Lesson 54 换元公式的应用

换元公式的应用

设 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Φ:D′→D\\Phi:D'\\to DΦ:D′→D 是 C1C^1C1 的双射，则对于 DDD 上可积的 fff 有：\n∫Df=∫D′(f∘Φ)∣det⁡JΦ∣\\int_Df=\\int_{D'}(f\\circ\\Phi)|\\det J_\\Phi|\n∫Df=∫D′(f∘Φ)∣detJΦ∣\n也就是权重拉回 (f∘Φf\\circ\\Phif∘Φ) 乘上 Φ\\PhiΦ 引起体积微元的放大倍数.\n/Example/ (极坐标计算二重积分)\nΦ:[0,+∞)×[0,2π]→R2\\Phi:[0,+\\infty)\\times[0,2\\pi]\\to\\R^2Φ:[0,+∞)×[0,2π]→R2，Φ=Φ(rcos⁡θ,rsin⁡θ)\\Phi=\\Phi(r\\cos\\theta,r\\sin\\theta)Φ=Φ(rcosθ,rsinθ).\n在这里，Φ\\PhiΦ 虽然是 C1C^1C1 的，但是 r=0r=0r=0 这一条线段被映射到一个点 (坐标原点)、θ=2π,θ=0\\theta=2\\pi,\\theta=0θ=2π,θ=0 两条射线被映射为一条射线 (xxx 轴正方向)，所以 Φ\\PhiΦ 并非单射.\n但是我们仍然可以处理这个问题：对于 r=0r=0r=0 映射到一个点的情况，\n∣∫Df−∫D∩(R2∣Bε(0⃗))f∣=∣∫D∩Bε(0⃗)f∣≤Mπ2ε→0\\left|\\int_Df-\\int_{D\\cap(\\R^2|B_{\\varepsilon}(\\vec{0}))}f\\right|=\\left|\\int_{D\\cap B_\\varepsilon(\\vec{0})}f\\right|\\leq M\\pi^2\\varepsilon\\to0\n\n∫Df−∫D∩(R2∣Bε(0\n))f\n=\n∫D∩Bε(0\n)f\n≤Mπ2ε→0\n可见\n∫Df=lim⁡ε→0+∫D∩(R2∣Bε(0⃗))f\\int_Df=\\lim_{\\varepsilon\\to0+}\\int_{D\\cap(\\R^2|B_\\varepsilon(\\vec{0}))}f\n∫Df=ε→0+lim∫D∩(R2∣Bε(0\n))f\n只需要对 D∣Bε(0⃗)D|B_\\varepsilon(\\vec{0})D∣Bε(0\n) (记为 D~\\tilde DD~) 进行换元. 对于 θ=0,θ=2π\\theta=0,\\theta=2\\piθ=0,θ=2π 重合的情况，\n∫D~f=∫D~∩R+2f+∫D~∩R−2f\\int_{\\tilde D}f=\\int_{\\tilde D\\cap\\R^2_+}f+\\int_{\\tilde D\\cap\\R^2_-}f\n∫D~f=∫D~∩R+2f+∫D~∩R−2f\n也就是将 R2\\R^2R2 从 +x+x+x 无穷远处剪开剪到原点处，分为上半和下半两个平面来换元. 这样我们得到，\n∬Df(x,y)dxdy=∬Φ−1(D)f(rcos⁡θ,rsin⁡θ)rdrdθ\\iint_Df(x,y)\\text{d}x\\text{d}y=\\iint_{\\Phi^{-1}(D)}f(r\\cos\\theta,r\\sin\\theta)r\\text{d}r\\text{d}\\theta\n∬Df(x,y)dxdy=∬Φ−1(D)f(rcosθ,rsinθ)rdrdθ\n这是因为 det⁡JΦ=r\\det J_\\Phi=rdetJΦ=r.\n/Remark/\n在某些古老的积分教材中，可能会画一个示意图，一个曲边四边形的面积会约等于 rΔθ⋅Δrr\\Delta\\theta\\cdot\\Delta rrΔθ⋅Δr.\n实际上我们的推导在几何意义上也是这样，解读为 Φ\\PhiΦ 把标准的矩体变为曲边体，近似为平行体，之后找到其 Vol\\mathcal{Vol}Vol：\nVol≈∣det⁡JΦ∣(∏i=1nΔui)\\mathcal{Vol}\\approx|\\det J_\\Phi|\\left(\\prod_{i=1}^n\\Delta u_i\\right)\nVol≈∣detJΦ∣(i=1∏nΔui)\n/Example/ (接着上面的例子，Gauss 积分)\n求：\n∬x2+y2≤R2e−x2−y2dxdy\\iint_{x^2+y^2\\leq R^2}e^{-x^2-y^2}\\text{d}x\\text{d}y\n∬x2+y2≤R2e−x2−y2dxdy"])self.__next_f.push([1,"19:Tadc,"])self.