目录
6.1重积分例9计算二重积分∬Dy2dσ\displaystyle\iint\limits_{D}y^2\mathrm{d}\sigmaD∬y2dσ,其中DDD由{x=a(t−sint)y=a(1−cost)(0⩽t⩽2π)\begin{cases}x=a(t-\sin t)\\y=a(1-\cos t)\end{cases}(0\leqslant t\leqslant2\pi){x=a(t−sint)y=a(1−cost)(0⩽t⩽2π)与y=0y=0y=0围成。例11计算二重积分∬D∣x2+y2−2y∣dσ\displaystyle\iint\limits_{D}|x^2+y^2-2y|\mathrm{d}\sigmaD∬∣x2+y2−2y∣dσ,其中DDD由不等式x2+y2⩽4x^2+y^2\leqslant4x2+y2⩽4所确定。例12计算二重积分∬Dmin{x,y}e−(x2+y2)dσ\displaystyle\iint\limits_{D}\min\{x,y\}e^{-(x^2+y^2)}\mathrm{d}\sigmaD∬min{x,y}e−(x2+y2)dσ,其中DDD为全平面。例13设平面域D={(x,y)∣1⩽x2+y2⩽4,x⩾0,y⩾0}D=\{(x,y)|1\leqslant x^2+y^2\leqslant4,x\geqslant0,y\geqslant0\}D={(x,y)∣1⩽x2+y2⩽4,x⩾0,y⩾0},计算∬Dxsin(πx2+y2)x+ydxdy\displaystyle\iint\limits_{D}\cfrac{x\sin(\pi\sqrt{x^2+y^2})}{x+y}\mathrm{d}x\mathrm{d}yD∬x+yxsin(πx2+y2)dxdy。例20设f(t)f(t)f(t)在[0,+∞)[0,+\infty)[0,+∞)上连续,且满足f(t)=e4πt2+∬x2+y2⩽4t2f(12x2+y2)dxdyf(t)=e^{4\pi t^2}+\displaystyle\iint\limits_{x^2+y^2\leqslant4t^2}f\left(\cfrac{1}{2}\sqrt{x^2+y^2}\right)\mathrm{d}x\mathrm{d}yf(t)=e4πt2+x2+y2⩽4t2∬f(21x2+y2)dxdy,求f(t)f(t)f(t)。例21设f(x,y)f(x,y)f(x,y)是定义在0⩽x⩽1,0⩽y⩽10\leqslant x\leqslant1,0\leqslant y\leqslant10⩽x⩽1,0⩽y⩽1上的连续函数,f(0,0)=1f(0,0)=1f(0,0)=1,求极限limx→0+∫0x2dt∫xtf(t,u)du1−e−x3\lim\limits_{x\to0^+}\cfrac{\displaystyle\int^{x^2}_0\mathrm{d}t\displaystyle\int^{\sqrt{t}}_xf(t,u)\mathrm{d}u}{1-e^{-x^3}}x→0+lim1−e−x3∫0x2dt∫xtf(t,u)du。例25设函数f(x)f(x)f(x)在区间[a,b][a,b][a,b]上连续,且恒大于零,证明:∫abf(x)dx∫ab1f(x)dx⩾(b−a)2\displaystyle\int^b_af(x)\mathrm{d}x\displaystyle\int^b_a\cfrac{1}{f(x)}\mathrm{d}x\geqslant(b-a)^2∫abf(x)dx∫abf(x)1dx⩾(b−a)2。例26设f(x)f(x)f(x)在区间[0,1][0,1][0,1]上单调递减且为正值的连续函数,证明∫01xf2(x)dx∫01xf(x)dx⩽∫01f2(x)dx∫01f(x)dx\cfrac{\displaystyle\int^1_0xf^2(x)\mathrm{d}x}{\displaystyle\int^1_0xf(x)\mathrm{d}x}\leqslant\cfrac{\displaystyle\int^1_0f^2(x)\mathrm{d}x}{\displaystyle\int^1_0f(x)\mathrm{d}x}∫01xf(x)dx∫01xf2(x)dx⩽∫01f(x)dx∫01f2(x)dx。