題目
Problem
(Q) Let S S S z = x 2 + y 2 2 z=x^2+\dfrac{y^2}{2} z = x 2 + 2 y 2 4 x 2 + y 2 ≤ 1 4x^2+y^2\le 1 4 x 2 + y 2 ≤ 1 k k k F ( x , y , z ) = x i − y j + k \mathbf{F}(x,y,z)=x\mathbf{i}-y\mathbf{j}+\mathbf{k} F ( x , y , z ) = x i − y j + k ∬ S F ⋅ d S \iint_S \mathbf{F}\cdot \mathrm{d}\mathbf{S} ∬ S F ⋅ d S ( 18 ) ‾ \underline{\quad(18)\quad} ( 18 ) ∬ S d S \iint_S \mathrm{d}S ∬ S d S ( 19 ) ‾ \underline{\quad(19)\quad} ( 19 )
解答
解法一
思路
展開
將曲面視為圖形，用向上面積向量求通量；橢圓投影再線性化為單位圓盤。
答題過程
展開
向上面積向量為 ( − 2 x , − y , 1 ) d A (-2x,-y,1)\,\mathrm{d}A ( − 2 x , − y , 1 ) d A 1 − 2 x 2 + y 2 1-2x^2+y^2 1 − 2 x 2 + y 2 u = 2 x , v = y u=2x,v=y u = 2 x , v = y
∬ S F ⋅ d S = 9 π 16 . \iint_S\mathbf F\cdot\mathrm{d}\mathbf S=\boxed{\frac{9\pi}{16}}. ∬ S F ⋅ d S = 16 9 π .
同一代換下，曲面積為
∬ S d S = 1 2 ∬ u 2 + v 2 ≤ 1 1 + u 2 d u d v = ∫ − 1 1 1 − u 4 d u = 1 2 B ( 1 4 , 3 2 ) . \begin{align*}
\iint_S\mathrm{d}S
=&\,\frac12\iint_{u^2+v^2\le1}\sqrt{1+u^2}\,\mathrm{d}u\mathrm{d}v\\[4mm]
=&\,\int_{-1}^1\sqrt{1-u^4}\,\mathrm{d}u\\[4mm]
=&\,\boxed{\frac12B\left(\frac14,\frac32\right)}.
\end{align*} ∬ S d S = = = 2 1 ∬ u 2 + v 2 ≤ 1 1 + u 2 d u d v ∫ − 1 1 1 − u 4 d u 2 1 B ( 4 1 , 2 3 ) . 其中 B B B π Γ ( 1 / 4 ) 3 Γ ( 3 / 4 ) \displaystyle\frac{\sqrt\pi\,\Gamma(1/4)}{3\Gamma(3/4)} 3Γ ( 3/4 ) π Γ ( 1/4 )