__next_f.push([1,"订阅更新(RSS/Atom): 换元公式的应用\n设 Φ:D′→D\\Phi:D'\\to DΦ:D′→D 是 C1C^1C1 的双射，则对于 DDD 上可积的 fff 有：\n∫Df=∫D′(f∘Φ)∣det⁡JΦ∣\\int_Df=\\int_{D'}(f\\circ\\Phi)|\\det J_\\Phi|\n∫Df=∫D′(f∘Φ)∣detJΦ∣\n也就是权重拉回 (f∘Φf\\circ\\Phif∘Φ) 乘上 Φ\\PhiΦ 引起体积微元的放大倍数.\n/Example/ (极坐标计算二重积分)\nΦ:[0,+∞)×[0,2π]→R2\\Phi:[0,+\\infty)\\times[0,2\\pi]\\to\\R^2Φ:[0,+∞)×[0,2π]→R2，Φ=Φ(rcos⁡θ,rsin⁡θ)\\Phi=\\Phi(r\\cos\\theta,r\\sin\\theta)Φ=Φ(rcosθ,rsinθ).\n在这里，Φ\\PhiΦ 虽然是 C1C^1C1 的，但是 r=0r=0r=0 这一条线段被映射到一个点 (坐标原点)、θ=2π,θ=0\\theta=2\\pi,\\theta=0θ=2π,θ=0 两条射线被映射为一条射线 (xxx 轴正方向)，所以 Φ\\PhiΦ 并非单射.\n但是我们仍然可以处理这个问题：对于 r=0r=0r=0 映射到一个点的情况，\n∣∫Df−∫D∩(R2∣Bε(0⃗))f∣=∣∫D∩Bε(0⃗)f∣≤Mπ2ε→0\\left|\\int_Df-\\int_{D\\cap(\\R^2|B_{\\varepsilon}(\\vec{0}))}f\\right|=\\left|\\int_{D\\cap B_\\varepsilon(\\vec{0})}f\\right|\\leq M\\pi^2\\varepsilon\\to0\n\n∫Df−∫D∩(R2∣Bε(0\n))f\n=\n∫D∩Bε(0\n)f\n≤Mπ2ε→0\n可见\n∫Df=lim⁡ε→0+∫D∩(R2∣Bε(0⃗))f\\int_Df=\\lim_{\\varepsilon\\to0+}\\int_{D\\cap(\\R^2|B_\\varepsilon(\\vec{0}))}f\n∫Df=ε→0+lim∫D∩(R2∣Bε(0\n))f\n只需要对 D∣Bε(0⃗)D|B_\\varepsilon(\\vec{0})D∣Bε(0\n) (记为 D~\\tilde DD~) 进行换元. 对于 θ=0,θ=2π\\theta=0,\\theta=2\\piθ=0,θ=2π 重合的情况，\n∫D~f=∫D~∩R+2f+∫D~∩R−2f\\int_{\\tilde D}f=\\int_{\\tilde D\\cap\\R^2_+}f+\\int_{\\tilde D\\cap\\R^2_-}f\n∫D~f=∫D~∩R+2f+∫D~∩R−2f\n也就是将 R2\\R^2R2 从 +x+x+x 无穷远处剪开剪到原点处，分为上半和下半两个平面来换元. 这样我们得到，\n∬Df(x,y)dxdy=∬Φ−1(D)f(rcos⁡θ,rsin⁡θ)rdrdθ\\iint_Df(x,y)\\text{d}x\\text{d}y=\\iint_{\\Phi^{-1}(D)}f(r\\cos\\theta,r\\sin\\theta)r\\text{d}r\\text{d}\\theta\n∬Df(x,y)dxdy=∬Φ−1(D)f(rcosθ,rsinθ)rdrdθ\n这是因为 det⁡JΦ=r\\det J_\\Phi=rdetJΦ=r.\n/Remark/\n在某些古老的积分教材中，可能会画一个示意图，一个曲边四边形的面积会约等于 rΔθ⋅Δrr\\Delta\\theta\\cdot\\Delta rrΔθ⋅Δr.\n实际上我们的推导在几何意义上也是这样，解读为 Φ\\PhiΦ 把标准的矩体变为曲边体，近似为平行体，之后找到其 Vol\\mathcal{Vol}Vol：\nVol≈∣det⁡JΦ∣(∏i=1nΔui)\\mathcal{Vol}\\approx|\\det J_\\Phi|\\left(\\prod_{i=1}^n\\Delta u_i\\right)\nVol≈∣detJΦ∣(i=1∏nΔui)\n/Example/ (接着上面的例子，Gauss 积分)\n求：\n∬x2+y2≤R2e−x2−y2dxdy\\iint_{x^2+y^2\\leq R^2}e^{-x^2-y^2}\\text{d}x\\text{d}y\n∬x2+y2≤R2e−x2−y2dxdy"])self.