(a) 证明直角坐标方程为

的曲线，其极坐标方程为 $r=\frac12\sin2\theta$。

曲线 $C_1$ 和 $C_2$ 的极坐标方程分别为

其中 $0\leqslant\theta\leqslant\frac12\pi$。

(b) 证明 $C_1$ 与 $C_2$ 的交点满足

并通过计算验证此方程在 $\theta=0.5$ 与 $\theta=0.6$ 之间有一个根 $\alpha$。

(c) 在同一图中画出 $C_1$ 和 $C_2$ 的草图，清楚标明每条曲线及所有对称轴。

(d) 区域 $R$ 由 $C_1$ 和 $C_2$ 围成，且包含极点。证明 $R$ 的面积为

其中 $\alpha$ 是 7(b) 中求得的根。

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。

代入x=r cosθ与y=r sinθ，注意r取非负值；对非极点约去r平方，极点则单独保留。

With $x=r\cos\theta$, $y=r\sin\theta$ and $r\ge0$,

Thus

For non-pole points, $r>0$ and division by $r^2$ gives $\boxed{r=\frac12\sin2\theta}$. The pole is also represented by this equation, for example at $\theta=0$ or $\pi/2$, so it is retained.

交点处两个半径相同，先平方整理出指定方程，再代入两个角度验证变号。

At an intersection,

Multiplying by the positive denominator gives

as required.

Let $F(\theta)$ be the left-hand side minus $4$. In radians,

Since $F$ is continuous, this change of sign verifies a root $\boxed{0.5<\alpha<0.6}$.

两条极坐标式都关于θ=π/4对称。C1是从极点出发又回到极点的花瓣，C2是与两轴相交的外侧弧线，两曲线交两次。

Both equations are unchanged by replacing $\theta$ with $\pi/2-\theta$, so both curves are symmetric about $\boxed{\theta=\pi/4}$.

Sketch only the first-quadrant portions:

• $C_1$: a rounded petal, starting and ending at the pole, tangent there to the two coordinate axes. Its greatest radius is $1/2$ at $\theta=\pi/4$.
• $C_2$: a symmetric convex arc with the pole on its concave side. It meets both axes at radius $\big(5+\pi^2/16\big)^{-1/2}$ and has radius $1/\sqrt5$ on the symmetry line.
• Mark the two intersections at angles $\alpha$ and $\pi/2-\alpha$. Near either axis $C_1$ is closer to the pole; on the symmetry line $C_2$ is closer.

Label both curves, the pole, the initial line and the symmetry line.

所求是包含极点的公共区域，径向边界取两曲线中较小的半径。利用对称只算0到π/4，并在交角α处分段。

By symmetry about $\theta=\pi/4$, the area of $R$ is twice its area for $0\le\theta\le\pi/4$. On $[0,\alpha]$ the boundary is $C_1$; on $[\alpha,\pi/4]$ it is $C_2$. Hence

Using $\sin^22\theta=(1-\cos4\theta)/2$, the first contribution is

For the second contribution,

Therefore

as required.