A region bounded by two polar curves is found with the same sectors as a region bounded by one. Where a region lies between two curves, each thin sector of the outer curve has a sector of the inner curve cut out of it.
Inside one curve and outside another Each radius is squared before subtracting, since the area of a sector grows with the square of its radius. The limits are usually the angles where the two curves meet, found by setting the two formulas for
r r r equal.
Example 1
Find the total area of the regions inside
r = 2 + cos 2 θ r = 2 + \cos 2\theta r = 2 + cos 2 θ and outside the circle
r = 3 2 r = \frac{3}{2} r = 2 3 .
The curves meet where
cos 2 θ = − 1 2 \cos 2\theta = -\frac{1}{2} cos 2 θ = − 2 1 , at
θ = ± π 3 \theta = \pm\frac{\pi}{3} θ = ± 3 π and
θ = 2 π 3 , 4 π 3 \theta = \frac{2\pi}{3}, \frac{4\pi}{3} θ = 3 2 π , 3 4 π . The peanut is outside the circle where
cos 2 θ > − 1 2 \cos 2\theta > -\frac{1}{2} cos 2 θ > − 2 1 , on
− π 3 < θ < π 3 -\frac{\pi}{3} < \theta < \frac{\pi}{3} − 3 π < θ < 3 π and on the matching interval on the left. The two pieces have the same area, so
A = 2 ⋅ 1 2 ∫ − π / 3 π / 3 ( ( 2 + cos 2 θ ) 2 − 9 4 ) d θ = ∫ − π / 3 π / 3 ( 7 4 + 4 cos 2 θ + cos 2 2 θ ) d θ . \begin{aligned}
A &= 2\cdot\frac{1}{2}\int_{-\pi/3}^{\pi/3} \Big((2 + \cos 2\theta)^2 \\
&\qquad\qquad - \tfrac{9}{4}\Big)\,d\theta \\[4pt]
&= \int_{-\pi/3}^{\pi/3} \Big(\tfrac{7}{4} + 4\cos 2\theta \\
&\qquad\qquad + \cos^2 2\theta\Big)\,d\theta.
\end{aligned} A = 2 ⋅ 2 1 ∫ − π /3 π /3 ( ( 2 + cos 2 θ ) 2 − 4 9 ) d θ = ∫ − π /3 π /3 ( 4 7 + 4 cos 2 θ + cos 2 2 θ ) d θ . The constant gives
7 4 ⋅ 2 π 3 = 7 π 6 \frac{7}{4}\cdot\frac{2\pi}{3} = \frac{7\pi}{6} 4 7 ⋅ 3 2 π = 6 7 π . The middle term gives
2 sin 2 θ 2\sin 2\theta 2 sin 2 θ , which changes by
2 3 2\sqrt{3} 2 3 . With
cos 2 2 θ = 1 2 ( 1 + cos 4 θ ) \cos^2 2\theta = \frac{1}{2}(1 + \cos 4\theta) cos 2 2 θ = 2 1 ( 1 + cos 4 θ ) , the last term gives
π 3 − 3 8 \frac{\pi}{3} - \frac{\sqrt{3}}{8} 3 π − 8 3 .
A = 3 π 2 + 15 3 8 ≈ 7.960. A = \frac{3\pi}{2} + \frac{15\sqrt{3}}{8} \approx 7.960. A = 2 3 π + 8 15 3 ≈ 7.960.
The region inside r = 2 + cos 2θ and outside the circle r = 3/2. It has two pieces: one between the rays θ = ±π/3, and its match on the left. Inside both curves A point is inside both curves when it's inside the one nearer the origin. So split the angles at the intersections, and on each arc integrate the square of whichever curve is smaller there.
Example 2
A calculator is allowed. Find the area of the region inside both
r = 2 + cos 2 θ r = 2 + \cos 2\theta r = 2 + cos 2 θ and
r = 2 + 3 2 sin θ r = 2 + \frac{3}{2}\sin\theta r = 2 + 2 3 sin θ , to three decimal places.
The curves meet where
cos 2 θ = 3 2 sin θ \cos 2\theta = \frac{3}{2}\sin\theta cos 2 θ = 2 3 sin θ . With
cos 2 θ = 1 − 2 sin 2 θ \cos 2\theta = 1 - 2\sin^2\theta cos 2 θ = 1 − 2 sin 2 θ , this is
4 sin 2 θ + 3 sin θ − 2 = 0 4\sin^2\theta + 3\sin\theta - 2 = 0 4 sin 2 θ + 3 sin θ − 2 = 0 , so
sin θ = − 3 ± 41 8 \sin\theta = \frac{-3 \pm \sqrt{41}}{8} sin θ = 8 − 3 ± 41 . The minus sign gives a value below
− 1 -1 − 1 , so
sin θ = 41 − 3 8 \sin\theta = \frac{\sqrt{41} - 3}{8} sin θ = 8 41 − 3 . The two angles in
[ 0 , 2 π ) [0, 2\pi) [ 0 , 2 π ) are
a ≈ 0.439 a \approx 0.439 a ≈ 0.439 and
π − a ≈ 2.702 \pi - a \approx 2.702 π − a ≈ 2.702 .
At
θ = π 2 \theta = \frac{\pi}{2} θ = 2 π the peanut gives
r = 1 r = 1 r = 1 and the other curve gives
r = 7 2 r = \frac{7}{2} r = 2 7 , so the peanut is nearer the origin on
( a , π − a ) (a, \pi - a) ( a , π − a ) . At
θ = 3 π 2 \theta = \frac{3\pi}{2} θ = 2 3 π they give 1 and
1 2 \frac{1}{2} 2 1 , so the other curve is nearer on the rest of the turn, from
π − a \pi - a π − a around to
2 π + a 2\pi + a 2 π + a .
A = 1 2 ∫ a π − a ( 2 + cos 2 θ ) 2 d θ + 1 2 ∫ π − a 2 π + a ( 2 + 3 2 sin θ ) 2 d θ ≈ 3.429 + 4.439 = 7.868. \begin{aligned}
A &= \frac{1}{2}\int_{a}^{\pi - a} (2 + \cos 2\theta)^2\,d\theta \\[4pt]
&\qquad + \frac{1}{2}\int_{\pi - a}^{2\pi + a} \big(2 + \tfrac{3}{2}\sin\theta\big)^2\,d\theta \\[4pt]
&\approx 3.429 + 4.439 = 7.868.
\end{aligned} A = 2 1 ∫ a π − a ( 2 + cos 2 θ ) 2 d θ + 2 1 ∫ π − a 2 π + a ( 2 + 2 3 sin θ ) 2 d θ ≈ 3.429 + 4.439 = 7.868.
The region inside both curves. Across the top it's bounded by the peanut, and everywhere else by r = 2 + (3/2) sin θ.