在很多计算题中,当我们遇到两个式子相等时,一般的做法就是两者加起来再除以 2,一般相加后表达式会得到极大的简化
例如
∬D(x2a2+y2b2)dxdy=∬D(y2a2+x2b2)dxdy=12(1a2+1b2)∬D(x2+y2)dxdy \iint_{D}\left(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\right) \mathrm{d} x \mathrm{d} y =\iint_{D}\left(\frac{y^{2}}{a^{2}}+\frac{x^{2}}{b^{2}}\right) \mathrm{d} x \mathrm{d} y=\frac{1}{2}\left(\frac{1}{a^{2}}+\frac{1}{b^{2}}\right) \iint_{D}\left(x^{2}+y^{2}\right) \mathrm{d} x \mathrm{d} y ∬D(a2x2+b2y2)dxdy=∬D(a2y2+b2x2)dxdy=21(a21+b21)∬D(x2+y2)dxdy
=12(1a2+1b2)∫02πdθ∫0R(r2cos2θ+r2sin2θ)rdr=πR4(a2+b2)4a2b2. =\frac{1}{2}\left(\frac{1}{a^{2}}+\frac{1}{b^{2}}\right) \int_{0}^{2 \pi} \mathrm{d} \theta \int_{0}^{R}\left(r^{2} \cos ^{2} \theta+r^{2} \sin ^{2} \theta\right) r \mathrm{d} r=\frac{\pi R^{4}\left(a^{2}+b^{2}\right)}{4 a^{2} b^{2}} . =21(a21+b21)∫02πdθ∫0R(r2cos2θ+r2sin2θ)rdr=4a2b2πR4(a2+b2).
其中 区域 D={(x,y)∣x2+y2⩽R2}D=\left\{(x, y) | x^{2}+y^{2} \leqslant R^{2}\right\}D={(x,y)∣x2+y2⩽R2}