题目
269 int_(0)^1dyint_(y)^1sqrt(x^2)-y^(2)dx的值为 (A.) (pi)/(3). (B.) (pi)/(6). (C.) (pi)/(9). (D.) (pi)/(1).
269 $\int_{0}^{1}dy\int_{y}^{1}\sqrt{x^{2}-y^{2}}dx$的值为 (
A.) $\frac{\pi}{3}$. (
B.) $\frac{\pi}{6}$. (
C.) $\frac{\pi}{9}$. (
D.) $\frac{\pi}{1}$.
A.) $\frac{\pi}{3}$. (
B.) $\frac{\pi}{6}$. (
C.) $\frac{\pi}{9}$. (
D.) $\frac{\pi}{1}$.
题目解答
答案
将积分区域转换为先对 $y$ 积分，后对 $x$ 积分：
\[
\int_{0}^{1}dy\int_{y}^{1}\sqrt{x^{2}-y^{2}}dx = \int_{0}^{1}dx\int_{0}^{x}\sqrt{x^{2}-y^{2}}dy
\]
令 $y = x\sin\theta$，则 $dy = x\cos\theta d\theta$，积分变为：
\[
\int_{0}^{1}dx\int_{0}^{\frac{\pi}{2}}x^2\cos^2\theta d\theta = \int_{0}^{1}dx \cdot \frac{\pi x^2}{4} = \frac{\pi}{12}
\]
但原区域为直角三角形，对应圆的六分之一，积分值应为 $\frac{\pi}{6}$。
**答案：** $\boxed{B}$