題目
Problem
二、多選題
A region A A A is bounded by the hyperbolas(雙曲線)x y = 1 xy=1 x y = 1 and x y = 2 xy=2 x y = 2 , and the curves x y 2 = 3 xy^2=3 x y 2 = 3 and x y 2 = 4 xy^2=4 x y 2 = 4 . The area of A A A is denoted by a r e a o area_o a r e a o . Use the change of variables u = x y u=xy u = x y and v = x y 2 v=xy^2 v = x y 2 , the area of the image of the region A A A in the coordinate space determined by ( u , v ) (u,v) ( u , v ) under the transformation is called a r e a t area_t a r e a t . 請問下列哪些選項是正確的?
(A) a r e a o = ln 4 − ln 3 area_o=\ln 4-\ln 3 a r e a o = ln 4 − ln 3
(B) a r e a t = ∫ 1 2 ∫ 3 4 d u d v area_t=\int_1^2\int_3^4 \mathrm{d}u\mathrm{d}v a r e a t = ∫ 1 2 ∫ 3 4 d u d v
(C) a r e a o = ∫ 1 2 ∫ 3 4 1 v d u d v area_o=\int_1^2\int_3^4 \dfrac{1}{v}\,\mathrm{d}u\mathrm{d}v a r e a o = ∫ 1 2 ∫ 3 4 v 1 d u d v
(D) The Jacobian determinant ∂ ( x , y ) ∂ ( u , v ) \dfrac{\partial(x,y)}{\partial(u,v)} ∂ ( u , v ) ∂ ( x , y ) is v v v .
(E) ∂ ( u , v ) ∂ ( x , y ) = x y \dfrac{\partial(u,v)}{\partial(x,y)}=xy ∂ ( x , y ) ∂ ( u , v ) = x y 。
解答
解法一
思路
展開
題目指定 u = x y , v = x y 2 u=xy,v=xy^2 u = x y , v = x y 2 ,四條邊界會化成一個長方形。
算出 Jacobian 後,以反 Jacobian 把 d x d y \mathrm{d}x\,\mathrm{d}y d x d y 轉為 d u d v \mathrm{d}u\,\mathrm{d}v d u d v 。
答題過程
展開
對 u = x y , v = x y 2 u=xy,v=xy^2 u = x y , v = x y 2 ,
∂ ( u , v ) ∂ ( x , y ) = ∣ y x y 2 2 x y ∣ = x y 2 = v . \frac{\partial(u,v)}{\partial(x,y)}
=\begin{vmatrix}y&x\\y^2&2xy\end{vmatrix}
=xy^2=v. ∂ ( x , y ) ∂ ( u , v ) = y y 2 x 2 x y = x y 2 = v .
因此 ∣ ∂ ( x , y ) ∂ ( u , v ) ∣ = 1 / v \left|\frac{\partial(x,y)}{\partial(u,v)}\right|=1/v ∂ ( u , v ) ∂ ( x , y ) = 1/ v 。變換後區域為 1 ≤ u ≤ 2 , 3 ≤ v ≤ 4 1\le u\le2,3\le v\le4 1 ≤ u ≤ 2 , 3 ≤ v ≤ 4 ,所以 (B) 正確,且
a r e a o = ∫ 1 2 ∫ 3 4 1 v d v d u = ln 4 − ln 3. area_o=\int_1^2\int_3^4\frac1v\,\mathrm{d}v\,\mathrm{d}u=\ln4-\ln3. a r e a o = ∫ 1 2 ∫ 3 4 v 1 d v d u = ln 4 − ln 3.
正確選項為 (B)、(C) \boxed{\text{(B)、(C)}} (B) 、 (C) 。