次积分,其中\Omega是由2z=x^2+y^2,z=2,z=1所围成的立体I=Ωf(x,y,z)dxdydzΩ2z=x2+y2,z=2,z=1.
    解:
    I=∫02π dθ∫0π4sin⁡φ dφ∫1cos⁡φ2cos⁡φf(rsin⁡φcos⁡θ,rsin⁡φsin⁡θ,rcos⁡φ)r2 dr+∫02π dθ∫π4arctan⁡2sin⁡φ dφ∫1cos⁡φ2cos⁡φsin⁡2φf(rsin⁡φcos⁡θ,rsin⁡φsin⁡θ,rcos⁡φ)r2 dr. I=\int_0^{2\pi}\,d\theta \int_0^{\pi\over4}\sin\varphi\,d\varphi \int_{1\over{\cos\varphi}}^{2\over{\cos\varphi}}f(r\sin\varphi\cos\theta, r\sin\varphi\sin\theta, r\cos\varphi)r^2\,dr \\+ \int_0^{2\pi}\,d\theta \int_{\pi\over4}^{\arctan{\sqrt2}}\sin\varphi\,d\varphi {\color{red}\int_{1\over{\cos\varphi}}^{2\cos\varphi \over \sin^2\varphi}} f(r\sin\varphi\cos\theta, r\sin\varphi\sin\theta, r\cos\varphi)r^2\,dr. I=02πdθ04πsinφdφcosφ1cosφ2f(rsinφcosθ,rsinφsinθ,rcosφ)r2dr+02πdθ4πarctan2sinφdφcosφ1sin2φ2cosφf(rsinφcosθ,rsinφsinθ,rcosφ)r2dr.
    在这里插入图片描述

4 重积分的应用

一、曲面面积

  • 由定义可知:
    • ∬D1 dσ=\iint\limits_D 1\,d\sigma =D1dσ=平面区域面积
    • ∭Ω1 dV=∬Dxyz dσ=\iiint\limits_{\Omega} 1\,dV= \iint\limits_{D_{xy}}z\,d\sigma =Ω1dV=Dxyzdσ=空间区域体积
  • 二重积分计算曲面区域z=f(x,y)z=f(x,y)z=f(x,y)的面积:
    由于n⃗=(fx,fy,−1)\vec n=(f_x,f_y,-1)n=(fx,fy,1)∣cos⁡α∣=n⃗×(0,0,1)|\cos \alpha|= \vec n\times(0,0,1)cosα=n×(0,0,1),故∣cos⁡α∣=11+fx2(x,y)+fy2(x,y)|\cos \alpha|= \dfrac{1}{\sqrt{1+f_x^2(x,y)+f_y^2(x,y)}}cosα=1+fx2(x,y)+fy2(x,y)1
    得到 S=∬D1+fx2(x,y)+fy2(x,y) dσS=\iint\limits_ D\sqrt{1+f_x^2(x,y)+f_y^2(x,y)}\,d\sigmaS=D1+fx2(x,y)+fy2(x,y)dσ.

二、质心

  • 质心:使物体上的点PPP对空间任意一点P0P_0P0力矩积分为0,即∭Ω(OP→−OP0→)ρ(x,y,z) dV=0\iiint\limits_{\Omega}(\overrightarrow{OP}- \overrightarrow{OP_0})\rho(x,y,z)\,dV=0Ω(OPOP0)ρ(x,y,z)dV=0
  • 故质心坐标(x0,y0,z0)(x_0,y_0,z_0)(x0,y0,z0)满足:{∭Ω(x−x0)ρ(x,y,z) dV=0∭Ω(y−y0)ρ(x,y,z) dV=0∭Ω(z−z0)ρ(x,y,z) dV=0\begin{cases}\iiint\limits_{\Omega}(x-x_0)\rho(x,y,z)\,dV=0 \\ \iiint\limits_{\Omega}(y-y_0)\rho(x,y,z)\,dV=0 \\ \iiint\limits_{\Omega}(z-z_0)\rho(x,y,z)\,dV=0 \end{cases}Ω(xx0)ρ(x,y,z)dV=0Ω(yy0)ρ(x,y,z)dV=0Ω(zz0)ρ(x,y,z)dV=0

