两道二重积分

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Ⅰ 设闭区域$D$是由直线$x+y=1,x=0,y=0$所围成,求证\[\iint_{D} \cos \left(\frac{x-y}{x+y}\right) \mathrm{d} x \mathrm{~d} y=\frac{1}{2} \sin 1\]

证:换元$\cases{u=x−y\\v=x+y}$, $\frac{\partial(x,y)}{\partial(u,v)}=\frac1{\begin{vmatrix}1&1\\-1&1\end{vmatrix}}=\frac12$
所以$\int_0^1 \int_0^{1-y} \cos\Big( \frac{x-y}{x+y} \Big)\mathrm dx\mathrm dy=\frac12\int_0^1\mathrm dv \int_{-v}^{v} \cos\frac{u}{v}\mathrm du=\int_0^1v\sin1\mathrm{d}v=\frac12\sin1$.

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Ⅱ 设$D=\{(x,y)|x^2+y^2\le1\}$,且$a^2+b^2\ne0$.\[\iint_{D} f(a x+b y+c) \mathrm{d} x \mathrm{~d} y=2 \int_{-1}^{1} \sqrt{1-u^{2}} f\left(u \sqrt{a^{2}+b^{2}}+c\right) \mathrm{d} u\]

证:换元$\begin{cases}u=\frac{ax+by}{\sqrt{a^2+b^2}}\\v=\frac{bx-ay}{\sqrt{a^2+b^2}}\end{cases}$, $u^2+v^2=1$, $\frac{\partial(x,y)}{\partial(u,v)}=1$.
所以$\iint_Df(ax+by+c)\mathrm dx\mathrm{~d}y=\int_{-1}^1\int_{-\sqrt{1-u^2}}^{\sqrt{1-u^2}}f(\sqrt{a^2+b^2}u+c)\mathrm dv\mathrm du=\int_{-1}^12\sqrt{1-u^2}f(\sqrt{a^2+b^2}u+c)\mathrm du$