2011年考研数二第21题解析:二重积分、分部积分

题目

已知函数 $f(x,y)$ 具有二阶连续偏导数,且 $f(1,y) = 0$, $f(x,1)=0$, $\iint_{D} f(x,y) dx dy = a$, 其中,$D={(x,y) | 0 \leqslant x \leqslant 1, 0 \leqslant y \leqslant 1}$, 计算二重积分 $I = \iint_{D} xy f_{xy}^{”}(x,y) dxdy$.

解析

首先,由:

$$
\iint_{D} f(x,y) dxdy = a
$$

可得:

$$
\int_{0}^{1} dx \int_{0}^{1}f(x,y) dy = a.
$$

接着,我们对 $I = \iint_{D} xy f_{xy}^{”}(x,y) dxdy$ 进行如下转化:

$$
I = \iint_{D} xy f_{xy}^{”}(x,y) dxdy =
$$

$$
\int_{0}^{1} x dx \int_{0}^{1} y f_{xy}^{”}(x,y) dy =
$$

$$
\int_{0}^{1} x dx \int_{0}^{1} y \cdot d[f_{x}^{‘}(x,y)].
$$

注:分部积分可以用来减少导数的“层数”,例如上面这个过程就是利用分部积分将二阶偏导 $f_{xy}^{”}$ 化为了一阶偏导 $f_{x}^{‘}$.

又:

$$
\int_{0}^{1} y \cdot d[f_{x}^{‘}(x,y)] = ①
$$

$$
y f_{x}^{‘}(x,y) |_{0}^{1} – \int_{0}^{1} f_{x}^{‘}(x,y) dy = ②
$$

$$
1 \cdot f_{x}^{‘} (x,1) – \int_{0}^{1} f_{x}^{‘} (x,y) dy = ③
$$

$$
f_{x}^{‘} (x,1) – \int_{0}^{1} f_{x}^{‘} (x,y) dy.
$$

注:由于 $①$ 式是在对变量 $y$ 进行积分,因此在将 $②$ 式中的数字代入,计算出 ③ 式的过程中,我们代入数字的是变量 $y$ 而不是变量 $x$.

于是:

$$
\int_{0}^{1} x dx \int_{0}^{1} y \cdot d[f_{x}^{‘}(x,y)] =
$$

$$
\int_{0}^{1} x dx [f_{x}^{‘}(x,1) – \int_{0}^{1} f_{x}^{‘} (x,y) dy] =
$$

$$
\int_{0}^{1} x f_{x}^{‘} (x,1) dx – \int_{0}^{1} x dx \int_{0}^{1} f_{x}^{‘}(x,y) dy.
$$

又:

$$
\int_{0}^{1} x f_{x}^{‘}(x,1) dx =
$$

$$
x f(x,1)|_{0}^{1} – \int_{0}^{1} f(x,1) dx =
$$

$$
x \cdot 0|_{0}^{1} – \int_{0}^{1} f(x,1) dx =
$$

$$
(-1) \cdot \int_{0}^{1} f(x,1) dx = 0.
$$

于是:

$$
\int_{0}^{1} x f_{x}^{‘} (x,1) dx – \int_{0}^{1} x dx \int_{0}^{1} f_{x}^{‘}(x,y) dy =
$$

$$
(-1) \cdot \int_{0}^{1} x dx \int_{0}^{1} f_{x}^{‘} (x,y) dy =
$$

$$
(-1) \cdot \int_{0}^{1} dy \int_{0}^{1} x f_{x}^{‘} (x,y) dx.
$$

又:

$$
\int_{0}^{1} x f_{x}^{‘} (x,y) dx =
$$

$$
x f(x,y)|_{0}^{1} – \int_{0}^{1} f(x,y) dx =
$$

$$
1 \cdot f(1,y) – 0 – \int_{0}^{1} f(x,y) dx =
$$

$$
(-1) \cdot \int_{0}^{1} f(x,y) dx.
$$

于是:

$$
(-1) \cdot \int_{0}^{1} dy \int_{0}^{1} x f_{x}^{‘} (x,y) dx =
$$

$$
(-1) \cdot (-1) \cdot \int_{0}^{1} dy \int_{0}^{1} f(x,y) dx =
$$

$$
\int_{0}^{1} dy \int_{0}^{1} f(x,y) dx =
$$

$$
\int_{0}^{1} dx \int_{0}^{1} f(x,y) dy = a.
$$

即:

$$
I = \iint_{D} xy f_{xy}^{”}(x,y) dxdy = a.
$$