    {x0=∭Ωxρ(x,y,z) dVmy0=∭Ωyρ(x,y,z) dVmz0=∭Ωzρ(x,y,z) dVm \color{red}\begin{cases}x_0=\dfrac{\iiint\limits_{\Omega}x\rho(x,y,z)\,dV}{m} \\ y_0=\dfrac{\iiint\limits_{\Omega}y\rho(x,y,z)\,dV}{m} \\ z_0=\dfrac{\iiint\limits_{\Omega}z\rho(x,y,z)\,dV}{m} \end{cases} x0=mΩxρ(x,y,z)dVy0=mΩyρ(x,y,z)dVz0=mΩzρ(x,y,z)dV
    其中m=∭Ωρ(x,y,z) dVm=\iiint\limits_{\Omega}\rho(x,y,z)\,dVm=Ωρ(x,y,z)dV即物体的质量。
  • 假如物体为一薄板,且水平放置在Oxy面上,则
    {x0=∬Dxρ(x,y) dσmy0=∭Dyρ(x,y) dσm \begin{cases}x_0=\dfrac{\iint\limits_ D x\rho(x,y)\,d\sigma}{m} \\ y_0=\dfrac{\iiint\limits_ D y\rho(x,y)\,d\sigma}{m} \end{cases} x0=mDxρ(x,y)dσy0=mDyρ(x,y)dσ

三、转动惯量

  • 整个物体对于坐标轴的转动变量:
    例如对于zzz轴:Jz=∭Ω(x2+y2)ρ(x,y,z) dVJ_z=\iiint\limits_{\Omega}(x^2+y^2)\rho(x,y,z)\,dVJz=Ω(x2+y2)ρ(x,y,z)dV.
  • 整个物体对于一点P0(x0,y0,z0)P_0(x_0,y_0,z_0)P0(x0,y0,z0)的转动变量:JP0=∭Ω[(x−x0)2+(y−y0)2+(z−z0)2]ρ(x,y,z) dVJ_{P_0}=\iiint\limits_{\Omega}\left[(x-x_0)^2+(y-y_0)^2+(z-z_0)^2 \right]\rho(x,y,z)\,dVJP0=Ω[(xx0)2+(yy0)2+(zz0)2]ρ(x,y,z)dV
    特别的,物体关于
    原点
    的转动惯量为J0=∭Ω(x2+y2+z2)ρ(x,y,z) dVJ_0=\iiint\limits_{\Omega}(x^2+y^2+z^2)\rho(x,y,z)\,dVJ0=Ω(x2+y2+z2)ρ(x,y,z)dV.
  • 由于一个质点关于一个平面的转动惯量为质量乘以质点到平面最短距离之平方,故整个物体对于坐标面的转动惯量:
    例如对于OxyOxyOxy面:Jxy=∭Ωz2ρ(x,y,z) dVJ_{xy}=\iiint\limits_{\Omega}z^2\rho(x,y,z)\,dVJxy=Ωz2ρ(x,y,z)dV.

四、引力

  • 物体对于质点P0P_0P0的总引力为
    F=G∭ΩM0 dmr2⋅r0=G∭ΩM0⋅r⃗(x,y,z)⋅ρ(x,y,z) dVr3 F=G\iiint\limits_{\Omega}\dfrac{M_0\,dm}{r^2}\cdot r_0 = G\iiint\limits_{\Omega}\dfrac{M_0\cdot \vec r(x,y,z)\cdot \rho(x,y,z)\,dV}{r^3} F=GΩr2M0dmr0=GΩr3M0r(x,y,z)ρ(x,y,z)dV
    其中r0=r⃗(x,y,z)rr_0=\dfrac{\vec r(x,y,z)}{r}r0=rr(x,y,z)为单位向量,r=(x−x0)2+(y−y0)2+(z−z0)2r=\sqrt{(x-x_0)^2+(y-y_0)^2+(z-z_0)^2}r=(xx0)2+(yy0)2+(zz0)2r⃗(x,y,z)=(x−x0,y−y0,z−z0)\vec r(x,y,z)=(x-x_0,y-y_0,z-z_0)r(x,y,z)=(xx0,yy0,zz0),则坐标分量为
    {Fx=GM0∭Ωx−x0r3ρ(x,y,z) dVFy=GM0∭Ωy−y0r3ρ(x,y,z) dVFz=GM0∭Ωz−z0r3ρ(x,y,z) dV \color{red}\begin{cases}F_x=GM_0\iiint\limits_{\Omega}\dfrac{x-x_0}{r^3}\rho(x,y,z)\,dV \\ F_y=GM_0\iiint\limits_{\Omega}\dfrac{y-y_0}{r^3}\rho(x,y,z)\,dV \\ F_z=GM_0\iiint\limits_{\Omega}\dfrac{z-z_0}{r^3}\rho(x,y,z)\,dV \end{cases} Fx=GM0Ωr3xx0ρ(x,y,z)dVFy=GM0Ωr3yy0ρ(x,y,z)dVFz=GM0Ωr3zz0ρ(x,y,z)dV
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