心得

随想

$$\begin{aligned} \iint_R f(x,y)\,\mathrm{d} x\,\mathrm{d} y &= \iint_S f(aR\cos(\theta),bR\sin(\theta))\left|\frac{\partial(x,y)}{\partial(R,\theta)}\right|\,\mathrm{d} R\,\mathrm{d} \theta \\ &= \iint_S f(aR\cos(\theta),bR\sin(\theta))\left| \begin{vmatrix} \frac{\partial x}{\partial R} & \frac{\partial x}{\partial \theta} \\ \frac{\partial y}{\partial R} & \frac{\partial y}{\partial \theta} \end{vmatrix} \right|\,\mathrm{d} R\,\mathrm{d} \theta \\ &= \iint_S f(aR\cos(\theta),bR\sin(\theta))\left| \begin{vmatrix} a\cos(\theta) & -aR\sin(\theta) \\ b\sin(\theta) & bR\cos(\theta) \end{vmatrix} \right|\,\mathrm{d} R\,\mathrm{d} \theta \\ &= \iint_S f(aR\cos(\theta),bR\sin(\theta))\left| (a\cos(\theta))(bR\cos(\theta)) - (-aR\sin(\theta))(b\sin(\theta)) \right|\,\mathrm{d} R\,\mathrm{d} \theta \\ &= \iint_S f(aR\cos(\theta),bR\sin(\theta))\left| abR\cos^2(\theta) + abR\sin^2(\theta) \right|\,\mathrm{d} R\,\mathrm{d} \theta \\ &= \iint_S f(aR\cos(\theta),bR\sin(\theta))\left| abR(\cos^2(\theta) + \sin^2(\theta)) \right|\,\mathrm{d} R\,\mathrm{d} \theta \\ &= \iint_S f(aR\cos(\theta),bR\sin(\theta)) abR \, \mathrm{d} R\,\mathrm{d} \theta \quad (\text{假设 } a,b,R \ge 0) \end{aligned}$$

这里可以理解为坐标轴改变，即假设以($\theta,R$)为新坐标轴，原先的$x,y$就变成了$\partial x,\partial y$的新坐标，新的面积即为他们的叉乘（即二阶行列式）

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当函数$f(x)$满足$\forall a\leq x \leq b,f(x)=f(a+b-x)$(即积分区间内函数对称)时，有

$$\begin{aligned} \int_a^b xf(x)\,\mathrm{d} x &=\int_a^b(a+b-x)f(x)\,\mathrm{d} x\\ &=\frac{a+b}{2}\int_a^b f(x)\,\mathrm{d} x \\ \text{e.g.} \quad \int_{0}^{\pi} x\sin(x)\,\mathrm{d} x &=\frac{\pi}{2}\int_{0}^\pi \sin(x)\,\mathrm{d} x \end{aligned}$$