# 2012年考研数二第06题解析

## 题目

$$A. \pi$$

$$B. 2$$

$$C. -2$$

$$D. -\pi$$

## 解析

### 方法一

$$\iint_{D} (xy^{5} – 1) dxdy =$$

$$\iint_{D} (xy^{5}) dxdy – \iint_{D} 1 dxdy \Rightarrow$$

$$\iint 1 dxdy = 1 \times (\frac{\pi}{2} + \frac{\pi}{2}) = \pi.$$

P.S: $\iint 1 dxdy$ 中的 $1$ 表示的是 $z=1$, 并且，在三维直角坐标系中，$z=1$ 位于 $z$ 轴的正半轴，因此，$\iint 1 dxdy$ 所表示的曲顶柱体的有向体积一定是个正数。

$$\iint_{D} (xy^{5}) dxdy – \pi =$$

$$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x dx \int_{\sin x}^{1} y^{5} dy – \pi =$$

$$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} [\frac{1}{6} x – \frac{1}{6} x (\sin x)^{6}] dx – \pi =$$

$$\frac{1}{6} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} xdx – \frac{1}{6} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x (\sin x)^{6} dx – \pi \Rightarrow$$

$$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} xdx = 0;$$

$$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x (\sin x)^{6} dx = 0.$$

$$原式 = 0 – 0 – \pi = – \pi.$$

### 方法二

$$\iint_{D} (xy^{5} – 1) dxdy =$$

$$\iint_{D_{1}} (xy^{5} – 1) dxdy + \iint_{D_{2}} (xy^{5} – 1) dxdy =$$

$$\iint_{D_{1}} (xy^{5}) dxdy – \iint_{D_{1}} 1 dxdy +$$

$$\iint_{D_{2}} (xy^{5}) dxdy – \iint_{D_{2}} 1 dxdy =$$

$$\iint_{D_{1}} (xy^{5}) dxdy = 0;$$

$$\iint_{D_{2}} (xy^{5}) dxdy = 0.$$

$$0 – \iint_{D_{1}} 1 dxdy + 0 – \iint_{D_{2}} 1 dxdy =$$

$$-\iint_{D} 1 dxdy =$$

$$-\int_{\frac{\pi}{2}}^{-\frac{\pi}{2}} 1 dx \int_{\sin x}^{1} 1 dy =$$

$$-\int_{\frac{\pi}{2}}^{-\frac{\pi}{2}} (1 – \sin x) dx =$$

$$-(x + \cos x) |_{-\frac{\pi}{2}}^{\frac{\pi}{2}} =$$

$$-(\frac{\pi}{2} + 0 + \frac{\pi}{2} – 0) =$$

$$-\pi.$$

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