IAL 2025 June Q7 | 國際數學站

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 $C_1$ and $C_2$

The circle $C_1$ has polar equation

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

The circle $C_2$ has polar equation

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

Circles $C_1$ and $C_2$ 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 $C_1$ and $C_2$ and is shown shaded in Figure 1.

(b) Use algebraic integration to determine the exact area of $R$ , giving your answer in the form $a\pi + b\sqrt{3}$ where $a$ and $b$ are simplified rational numbers.

(6)

解答

(a)

To find the intersection $P$ , equate the two polar equations:

\begin{align*} \sqrt{3}\sin\theta = &\,3\cos\theta\\[4mm] \frac{\sin\theta}{\cos\theta} = &\,\frac{3}{\sqrt{3}}\\[4mm] \tan\theta = &\,\sqrt{3} \end{align*}

Since $0 \le \theta \le \pi/2$ at the intersection $P$ , we have:

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

Substitute $\theta$ back to find $r$ :

\begin{align*} r = &\,\sqrt{3}\sin\left(\frac{\pi}{3}\right)\\[4mm] = &\,\sqrt{3}\left(\frac{\sqrt{3}}{2}\right)\\[4mm] = &\,\frac{3}{2} \end{align*}

The polar coordinates of $P$ are $\left(\frac{3}{2}, \frac{\pi}{3}\right)$ .

(b)

The area $R$ is split into two parts by the line $OP$ ( $\theta = \pi/3$ ). We integrate $C_1$ from $0$ to $\pi/3$ , and $C_2$ from $\pi/3$ to $\pi/2$ :

\begin{align*} \text{Area} = \frac{1}{2} \int_{0}^{\frac{\pi}{3}} r_1^2 \,\mathrm{d}\theta + \frac{1}{2} \int_{\frac{\pi}{3}}^{\frac{\pi}{2}} r_2^2 \,\mathrm{d}\theta \end{align*}

For the first integral:

\begin{align*} \frac{1}{2} \int_{0}^{\frac{\pi}{3}} (\sqrt{3}\sin\theta)^2 \,\mathrm{d}\theta = &\,\frac{1}{2} \int_{0}^{\frac{\pi}{3}} 3\sin^2\theta \,\mathrm{d}\theta\\[4mm] = &\,\frac{3}{2} \int_{0}^{\frac{\pi}{3}} \left(\frac{1}{2} - \frac{1}{2}\cos 2\theta\right) \,\mathrm{d}\theta\\[4mm] = &\,\frac{3}{4} \left[ \theta - \frac{1}{2}\sin 2\theta \right]_{0}^{\frac{\pi}{3}}\\[4mm] = &\,\frac{3}{4} \left( \left(\frac{\pi}{3} - \frac{1}{2}\sin\frac{2\pi}{3}\right) - 0 \right)\\[4mm] = &\,\frac{3}{4} \left( \frac{\pi}{3} - \frac{\sqrt{3}}{4} \right)\\[4mm] = &\,\frac{\pi}{4} - \frac{3\sqrt{3}}{16} \end{align*}

For the second integral:

\begin{align*} \frac{1}{2} \int_{\frac{\pi}{3}}^{\frac{\pi}{2}} (3\cos\theta)^2 \,\mathrm{d}\theta = &\,\frac{1}{2} \int_{\frac{\pi}{3}}^{\frac{\pi}{2}} 9\cos^2\theta \,\mathrm{d}\theta\\[4mm] = &\,\frac{9}{2} \int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \left(\frac{1}{2} + \frac{1}{2}\cos 2\theta\right) \,\mathrm{d}\theta\\[4mm] = &\,\frac{9}{4} \left[ \theta + \frac{1}{2}\sin 2\theta \right]_{\frac{\pi}{3}}^{\frac{\pi}{2}}\\[4mm] = &\,\frac{9}{4} \left( \left(\frac{\pi}{2} + 0\right) - \left(\frac{\pi}{3} + \frac{1}{2}\sin\frac{2\pi}{3}\right) \right)\\[4mm] = &\,\frac{9}{4} \left( \frac{\pi}{2} - \frac{\pi}{3} - \frac{\sqrt{3}}{4} \right)\\[4mm] = &\,\frac{9}{4} \left( \frac{\pi}{6} - \frac{\sqrt{3}}{4} \right)\\[4mm] = &\,\frac{3\pi}{8} - \frac{9\sqrt{3}}{16} \end{align*}

Adding the two areas together:

\begin{align*} \text{Total Area} = &\,\left(\frac{\pi}{4} - \frac{3\sqrt{3}}{16}\right) + \left(\frac{3\pi}{8} - \frac{9\sqrt{3}}{16}\right)\\[4mm] = &\,\frac{5\pi}{8} - \frac{12\sqrt{3}}{16}\\[4mm] = &\,\frac{5\pi}{8} - \frac{3\sqrt{3}}{4} \end{align*}