6.2曲线积分例42计算∮Lx2ds\displaystyle\oint_Lx^2\mathrm{d}s∮Lx2ds,其中L:{x2+y2+z2=R2,x+y+z=0.L:\begin{cases}x^2+y^2+z^2=R^2,\\x+y+z=0.\end{cases}L:{x2+y2+z2=R2,x+y+z=0.6.3曲面积分例56计算∬ΣdSx2+y2+z2\displaystyle\iint\limits_{\Sigma}\cfrac{\mathrm{d}S}{x^2+y^2+z^2}Σ∬x2+y2+z2dS,其中Σ\SigmaΣ为柱面x2+y2=R2x^2+y^2=R^2x2+y2=R2夹在z=0z=0z=0和z=Rz=Rz=R之间的部分。写在最后6.1重积分
例9计算二重积分∬Dy2dσ\displaystyle\iint\limits_{D}y^2\mathrm{d}\sigmaD∬y2dσ,其中DDD由{x=a(t−sint)y=a(1−cost)(0⩽t⩽2π)\begin{cases}x=a(t-\sin t)\\y=a(1-\cos t)\end{cases}(0\leqslant t\leqslant2\pi){x=a(t−sint)y=a(1−cost)(0⩽t⩽2π)与y=0y=0y=0围成。
解不妨设曲线{x=a(t−sint)y=a(1−cost)\begin{cases}x=a(t-\sin t)\\y=a(1-\cos t)\end{cases}{x=a(t−sint)y=a(1−cost)的直角坐标方程为y=y(x)y=y(x)y=y(x),则
∬Dy2dσ=∫02πadx∫0y(x)y2dy=13∫02πay3(x)dx=13∫02πa3(1−cost)3a(1−cost)dt=16a43∫02πsin8t2dt=t2=u32a43∫0πsin8udu=64a43∫0π2sin8udu=64a43⋅78⋅56⋅34⋅12⋅π2=3512πa4.\begin{aligned} \displaystyle\iint\limits_{D}y^2\mathrm{d}\sigma&=\displaystyle\int^{2\pi a}_0\mathrm{d}x\displaystyle\int^{y(x)}_0y^2\mathrm{d}y=\cfrac{1}{3}\displaystyle\int^{2\pi a}_0y^3(x)\mathrm{d}x\\ &=\cfrac{1}{3}\displaystyle\int^{2\pi}_0a^3(1-\cos t)^3a(1-\cos t)\mathrm{d}t\\ &=\cfrac{16a^4}{3}\displaystyle\int^{2\pi}_0\sin^8\cfrac{t}{2}\mathrm{d}t\xlongequal{\frac{t}{2}=u}\cfrac{32a^4}{3}\displaystyle\int^{\pi}_0\sin^8u\mathrm{d}u\\ &=\cfrac{64a^4}{3}\displaystyle\int^{\frac{\pi}{2}}_0\sin^8u\mathrm{d}u=\cfrac{64a^4}{3}\cdot\cfrac{7}{8}\cdot\cfrac{5}{6}\cdot\cfrac{3}{4}\cdot\cfrac{1}{2}\cdot\cfrac{\pi}{2}=\cfrac{35}{12}\pi a^4. \end{aligned} D∬y2dσ=∫02πadx∫0y(x)y2dy=31∫02πay3(x)dx=31∫02πa3(1−cost)3a(1−cost)dt=316a4∫02πsin82tdt2t=u332a4∫0πsin8udu=364a4∫02πsin8udu=364a4⋅87⋅65⋅43⋅21⋅2π=1235πa4.
