Area of a polar curve - no option but draw it out?

Main Question or Discussion Point

Polar curves absolutely drives me nuts. I can do the integration to find the area of it no problem. The problem is.. I don't know the limits of the integration is! Sometimes is from 0 to $$\pi$$ other times it is from 0 to $$2\pi$$..

Is there no other way but to graph the curve out? Believe me, if I have time I would sketch out the curve but in my calc exams, every second counts. They literally swarm us with questions and our goal is to solve as many as possible and hope the curve carry us up. Is there no faster way? If graph it out is the only way to do it, I will practice it but I am just curious to see if there is any other tricks out there. Thanks.

Not a trick but an non-graphical way to find the limits.

Will edit with it since it might be a bit long.

Teacher taught with examples so:

You have 2 lines that you're looking to find the area within, you want to find the points of intersection.

I.$$r=4-4cos\theta$$
II.$$4r=5+6cos\theta$$

Find points of intersection by equating the equations.
multiply I by 4 and equate it to II.
$$16-16cos\theta=5+6cos\theta$$
$$11=22cos\theta$$
$$\frac{1}{2}=cos\theta$$

once you find what cos or sin equals to change to Cartesian coordinates.
plug-in cos in I.
r=4-4(.5)
r=2
$$x=rcos\theta=2*\frac{1}{2}=1$$
$$sin\theta=\pm\sqrt{1-cos^2\theta}$$
$$sin\theta=\pm\sqrt{1-\frac{1}{4}}=\pm\sqrt{\frac{3}{4}}=\pm\frac{\sqrt{3}}{2}$$
$$y=rsin\theta=2*\pm\frac{\sqrt{3}}{2}=\pm\sqrt{3}$$

2 points of intersection found so far:
$$P_1=(1,\sqrt{3})$$
$$P_2=(1,-\sqrt{3})$$

Now to check for the origin..

Plug in r=0 into equations I and II and you will get an equation for cos or sin, check that cos or sin are within the values that cos and sin should have. [-1,1]

I, r=0 => $$0=4-4cos\theta$$
$$cos\theta=1$$ Acceptable Value
$$0=5+6cos\theta$$
$$cos\theta=-\frac{5}{6}$$ Another Acceptable Value
So the origin is also a point of intersection.

Edit: Figured it out..

You have a value for Cos, use that to find a value for theta.
$$\theta_1=\frac{\pi}{3}$$
$$\theta_2=-\frac{\pi}{3}$$

If you have these points on a circle you split up the circle into 2 portions. [-pi/3,pi/3] (Goes through 0) and [pi/3,-pi/3] (Goes through pi).
You may have many other values for theta. You'll have to check where I is above II in all of the subsets.

Now say you want to find the area of I that is above II.

So you'll plug in some value for theta in those subsets and solve for r, and check where I is > II.

$$\theta=0$$
I.$$r=4-4cos0=0$$
II.$$4r=5+6cos0=> r=11/4$$
Line II is above line I so you don't want this section.

$$\theta=pi$$
I.$$r=4+4=8$$
II.$$4r=5-6=>r=-1/4$$
Line I is above line II so you want this section.

Your first limit will be pi/3. you second limit will be at -pi/3. But you want to go counterclockwise around the circle through pi instead of through 0.

So you take -pi/3 + 2pi to find the another value for this location. This gives you 5pi/3.

Your limits will be $[\frac{\pi}{3},\frac{5pi}{3}]$

and A:

$$A=\frac{1}{2}\int_\frac{\pi}{3}^\frac{5\pi}{3}((4-4cos\theta)^2-(\frac{5+6cos\theta}{4})^2)d\theta$$

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