Find the area between $r=1$ and $r=3\cos\theta$

Question

Find the area between $r=1$ and $r=3\cos\theta$.

I squared both sides to get $r^2 = 1$, then did $r^2(\cos^2 \theta + \sin^2 \theta) = (r \cos \theta)^2 + (r \sin \theta)^2$$ = x^2+y^2 = 1$ to get $x^2+y^2=1$.

For $r = 3 \cos \theta$, I multiplied by $r$ on both sides to get $r^2 = 3r \cos \theta$, then substituted $x = r \cos \theta$ to get $x^2+y^2 =3x$. However, I don't know if it is easier to do it this way.

If not, how can I find this area?

Answer 1

Please always try and draw a sketch. It helps in ensuring you have the right bounds and also understand the easy way to find the area.

Now in the sketch, you can see that there are two circles -

i) Centered at $(0,0)$ with radius $1$
ii) Centered at $(3/2, 0)$ with radius $3/2$

As they are both symmetric to X-axis, they will intersect at the same angle in both first and fourth quadrant. Say that angle is $\alpha$. The Intersection points of both circles will be given by equating -

$3 \cos \alpha = 1$

$\alpha = cos^{-1} ({\frac{1}{3}}) \approx \frac{2\pi}{5}$ (I have taken as $2 \pi / 5$ but it is closer $1.231$. Use $1.231$ for more accurate area).

We have to find the area between two curves (thru points $O, A, B, C$).

a) Integrating the curve $r = 1$ over $\angle AOC$ will give us area bound by radii $OA, OC$ and arc $ABC$ (sector $OAC$ for circle $r = 1$).

b) Integrating the curve $r = 3 \cos \theta$ over angle between $Y$ axis and $OA$ and between $Y$ axis and $OC$ will give us area bound by chord $OA, OC$ and arc $AOC$.

If we add both, we get the area we desire.

a) $A_1 = \displaystyle \frac{1}{2} \int_{-\alpha}^{\alpha}d\theta = \frac{2\pi}{5}$

b) $A_2 = \displaystyle 2 \times \frac{1}{2} \int_{\alpha}^{\pi/2}(3 \cos \theta)^2 d\theta$

$ = \displaystyle \frac{9}{2} \int_{\alpha}^{\pi/2} 2 \cos^2 \theta \, d\theta$

$ = \displaystyle \frac{9}{2} \int_{\alpha}^{\pi/2} (1 + \cos2 \theta) \, d\theta$

$ = \displaystyle \frac{9}{2} [\theta + \frac{1}{2} \sin 2 \theta)]_{2\pi/5}^{\pi/2}$

$ = \displaystyle \frac{9}{2} [\frac{\pi}{2} - \frac{2\pi}{5} - \frac{1}{2}\sin \frac{4\pi}{5}]$

$A = A_1 + A_2 \approx 1.257 + 0.063 = 1.32$

Answer 2

The area under a polar function $r(\theta)$ is:

$$\frac{1}{2} \int_a^b \big(r(\theta) \big)^2 \ d \theta$$

Here $r(\theta) = 3 \cos \theta - 1$. Now find the limits of integration, and use the identity $\cos^2 \theta = \frac{1}{2} (\cos 2 \theta+1)$.

Answer 3

Geometric solution:

Rewrite $x^2+y^2=3x$ as $x^2-3x+(\frac{3}{2})^2 + y^2 = (\frac{3}{2})^2 \Rightarrow (x-\frac{3}{2})^2 + y^2 = \frac{9}{4}$. Then solving for the intersections:

$$(x-\frac{3}{2})^2 + y^2 = \frac{9}{4} \tag{1}$$ $$x^2+y^2 = 1 \tag{2}$$

$(1) - (2)$ gives $-3x + \frac{9}{4} = \frac{5}{4} \Rightarrow x = \frac{1}{3}$. Substituting back into equation $(2)$, we get that $y = ±\frac{2 \sqrt2}{3}$.

The area of $\Delta AFC$ can be found in multiple ways, including the shoelace formula. Using this gives the area as $\frac{1}{\sqrt2}$.

Now the central angle of sector $ACF$ is just $\tan^{-1} \frac{2 \sqrt{2}/3}{3/2 - 1/3} = \tan^{-1} \frac{4 \sqrt2}{7}$ radians. Therefore its area is $\frac{1}{2} \cdot (\frac{3}{2})^2 \tan^{-1} \frac{4 \sqrt2}{7} = \frac{9}{8} \tan^{-1} \frac{4 \sqrt2}{7}$, so area $a$ is just $\frac{9}{8} \tan^{-1} \frac{4 \sqrt2}{7} - \frac{1}{\sqrt2}$.

Finally, the central angle of sector $AFB$ is $\tan^{-1} \frac{2 \sqrt{2}/3}{1/3} = \tan^{-1} 2 \sqrt{2}$, so the area of that is $\frac{1}{2} \tan^{-1} 2 \sqrt{2}$.

Therefore circular area $AFB$ is $\frac{9}{8} \tan^{-1} \frac{4 \sqrt2}{7}- \frac{1}{\sqrt2} + \frac{1}{2} \tan^{-1} 2 \sqrt{2}$, and by symmetry, the total area is just twice that, or:

$$\frac{9}{4} \tan^{-1} \frac{4 \sqrt2}{7} -\sqrt{2} + \tan^{-1} 2 \sqrt{2} \approx 1.346.$$