(这道题主要利用了参数方程解积分求解)
例11计算二重积分∬D∣x2+y2−2y∣dσ\displaystyle\iint\limits_{D}|x^2+y^2-2y|\mathrm{d}\sigmaD∬∣x2+y2−2y∣dσ,其中DDD由不等式x2+y2⩽4x^2+y^2\leqslant4x2+y2⩽4所确定。
解令x2+y2−2y=0x^2+y^2-2y=0x2+y2−2y=0,即x2+y2=2yx^2+y^2=2yx2+y2=2y,该曲线就将原积分域划为两部分,如下图,其中红色部分为D1D_1D1,橙色部分为D2D_2D2。在D1D_1D1上x2+y2−2yx^2+y^2-2yx2+y2−2y为负,在D2D_2D2上x2+y2−2yx^2+y^2-2yx2+y2−2y为正,则
∬D∣x2+y2−2y∣dσ=∬D1∣x2+y2−2y∣dσ+∬D2∣x2+y2−2y∣dσ=∬D1(2y−x2−y2)dσ+∬D2(x2+y2−2y)dσ=∬D1(2y−x2−y2)dσ+(∬D(x2+y2−2y)dσ−∬D1(x2+y2−2y)dσ)=∬D(x2+y2−2y)dσ+2∬D1(2y−x2−y2)dσ=∫02πdθ∫02ρ3dρ+2∫0πdθ∫02sinθ(2ρsinθ−ρ2)ρdρ=9π.\begin{aligned} &\displaystyle\iint\limits_{D}|x^2+y^2-2y|\mathrm{d}\sigma\\ =&\displaystyle\iint\limits_{D_1}|x^2+y^2-2y|\mathrm{d}\sigma+\displaystyle\iint\limits_{D_2}|x^2+y^2-2y|\mathrm{d}\sigma\\ =&\displaystyle\iint\limits_{D_1}(2y-x^2-y^2)\mathrm{d}\sigma+\displaystyle\iint\limits_{D_2}(x^2+y^2-2y)\mathrm{d}\sigma\\ =&\displaystyle\iint\limits_{D_1}(2y-x^2-y^2)\mathrm{d}\sigma+\left(\displaystyle\iint\limits_{D}(x^2+y^2-2y)\mathrm{d}\sigma-\displaystyle\iint\limits_{D_1}(x^2+y^2-2y)\mathrm{d}\sigma\right)\\ =&\displaystyle\iint\limits_{D}(x^2+y^2-2y)\mathrm{d}\sigma+2\displaystyle\iint\limits_{D_1}(2y-x^2-y^2)\mathrm{d}\sigma\\ =&\displaystyle\int^{2\pi}_0\mathrm{d}\theta\displaystyle\int^2_0\rho^3\mathrm{d}\rho+2\displaystyle\int^{\pi}_0\mathrm{d}\theta\displaystyle\int^{2\sin\theta}_0(2\rho\sin\theta-\rho^2)\rho\mathrm{d}\rho=9\pi. \end{aligned} =====D∬∣x2+y2−2y∣dσD1∬∣x2+y2−2y∣dσ+D2∬∣x2+y2−2y∣dσD1∬(2y−x2−y2)dσ+D2∬(x2+y2−2y)dσD1∬(2y−x2−y2)dσ+⎝⎛D∬(x2+y2−2y)dσ−D1∬(x2+y2−2y)dσ⎠⎞D∬(x2+y2−2y)dσ+2D1∬(2y−x2−y2)dσ∫02πdθ∫02ρ3dρ+2∫0πdθ∫02sinθ(2ρsinθ−ρ2)ρdρ=9π.
