题目
(int )_(dfrac {pi )(4)}^dfrac (pi {2)}dtheta (int )_(0)^cot theta (r)^2cos theta f(r)dr= ()A (int )_(dfrac {pi )(4)}^dfrac (pi {2)}dtheta (int )_(0)^cot theta (r)^2cos theta f(r)dr= ()B (int )_(dfrac {pi )(4)}^dfrac (pi {2)}dtheta (int )_(0)^cot theta (r)^2cos theta f(r)dr= ()C (int )_(dfrac {pi )(4)}^dfrac (pi {2)}dtheta (int )_(0)^cot theta (r)^2cos theta f(r)dr= ()D (int )_(dfrac {pi )(4)}^dfrac (pi {2)}dtheta (int )_(0)^cot theta (r)^2cos theta f(r)dr= ()
A 
B 
C 
D 
题目解答
答案
根据题目条件，所做的极坐标变换为：
根据极坐标系，知：
且由于
则：
故：
所以正确答案为D。
解析
步骤 1：极坐标变换
根据题目条件，所做的极坐标变换为：
$x=r\cos \theta $, $y=r\sin \theta $ $\theta \in [ \dfrac {\pi }{4},\dfrac {\pi }{2}] $ $r\in [ 0,\cot \theta ] $,
步骤 2：极坐标系下的变量关系
根据极坐标系，知：
$r=\sqrt {{x}^{2}+{y}^{2}}$
步骤 3：确定积分区域
由于$\theta \in [ \dfrac {\pi }{4},\dfrac {\pi }{2}] $,则：
$y\in [ x,1] $,
步骤 4：转换积分
${\int }_{\dfrac {\pi }{4}}^{\dfrac {\pi }{2}}d\theta {\int }_{0}^{\cot \theta }{r}^{2}\cos \theta f(r)dr$
$={\int }_{0}^{1}xdx{\int }_{x}^{1}f(\sqrt {{x}^{2}+{y}^{2}})dy$
根据题目条件，所做的极坐标变换为：
$x=r\cos \theta $, $y=r\sin \theta $ $\theta \in [ \dfrac {\pi }{4},\dfrac {\pi }{2}] $ $r\in [ 0,\cot \theta ] $,
步骤 2：极坐标系下的变量关系
根据极坐标系，知：
$r=\sqrt {{x}^{2}+{y}^{2}}$
步骤 3：确定积分区域
由于$\theta \in [ \dfrac {\pi }{4},\dfrac {\pi }{2}] $,则：
$y\in [ x,1] $,
步骤 4：转换积分
${\int }_{\dfrac {\pi }{4}}^{\dfrac {\pi }{2}}d\theta {\int }_{0}^{\cot \theta }{r}^{2}\cos \theta f(r)dr$
$={\int }_{0}^{1}xdx{\int }_{x}^{1}f(\sqrt {{x}^{2}+{y}^{2}})dy$