题目
Problem
Figure 1 shows the curve C 1 C_1 C 1 r = 2 a sin 2 θ r=2a\sin2\theta r = 2 a sin 2 θ 0 ⩽ θ ⩽ π 2 0\leqslant\theta\leqslant\dfrac{\pi}{2} 0 ⩽ θ ⩽ 2 π C 2 C_2 C 2 r = a r=a r = a 0 ⩽ θ ⩽ 2 π 0\leqslant\theta\leqslant2\pi 0 ⩽ θ ⩽ 2 π a a a
(a) Find, in terms of a a a C 1 C_1 C 1 C 2 C_2 C 2
(3)
The regions enclosed by the curve C 1 C_1 C 1 C 2 C_2 C 2 R R R
(b) Use algebraic integration to find the area of the shaded region R R R
1 12 a 2 ( p π + q 3 ) \begin{align*}
\frac{1}{12}a^2(p\pi+q\sqrt3)
\end{align*} 12 1 a 2 ( p π + q 3 ) where p p p q q q
(7)
解答
(a)
解法一
思路
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交点处两个极径相等，所以令 a = 2 a sin 2 θ a=2a\sin2\theta a = 2 a sin 2 θ a > 0 a>0 a > 0 a a a
答题过程
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At an intersection,
a = 2 a sin 2 θ \begin{align*}
a=2a\sin2\theta
\end{align*} a = 2 a sin 2 θ Since a > 0 a>0 a > 0
sin 2 θ = 1 2 \begin{align*}
\sin2\theta=\frac12
\end{align*} sin 2 θ = 2 1 For 0 ⩽ θ ⩽ π 2 0\leqslant\theta\leqslant\dfrac{\pi}{2} 0 ⩽ θ ⩽ 2 π
0 ⩽ 2 θ ⩽ π \begin{align*}
0\leqslant2\theta\leqslant\pi
\end{align*} 0 ⩽ 2 θ ⩽ π so
2 θ = π 6 or 2 θ = 5 π 6 \begin{align*}
2\theta=\frac{\pi}{6}
\quad\text{or}\quad
2\theta=\frac{5\pi}{6}
\end{align*} 2 θ = 6 π or 2 θ = 6 5 π Hence
θ = π 12 or θ = 5 π 12 \begin{align*}
\theta=\frac{\pi}{12}
\quad\text{or}\quad
\theta=\frac{5\pi}{12}
\end{align*} θ = 12 π or θ = 12 5 π The polar coordinates are
( a , π 12 ) , ( a , 5 π 12 ) \begin{align*}
\left(a,\frac{\pi}{12}\right),
\qquad
\left(a,\frac{5\pi}{12}\right)
\end{align*} ( a , 12 π ) , ( a , 12 5 π )
(b)
解法一
思路
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公共区域由中间的圆扇形和两侧相同的曲线小区域组成。
中间扇形的角度是 5 π 12 − π 12 = π 3 \dfrac{5\pi}{12}-\dfrac{\pi}{12}=\dfrac{\pi}{3} 12 5 π − 12 π = 3 π r = 2 a sin 2 θ r=2a\sin2\theta r = 2 a sin 2 θ 0 0 0 π 12 \dfrac{\pi}{12} 12 π
答题过程
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The circle sector between
θ = π 12 and θ = 5 π 12 \begin{align*}
\theta=\frac{\pi}{12}
\quad\text{and}\quad
\theta=\frac{5\pi}{12}
\end{align*} θ = 12 π and θ = 12 5 π has angle
5 π 12 − π 12 = π 3 \begin{align*}
\frac{5\pi}{12}-\frac{\pi}{12}
=
\frac{\pi}{3}
\end{align*} 12 5 π − 12 π = 3 π So its area is
1 2 a 2 ⋅ π 3 = a 2 π 6 \begin{align*}
\frac12a^2\cdot\frac{\pi}{3}
=
\frac{a^2\pi}{6}
\end{align*} 2 1 a 2 ⋅ 3 π = 6 a 2 π Now find one of the two equal side regions:
I = 1 2 ∫ 0 π / 12 ( 2 a sin 2 θ ) 2 d θ = 2 a 2 ∫ 0 π / 12 sin 2 2 θ d θ \begin{align*}
I
={}&
\frac12
\int_{0}^{\pi/12}
(2a\sin2\theta)^2\,\mathrm{d}\theta\\[3mm]
={}&
2a^2
\int_{0}^{\pi/12}
\sin^22\theta\,\mathrm{d}\theta
\end{align*} I = = 2 1 ∫ 0 π /12 ( 2 a sin 2 θ ) 2 d θ 2 a 2 ∫ 0 π /12 sin 2 2 θ d θ Using
sin 2 2 θ = 1 2 ( 1 − cos 4 θ ) , \begin{align*}
\sin^22\theta=\frac12(1-\cos4\theta),
\end{align*} sin 2 2 θ = 2 1 ( 1 − cos 4 θ ) , we get
I = a 2 ∫ 0 π / 12 ( 1 − cos 4 θ ) d θ = a 2 [ θ − 1 4 sin 4 θ ] 0 π / 12 = a 2 ( π 12 − 1 4 sin π 3 ) = a 2 ( π 12 − 3 8 ) \begin{align*}
I
={}&
a^2\int_{0}^{\pi/12}(1-\cos4\theta)\,\mathrm{d}\theta\\[3mm]
={}&
a^2
\left[
\theta-\frac14\sin4\theta
\right]_{0}^{\pi/12}\\[3mm]
={}&
a^2
\left(
\frac{\pi}{12}
-\frac14\sin\frac{\pi}{3}
\right)\\[3mm]
={}&
a^2
\left(
\frac{\pi}{12}
-\frac{\sqrt3}{8}
\right)
\end{align*} I = = = = a 2 ∫ 0 π /12 ( 1 − cos 4 θ ) d θ a 2 [ θ − 4 1 sin 4 θ ] 0 π /12 a 2 ( 12 π − 4 1 sin 3 π ) a 2 ( 12 π − 8 3 ) Therefore the shaded area is
R = 2 I + a 2 π 6 = 2 a 2 ( π 12 − 3 8 ) + a 2 π 6 = a 2 ( π 6 − 3 4 ) + a 2 π 6 = a 2 ( π 3 − 3 4 ) = 1 12 a 2 ( 4 π − 3 3 ) \begin{align*}
R
={}&
2I+\frac{a^2\pi}{6}\\[3mm]
={}&
2a^2
\left(
\frac{\pi}{12}
-\frac{\sqrt3}{8}
\right)
+\frac{a^2\pi}{6}\\[3mm]
={}&
a^2
\left(
\frac{\pi}{6}
-\frac{\sqrt3}{4}
\right)
+\frac{a^2\pi}{6}\\[3mm]
={}&
a^2
\left(
\frac{\pi}{3}
-\frac{\sqrt3}{4}
\right)\\[3mm]
={}&
\frac{1}{12}a^2(4\pi-3\sqrt3)
\end{align*} R = = = = = 2 I + 6 a 2 π 2 a 2 ( 12 π − 8 3 ) + 6 a 2 π a 2 ( 6 π − 4 3 ) + 6 a 2 π a 2 ( 3 π − 4 3 ) 12 1 a 2 ( 4 π − 3 3 ) Thus
p = 4 , q = − 3 \begin{align*}
p=4,\qquad q=-3
\end{align*} p = 4 , q = − 3 