(这道题主要利用了分类积分求解)
例12计算二重积分∬Dmin{x,y}e−(x2+y2)dσ\displaystyle\iint\limits_{D}\min\{x,y\}e^{-(x^2+y^2)}\mathrm{d}\sigmaD∬min{x,y}e−(x2+y2)dσ,其中DDD为全平面。
解将全平面用直线y=xy=xy=x划分为两部分,直线y=xy=xy=x以上的部分记为D1D_1D1,以下部分记为D2D_2D2。如下图。
∬Dmin{x,y}e−(x2+y2)dσ=∬D1xe−(x2+y2)dσ+∬D2ye−(x2+y2)dσ=∫−∞+∞dy∫−∞yxe−x2⋅e−y2dx+∫−∞+∞dx∫−∞xye−y2⋅e−x2dx=−12∫−∞+∞e−2y2dy−12∫−∞+∞e−2x2dx=−∫−∞+∞e−2x2dx=2x=t−12∫−∞+∞e−t2dt=−12π=−π2.\begin{aligned} &\displaystyle\iint\limits_{D}\min\{x,y\}e^{-(x^2+y^2)}\mathrm{d}\sigma\\ &=\displaystyle\iint\limits_{D_1}xe^{-(x^2+y^2)}\mathrm{d}\sigma+\displaystyle\iint\limits_{D_2}ye^{-(x^2+y^2)}\mathrm{d}\sigma\\ &=\displaystyle\int^{+\infty}_{-\infty}\mathrm{d}y\displaystyle\int^y_{-\infty}xe^{-x^2}\cdot e^{-y^2}\mathrm{d}x+\displaystyle\int^{+\infty}_{-\infty}\mathrm{d}x\displaystyle\int^x_{-\infty}ye^{-y^2}\cdot e^{-x^2}\mathrm{d}x\\ &=-\cfrac{1}{2}\displaystyle\int^{+\infty}_{-\infty}e^{-2y^2}\mathrm{d}y-\cfrac{1}{2}\displaystyle\int^{+\infty}_{-\infty}e^{-2x^2}\mathrm{d}x=-\displaystyle\int^{+\infty}_{-\infty}e^{-2x^2}\mathrm{d}x\\ &\xlongequal{\sqrt{2}x=t}-\cfrac{1}{\sqrt{2}}\displaystyle\int^{+\infty}_{-\infty}e^{-t^2}\mathrm{d}t=-\cfrac{1}{\sqrt{2}}\sqrt{\pi}=-\sqrt{\cfrac{\pi}{2}}. \end{aligned} D∬min{x,y}e−(x2+y2)dσ=D1∬xe−(x2+y2)dσ+D2∬ye−(x2+y2)dσ=∫−∞+∞dy∫−∞yxe−x2⋅e−y2dx+∫−∞+∞dx∫−∞xye−y2⋅e−x2dx=−21∫−∞+∞e−2y2dy−21∫−∞+∞e−2x2dx=−∫−∞+∞e−2x2dx2x=t−21∫−∞+∞e−t2dt=−21π=−2π.
(这道题主要利用了分类积分求解)
例13设平面域D={(x,y)∣1⩽x2+y2⩽4,x⩾0,y⩾0}D=\{(x,y)|1\leqslant x^2+y^2\leqslant4,x\geqslant0,y\geqslant0\}D={(x,y)∣1⩽x2+y2⩽4,x⩾0,y⩾0},计算∬Dxsin(πx2+y2)x+ydxdy\displaystyle\iint\limits_{D}\cfrac{x\sin(\pi\sqrt{x^2+y^2})}{x+y}\mathrm{d}x\mathrm{d}yD∬x+yxsin(πx2+y2)dxdy。
解由于积分域DDD关于直线y=xy=xy=x对称,则
∬Dxsin(πx2+y2)x+ydxdy=∬Dysin(πx2+y2)x+ydxdy=12[∬Dxsin(πx2+y2)x+ydxdy+∬Dysin(πx2+y2)x+ydxdy]=12∬Dsin(πx2+y2)dxdy=12∫0π2dθ∫12sin(πr)dr=−14∫12rd(πr)=−34.\begin{aligned} &\displaystyle\iint\limits_{D}\cfrac{x\sin(\pi\sqrt{x^2+y^2})}{x+y}\mathrm{d}x\mathrm{d}y\\ =&\displaystyle\iint\limits_{D}\cfrac{y\sin(\pi\sqrt{x^2+y^2})}{x+y}\mathrm{d}x\mathrm{d}y\\ =&\cfrac{1}{2}\left[\displaystyle\iint\limits_{D}\cfrac{x\sin(\pi\sqrt{x^2+y^2})}{x+y}\mathrm{d}x\mathrm{d}y+\displaystyle\iint\limits_{D}\cfrac{y\sin(\pi\sqrt{x^2+y^2})}{x+y}\mathrm{d}x\mathrm{d}y\right]\\ =&\cfrac{1}{2}\displaystyle\iint\limits_{D}\sin(\pi\sqrt{x^2+y^2})\mathrm{d}x\mathrm{d}y=\cfrac{1}{2}\displaystyle\int^{\frac{\pi}{2}}_0\mathrm{d}\theta\displaystyle\int^2_1\sin(\pi r)\mathrm{d}r\\ =&-\cfrac{1}{4}\displaystyle\int^2_1r\mathrm{d}(\pi r)=-\cfrac{3}{4}. \end{aligned} ====D∬x+yxsin(πx2+y2)dxdyD∬x+yysin(πx2+y2)dxdy21⎣⎡D∬x+yxsin(πx2+y2)dxdy+D∬x+yysin(πx2+y2)dxdy⎦⎤21D∬sin(πx2+y2)dxdy=21∫02πdθ∫12sin(πr)dr−41∫12rd(πr)=−43.
