曲线 $C$ 的极坐标方程为 $r^2=\theta\sec^2\theta$，其中 $0\leqslant\theta\leqslant\frac{1}{3}\pi$。

(a) 画出 $C$ 的草图，并写出 $C$ 上的点到极点的最大距离。

(b) 求 $C$ 与射线 $\theta=\frac{1}{3}\pi$ 所围成区域的面积的精确值。

(c) 证明 $C$ 的直角坐标方程为 $y=x\tan(x^2)$，其中 $0\leqslant x\leqslant\frac{1}{3}\sqrt{3\pi}$。

(d) 使用 5(b) 的答案，推得 $\displaystyle\int_0^{\frac{1}{3}\sqrt{3\pi}}x\tan(x^2)\,\mathrm{d}x$ 的精确值。

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。

在给定区间内r随θ递增，最大距离在θ=π/3。草图从极点水平起步，向第一象限上方弯曲。

For $0\le\theta\le\pi/3$, $r=\sqrt\theta\sec\theta$ is increasing, since both non-negative factors increase. Its greatest value is

Sketch a curve in the first quadrant starting at the pole with a horizontal tangent, bending upwards to its endpoint on $\theta=\pi/3$. The endpoint has Cartesian coordinates $(\sqrt{\pi/3},\sqrt\pi)$, and its tangent is not vertical. Mark the initial line.

用极坐标面积公式，积分θ sec平方θ时分部积分。注意原积分外面的二分之一。

The required area is

Integrating by parts with $u=\theta$ and $\mathrm{d}v=\sec^2\theta\,\mathrm{d}\theta$,

Therefore $\boxed{A=\frac{\pi\sqrt3}{6}-\frac12\ln2}$.

由r平方乘cos平方θ得到x平方等于θ，再把tanθ=y/x换成直角坐标关系，同时处理极点与区间。

Since $x=r\cos\theta$,

For $x>0$, $y/x=\tan\theta=\tan(x^2)$, so $y=x\tan(x^2)$. This also holds at the pole $(0,0)$ by continuity.

As $0\le\theta\le\pi/3$ and $x\ge0$,

with

as required.

必须承接(b)的面积：所求曲线下方面积是端点与坐标轴形成的三角形面积，扣除已求的极坐标区域面积。

By (c), the integral is the area under $C$ from $x=0$ to $x=\sqrt{\pi/3}$. The endpoint is $(\sqrt{\pi/3},\sqrt\pi)$, so the triangle under the half-line $\theta=\pi/3$ has area

The triangle is divided into the region from (b) and the area under the Cartesian curve. Hence, using (b),