__next_f.push([1,"9:{\"metadata\":[[\"$\",\"title\",\"0\",{\"children\":\"Lesson 54 换元公式的应用 | 菲兹克斯喵 | 柑橘 RSS 阅读器\"}],[\"$\",\"meta\",\"1\",{\"name\":\"description\",\"content\":\"$18\"}],[\"$\",\"meta\",\"2\",{\"name\":\"keywords\",\"content\":\":blog\"}],[\"$\",\"meta\",\"3\",{\"name\":\"robots\",\"content\":\"index, follow\"}],[\"$\",\"meta\",\"4\",{\"property\":\"og:title\",\"content\":\"Lesson 54 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(坐标原点)、θ=2π,θ=0\\theta=2\\pi,\\theta=0θ=2π,θ=0 两条射线被映射为一条射线 (xxx 轴正方向)，所以 Φ\\PhiΦ 并非单射.\n但是我们仍然可以处理这个问题：对于 r=0r=0r=0 映射到一个点的情况，\n∣∫Df−∫D∩(R2∣Bε(0⃗))f∣=∣∫D∩Bε(0⃗)f∣≤Mπ2ε→0\\left|\\int_Df-\\int_{D\\cap(\\R^2|B_{\\varepsilon}(\\vec{0}))}f\\right|=\\left|\\int_{D\\cap B_\\varepsilon(\\vec{0})}f\\right|\\leq M\\pi^2\\varepsilon\\to0\n\n∫Df−∫D∩(R2∣Bε(0\n))f\n=\n∫D∩Bε(0\n)f\n≤Mπ2ε→0\n可见\n∫Df=lim⁡ε→0+∫D∩(R2∣Bε(0⃗))f\\int_Df=\\lim_{\\varepsilon\\to0+}\\int_{D\\cap(\\R^2|B_\\varepsilon(\\vec{0}))}f\n∫Df=ε→0+lim∫D∩(R2∣Bε(0\n))f\n只需要对 D∣Bε(0⃗)D|B_\\varepsilon(\\vec{0})D∣Bε(0\n) (记为 D~\\tilde DD~) 进行换元. 对于 θ=0,θ=2π\\theta=0,\\theta=2\\piθ=0,θ=2π 重合的情况，\n∫D~f=∫D~∩R+2f+∫D~∩R−2f\\int_{\\tilde D}f=\\int_{\\tilde D\\cap\\R^2_+}f+\\int_{\\tilde D\\cap\\R^2_-}f\n∫D~f=∫D~∩R+2f+∫D~∩R−2f\n也就是将 R2\\R^2R2 从 +x+x+x 无穷远处剪开剪到原点处，分为上半和下半两个平面来换元. 这样我们得到，\n∬Df(x,y)dxdy=∬Φ−1(D)f(rcos⁡θ,rsin⁡θ)rdrdθ\\iint_Df(x,y)\\text{d}x\\text{d}y=\\iint_{\\Phi^{-1}(D)}f(r\\cos\\theta,r\\sin\\theta)r\\text{d}r\\text{d}\\theta\n∬Df(x,y)dxdy=∬Φ−1(D)f(rcosθ,rsinθ)rdrdθ\n这是因为 det⁡JΦ=r\\det J_\\Phi=rdetJΦ=r.\n/Remark/\n在某些古老的积分教材中，可能会画一个示意图，一个曲边四边形的面积会约等于 rΔθ⋅Δrr\\Delta\\theta\\cdot\\Delta rrΔθ⋅Δr.\n实际上我们的推导在几何意义上也是这样，解读为 Φ\\PhiΦ 把标准的矩体变为曲边体，近似为平行体，之后找到其 Vol\\mathcal{Vol}Vol：\nVol≈∣det⁡JΦ∣(∏i=1nΔui)\\mathcal{Vol}\\approx|\\det J_\\Phi|\\left(\\prod_{i=1}^n\\Delta u_i\\right)\nVol≈∣detJΦ∣(i=1∏nΔui)\n/Example/ (接着上面的例子，Gauss 积分)\n求：\n∬x2+y2≤R2e−x2−y2dxdy\\iint_{x^2+y^2\\leq R^2}e^{-x^2-y^2}\\text{d}x\\text{d}y\n∬x2+y2≤R2e−x2−y2dxdy"])self.__next_f.push([1,"1f:[\"$\",\"meta\",\"11\",{\"name\":\"twitter:description\",\"content\":\"$22\"}]\n20:[\"$\",\"link\",\"12\",{\"rel\":\"icon\",\"href\":\"/favicon.ico?favicon.0b3bf435.ico\",\"sizes\":\"256x256\",\"type\":\"image/x-icon\"}]\n21:[\"$\",\"$L23\",\"13\",{}]\n"])$