(这道题主要利用了函数积分的对称性求解)
例20设f(t)f(t)f(t)在[0,+∞)[0,+\infty)[0,+∞)上连续,且满足f(t)=e4πt2+∬x2+y2⩽4t2f(12x2+y2)dxdyf(t)=e^{4\pi t^2}+\displaystyle\iint\limits_{x^2+y^2\leqslant4t^2}f\left(\cfrac{1}{2}\sqrt{x^2+y^2}\right)\mathrm{d}x\mathrm{d}yf(t)=e4πt2+x2+y2⩽4t2∬f(21x2+y2)dxdy,求f(t)f(t)f(t)。
解显然f(0)=1f(0)=1f(0)=1,且∬x2+y2⩽4t2f(12x2+y2)dxdy=∫02πdθ∫02tf(12ρ)ρdρ=2π∫02tρf(12ρ)dρ\displaystyle\iint\limits_{x^2+y^2\leqslant4t^2}f\left(\cfrac{1}{2}\sqrt{x^2+y^2}\right)\mathrm{d}x\mathrm{d}y=\displaystyle\int^{2\pi}_0\mathrm{d}\theta\displaystyle\int^{2t}_0f\left(\cfrac{1}{2}\rho\right)\rho\mathrm{d}\rho=2\pi\displaystyle\int^{2t}_0\rho f\left(\cfrac{1}{2}\rho\right)\mathrm{d}\rhox2+y2⩽4t2∬f(21x2+y2)dxdy=∫02πdθ∫02tf(21ρ)ρdρ=2π∫02tρf(21ρ)dρ,则f(t)=e4πt2+2π∫02tρf(12ρ)dρf(t)=e^{4\pi t^2}+2\pi\displaystyle\int^{2t}_0\rho f\left(\cfrac{1}{2}\rho\right)\mathrm{d}\rhof(t)=e4πt2+2π∫02tρf(21ρ)dρ。
对ttt上式两端求导得f′(t)=8πte4πt2+8πtf(t)f'(t)=8\pi te^{4\pi t^2}+8\pi tf(t)f′(t)=8πte4πt2+8πtf(t)。由一阶线性微分方程通解公式得f(t)=e∫8πtdt[∫8πte4πt2e−∫8πtdtdt+C]=(4πt2+C)e4πt2f(t)=e^{\int8\pi t\mathrm{d}t}\left[\displaystyle\int8\pi te^{4\pi t^2}e^{-\int8\pi t\mathrm{d}t}\mathrm{d}t+C\right]=(4\pi t^2+C)e^{4\pi t^2}f(t)=e∫8πtdt[∫8πte4πt2e−∫8πtdtdt+C]=(4πt2+C)e4πt2。
由f(0)=1f(0)=1f(0)=1得C=1C=1C=1,因此f(t)=(4πt2+1)e4πt2f(t)=(4\pi t^2+1)e^{4\pi t^2}f(t)=(4πt2+1)e4πt2。
(这道题主要利用了微分方程求解)
例21设f(x,y)f(x,y)f(x,y)是定义在0⩽x⩽1,0⩽y⩽10\leqslant x\leqslant1,0\leqslant y\leqslant10⩽x⩽1,0⩽y⩽1上的连续函数,f(0,0)=1f(0,0)=1f(0,0)=1,求极限limx→0+∫0x2dt∫xtf(t,u)du1−e−x3\lim\limits_{x\to0^+}\cfrac{\displaystyle\int^{x^2}_0\mathrm{d}t\displaystyle\int^{\sqrt{t}}_xf(t,u)\mathrm{d}u}{1-e^{-x^3}}x→0+lim1−e−x3∫0x2dt∫xtf(t,u)du。
解
