题目
Problem
In this question you must show all stages of your working.
Solutions relying entirely on calculator technology are not acceptable.
Figure 1
Figure 1 shows a sketch of the circles C1 and C2
The circle C1 has polar equation
r=3sinθ0⩽θ⩽π
The circle C2 has polar equation
r=3cosθ−2π⩽θ⩽2π
Circles C1 and C2 intersect at the origin O and at the point P.
(a) Determine the polar coordinates of P.
(2)
The finite region R is bounded by C1 and C2 and is shown shaded in Figure 1.
(b) Use algebraic integration to determine the exact area of R, giving your answer in the form aπ+b3 where a and b are simplified rational numbers.
(6)
解答
(a)
解法一
思路
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交点 P 同时在两条曲线上,所以令两个 r 相等。非原点交点对应 r=0,因此可以解出 θ,再代回求 r。
答题过程
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At P,
3sinθ=3cosθ
So
tanθ=3
In the shown region,
θ=3π
Then
r===3sin3π3⋅2323
Therefore
P=(23,3π)
(b)
解法一
思路
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区域 R 分成两段:从 0 到 3π 用 C1,从 3π 到 2π 用 C2。极坐标面积公式是 21∫r2dθ。积分时用
sin2θ=21(1−cos2θ),cos2θ=21(1+cos2θ)
答题过程
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The area is
Area=21∫0π/3(3sinθ)2dθ+21∫π/3π/2(3cosθ)2dθ
For the first part,
21∫0π/33sin2θdθ=====23∫0π/321−cos2θdθ43[θ−21sin2θ]0π/343(3π−21sin32π)43(3π−43)4π−1633
For the second part,
21∫π/3π/29cos2θdθ=====29∫π/3π/221+cos2θdθ49[θ+21sin2θ]π/3π/249[2π−(3π+21sin32π)]49(6π−43)83π−1693
Add the two areas:
Area==(4π−1633)+(83π−1693)85π−433