特定正实数区域二重积分计算的结果疑问及推导修正咨询
Hey there! Let's break down this double integral problem step by step to figure out why your result differs from the book's answer.
First, let's restate the problem clearly:
We need to compute the double integral
$$\iint\limits_{G} \sqrt{x^{2} - y^{2}} , dx, dy,$$
where the region $G$ is defined as
$$G={(x,y) \in \mathbb{R_{+}^{2}}: 0 \leq x + y \leq 2, 0 \leq x - y \leq 2}$$
(Note: $\mathbb{R_{+}^{2}}$ means $x \geq 0$ and $y \geq 0$—this is the key detail you missed in your initial substitution!)
Your Initial Derivation Steps
Here's your original work for reference:
$$
\iint\limits_{G} \sqrt{x^{2} - y^{2}} , dx, dy \
= \iint\limits_{G} \sqrt{(x-y)(x+y)} , dx, dy \
= \iint\limits_{G} \frac{1}{2}\sqrt{(x-y)(x+y)} , 2 , dx, dy, \
= \iint\limits_{\widetilde{G}} \frac{1}{2}\sqrt{vu} , du, dv, \
= \frac{1}{2} \iint\limits_{\widetilde{G}} \sqrt{vu} , du, dv, \
= \frac{1}{2} \int_{0}^{2} \sqrt{v} \left( \int_{0}^{2} \sqrt{u} , du \right) ,dv \
= \frac{1}{2} \int_{0}^{2} \sqrt{v} \left(\left[\frac{2}{3}u{\frac{3}{2}}\right]_{0}{2} \right) ,dv \
= \frac{1}{3}2^{\frac{3}{2}} \int_{0}^{2} \sqrt{v} ,dv \
= \frac{1}{3}2^{\frac{3}{2}} \left[\frac{2}{3}v{\frac{3}{2}}\right]_{0}{2} \
= \frac{2}{9}2^3 \
= \frac{16}{9}
$$
Where the Mistake Happened
Your substitution $u = x + y$ and $v = x - y$ is correct, and you properly calculated the Jacobian determinant (which gives the factor of $\frac{1}{2}$, since $\frac{\partial(x,y)}{\partial(u,v)} = \frac{1}{2}$).
The error comes from misdefining the transformed region $\widetilde{G}$. You assumed it's the full rectangle $0 \leq u \leq 2$, $0 \leq v \leq 2$, but we have to account for the $\mathbb{R_{+}^{2}}$ condition ($x \geq 0$, $y \geq 0$):
- Since $x = \frac{u + v}{2} \geq 0$, and $u,v \geq 0$ (from $0 \leq x+y \leq 2$ and $0 \leq x-y \leq 2$), this condition is automatically satisfied.
- The critical condition is $y = \frac{u - v}{2} \geq 0$, which rearranges to $u \geq v$.
So the actual transformed region $\widetilde{G}$ is:
$$\widetilde{G} = {(u,v) : 0 \leq u \leq 2, 0 \leq v \leq u}$$
Corrected Calculation
Let's redo the integral with the correct region limits:
$$
\frac{1}{2} \iint\limits_{\widetilde{G}} \sqrt{vu} , du, dv = \frac{1}{2} \int_{0}^{2} \sqrt{u} \left( \int_{0}^{u} \sqrt{v} , dv \right) ,du
$$
First compute the inner integral with respect to $v$:
$$
\int_{0}^{u} \sqrt{v} , dv = \left[\frac{2}{3}v{\frac{3}{2}}\right]_{0}{u} = \frac{2}{3}u^{\frac{3}{2}}
$$
Now substitute back into the outer integral:
$$
\frac{1}{2} \int_{0}^{2} \sqrt{u} \cdot \frac{2}{3}u^{\frac{3}{2}} ,du = \frac{1}{3} \int_{0}^{2} u^{\frac{1}{2} + \frac{3}{2}} ,du = \frac{1}{3} \int_{0}^{2} u^2 ,du
$$
Calculate the remaining integral:
$$
\frac{1}{3} \left[\frac{1}{3}u3\right]_{0}{2} = \frac{1}{9} \cdot 2^3 = \frac{8}{9}
$$
This matches the book's answer!
Why This Makes Sense
Visually, the original region $G$ is a polygon in the first quadrant bounded by:
- $x + y = 0$ (but since $x,y \geq 0$, this is just the origin)
- $x + y = 2$
- $x - y = 0$ (the line $y = x$)
- $x - y = 2$
When you transform to $u,v$ coordinates, the line $y = x$ becomes $v = 0$, and $y \geq 0$ restricts $v$ to be at most $u$, which cuts the original rectangle in half—hence the factor of 2 difference in your result.
备注：内容来源于stack exchange，提问作者Guyon Van Rooij