limx→0+∫0x2dt∫xtf(t,u)du1−e−x3=limx→0+−∫0xdu∫0u2f(t,u)dtx3=−limx→0+∫0x2f(t,u)dt3x2=−limx→0+x2f(c,x)3x2=−13f(0,0)=13.\begin{aligned} &\lim\limits_{x\to0^+}\cfrac{\displaystyle\int^{x^2}_0\mathrm{d}t\displaystyle\int^{\sqrt{t}}_xf(t,u)\mathrm{d}u}{1-e^{-x^3}}\\ =&\lim\limits_{x\to0^+}\cfrac{-\displaystyle\int^x_0\mathrm{d}u\displaystyle\int^{u^2}_0f(t,u)\mathrm{d}t}{x^3}\\ =&-\lim\limits_{x\to0^+}\cfrac{\displaystyle\int^{x^2}_0f(t,u)\mathrm{d}t}{3x^2}\\ =&-\lim\limits_{x\to0^+}\cfrac{x^2f(c,x)}{3x^2}=-\cfrac{1}{3}f(0,0)=\cfrac{1}{3}. \end{aligned} ===x→0+lim1−e−x3∫0x2dt∫xtf(t,u)dux→0+limx3−∫0xdu∫0u2f(t,u)dt−x→0+lim3x2∫0x2f(t,u)dt−x→0+lim3x2x2f(c,x)=−31f(0,0)=31.
(这道题主要利用了积分换序求解)
例25设函数f(x)f(x)f(x)在区间[a,b][a,b][a,b]上连续,且恒大于零,证明:∫abf(x)dx∫ab1f(x)dx⩾(b−a)2\displaystyle\int^b_af(x)\mathrm{d}x\displaystyle\int^b_a\cfrac{1}{f(x)}\mathrm{d}x\geqslant(b-a)^2∫abf(x)dx∫abf(x)1dx⩾(b−a)2。
证由柯西积分不等式(∫abf(x)g(x)dx)2⩽∫abf2(x)dx⋅∫abg2(x)dx\left(\displaystyle\int^b_af(x)g(x)\mathrm{d}x\right)^2\leqslant\displaystyle\int^b_af^2(x)\mathrm{d}x\cdot\displaystyle\int^b_ag^2(x)\mathrm{d}x(∫abf(x)g(x)dx)2⩽∫abf2(x)dx⋅∫abg2(x)dx知
∫abf(x)dx∫ab1f(x)dx=∫ab(f(x))2dx∫ab(1f(x))2dx⩾(∫abf(x)⋅1f(x)dx)2=(b−a)2\begin{aligned} \displaystyle\int^b_af(x)\mathrm{d}x\displaystyle\int^b_a\cfrac{1}{f(x)}\mathrm{d}x&=\displaystyle\int^b_a(\sqrt{f(x)})^2\mathrm{d}x\displaystyle\int^b_a\left(\sqrt{\cfrac{1}{f(x)}}\right)^2\mathrm{d}x\\ &\geqslant\left(\displaystyle\int^b_a\sqrt{f(x)}\cdot\cfrac{1}{\sqrt{f(x)}}\mathrm{d}x\right)^2=(b-a)^2 \end{aligned} ∫abf(x)dx∫abf(x)1dx=∫ab(f(x))2dx∫ab⎝⎛f(x)1⎠⎞2dx⩾(∫abf(x)⋅f(x)1dx)2=(b−a)2
(这道题主要利用了柯西中值定理求解,另这道题的另一种解法见高等数学张宇18讲第九讲积分等式与积分不等式的例9.7,传送门在这里)
例26设f(x)f(x)f(x)在区间[0,1][0,1][0,1]上单调递减且为正值的连续函数,证明∫01xf2(x)dx∫01xf(x)dx⩽∫01f2(x)dx∫01f(x)dx\cfrac{\displaystyle\int^1_0xf^2(x)\mathrm{d}x}{\displaystyle\int^1_0xf(x)\mathrm{d}x}\leqslant\cfrac{\displaystyle\int^1_0f^2(x)\mathrm{d}x}{\displaystyle\int^1_0f(x)\mathrm{d}x}∫01xf(x)dx∫01xf2(x)dx⩽∫01f(x)dx∫01f2(x)dx。
证(这道题主要利用了积分的对称性求解,另这道题的解法见李永乐复习全书高等数学第三章一元函数积分学练习三的第十题,传送门在这里)
6.2曲线积分
例42计算∮Lx2ds\displaystyle\oint_Lx^2\mathrm{d}s∮Lx2ds,其中L:{x2+y2+z2=R2,x+y+z=0.L:\begin{cases}x^2+y^2+z^2=R^2,\\x+y+z=0.\end{cases}L:{x2+y2+z2=R2,x+y+z=0.
