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IAL 2025 June Q7

A Level / Edexcel / FP2

IAL 2025 June Paper · Question 7

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 C1C_1 and C2C_2

The circle C1C_1 has polar equation

r=3sinθ0θπ\begin{align*} r=\sqrt3\sin\theta \qquad 0\leqslant\theta\leqslant\pi \end{align*}

The circle C2C_2 has polar equation

r=3cosθπ2θπ2\begin{align*} r=3\cos\theta \qquad -\frac{\pi}{2}\leqslant\theta\leqslant\frac{\pi}{2} \end{align*}

Circles C1C_1 and C2C_2 intersect at the origin OO and at the point PP.

(a) Determine the polar coordinates of PP.

(2)

The finite region RR is bounded by C1C_1 and C2C_2 and is shown shaded in Figure 1.

(b) Use algebraic integration to determine the exact area of RR, giving your answer in the form aπ+b3a\pi+b\sqrt3 where aa and bb are simplified rational numbers.

(6)

解答

(a)

解法一

思路

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交点 PP 同时在两条曲线上,所以令两个 rr 相等。非原点交点对应 r0r\neq0,因此可以解出 θ\theta,再代回求 rr

答题过程

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At PP,

3sinθ=3cosθ\begin{align*} \sqrt3\sin\theta ={}& 3\cos\theta \end{align*}

So

tanθ=3\begin{align*} \tan\theta ={}& \sqrt3 \end{align*}

In the shown region,

θ=π3\begin{align*} \theta=\frac{\pi}{3} \end{align*}

Then

r=3sinπ3=332=32\begin{align*} r ={}& \sqrt3\sin\frac{\pi}{3}\\[2mm] ={}& \sqrt3\cdot\frac{\sqrt3}{2}\\[2mm] ={}& \frac32 \end{align*}

Therefore

P=(32,π3)\begin{align*} \boxed{ P=\left(\frac32,\frac{\pi}{3}\right) } \end{align*}

(b)

解法一

思路

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区域 RR 分成两段:从 00π3\frac{\pi}{3}C1C_1,从 π3\frac{\pi}{3}π2\frac{\pi}{2}C2C_2。极坐标面积公式是 12r2dθ\frac12\int r^2\,\mathrm{d}\theta。积分时用

sin2θ=12(1cos2θ),cos2θ=12(1+cos2θ)\begin{align*} \sin^2\theta=\frac12(1-\cos2\theta), \qquad \cos^2\theta=\frac12(1+\cos2\theta) \end{align*}

答题过程

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The area is

Area=120π/3(3sinθ)2dθ+12π/3π/2(3cosθ)2dθ\begin{align*} \text{Area} ={}& \frac12\int_0^{\pi/3} (\sqrt3\sin\theta)^2\,\mathrm{d}\theta +\frac12\int_{\pi/3}^{\pi/2} (3\cos\theta)^2\,\mathrm{d}\theta \end{align*}

For the first part,

120π/33sin2θdθ=320π/31cos2θ2dθ=34[θ12sin2θ]0π/3=34(π312sin2π3)=34(π334)=π43316\begin{align*} \frac12\int_0^{\pi/3}3\sin^2\theta\,\mathrm{d}\theta ={}& \frac32\int_0^{\pi/3} \frac{1-\cos2\theta}{2} \,\mathrm{d}\theta\\[3mm] ={}& \frac34 \left[ \theta-\frac12\sin2\theta \right]_0^{\pi/3}\\[3mm] ={}& \frac34 \left( \frac{\pi}{3} -\frac12\sin\frac{2\pi}{3} \right)\\[3mm] ={}& \frac34 \left( \frac{\pi}{3} -\frac{\sqrt3}{4} \right)\\[3mm] ={}& \frac{\pi}{4}-\frac{3\sqrt3}{16} \end{align*}

For the second part,

12π/3π/29cos2θdθ=92π/3π/21+cos2θ2dθ=94[θ+12sin2θ]π/3π/2=94[π2(π3+12sin2π3)]=94(π634)=3π89316\begin{align*} \frac12\int_{\pi/3}^{\pi/2}9\cos^2\theta\,\mathrm{d}\theta ={}& \frac92\int_{\pi/3}^{\pi/2} \frac{1+\cos2\theta}{2} \,\mathrm{d}\theta\\[3mm] ={}& \frac94 \left[ \theta+\frac12\sin2\theta \right]_{\pi/3}^{\pi/2}\\[3mm] ={}& \frac94 \left[ \frac{\pi}{2} -\left( \frac{\pi}{3} +\frac12\sin\frac{2\pi}{3} \right) \right]\\[3mm] ={}& \frac94 \left( \frac{\pi}{6} -\frac{\sqrt3}{4} \right)\\[3mm] ={}& \frac{3\pi}{8}-\frac{9\sqrt3}{16} \end{align*}

Add the two areas:

Area=(π43316)+(3π89316)=5π8334\begin{align*} \text{Area} ={}& \left(\frac{\pi}{4}-\frac{3\sqrt3}{16}\right) +\left(\frac{3\pi}{8}-\frac{9\sqrt3}{16}\right)\\[3mm] ={}& \boxed{ \frac{5\pi}{8}-\frac{3\sqrt3}{4} } \end{align*}