解由于L:{x2+y2+z2=R2,x+y+z=0L:\begin{cases}x^2+y^2+z^2=R^2,\\x+y+z=0\end{cases}L:{x2+y2+z2=R2,x+y+z=0对变量x,y,zx,y,zx,y,z有对称性,即三个变量x,y,zx,y,zx,y,z中任意两个对称方程不变,则
∮Lx2ds=∮Ly2ds=∮Lz2ds=13∮L(x2+y2+z2)ds=13∮LR2ds=2π3R3.\displaystyle\oint_Lx^2\mathrm{d}s=\displaystyle\oint_Ly^2\mathrm{d}s=\displaystyle\oint_Lz^2\mathrm{d}s=\cfrac{1}{3}\displaystyle\oint_L(x^2+y^2+z^2)\mathrm{d}s=\cfrac{1}{3}\displaystyle\oint_LR^2\mathrm{d}s=\cfrac{2\pi}{3}R^3. ∮Lx2ds=∮Ly2ds=∮Lz2ds=31∮L(x2+y2+z2)ds=31∮LR2ds=32πR3.
(这道题主要利用了积分的对称性求解)
6.3曲面积分
例56计算∬ΣdSx2+y2+z2\displaystyle\iint\limits_{\Sigma}\cfrac{\mathrm{d}S}{x^2+y^2+z^2}Σ∬x2+y2+z2dS,其中Σ\SigmaΣ为柱面x2+y2=R2x^2+y^2=R^2x2+y2=R2夹在z=0z=0z=0和z=Rz=Rz=R之间的部分。
解将柱面方程x2+y2=R2x^2+y^2=R^2x2+y2=R2代入积分式得∬ΣdSx2+y2+z2=∬ΣdSR2+z2\displaystyle\iint\limits_{\Sigma}\cfrac{\mathrm{d}S}{x^2+y^2+z^2}=\displaystyle\iint\limits_{\Sigma}\cfrac{\mathrm{d}S}{R^2+z^2}Σ∬x2+y2+z2dS=Σ∬R2+z2dS。由于被积函数中只有zzz,则面积微元dS\mathrm{d}SdS可取为柱面x2+y2=R2x^2+y^2=R^2x2+y2=R2夹在两平面z=0z=0z=0和z=Rz=Rz=R之间的部分,即dS=2πRdz\mathrm{d}S=2\pi R\mathrm{d}zdS=2πRdz,则∬ΣdSR2+z2=2πR∫0HdzR2+z2=2πarctanHR\displaystyle\iint\limits_{\Sigma}\cfrac{\mathrm{d}S}{R^2+z^2}=2\pi R\displaystyle\int^H_0\cfrac{\mathrm{d}z}{R^2+z^2}=2\pi\arctan\cfrac{H}{R}Σ∬R2+z2dS=2πR∫0HR2+z2dz=2πarctanRH。(这道题主要利用了积分定义求解)
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