Question

Find the area of the region common to the circles r=cos(theta) r=sqrt(3)sin(theta)

Answer 1

i know how to set up the equations i just need help finding the endpoints

Answer 2

@satellite73 can you help?

Answer 3

@Coolsector

Answer 4

You will first need the intersection point of the two circles. This is actually rather easy.

\(\sqrt{3}\sin(\theta) = \cos(\theta) \implies \tan(\theta) = \dfrac{1}{\sqrt{3}} \implies \theta = \dfrac{\pi}{6}\)

Do you see that?

Answer 5

no i don't...like i know how u got it but why did u go from cos to tan?

Answer 6

can u just explain it actually....im studying for an exam so being detailed about it would help me

Answer 7

It didn't go from cosine to tangent. The entire equation changed.

\(\sqrt{3}\sin(\theta) = \cos(\theta)\)

\(\sqrt{3}\dfrac{\sin(\theta)}{\cos(\theta)} = 1\)

\(\dfrac{\sin(\theta)}{\cos(\theta)} = \dfrac{1}{\sqrt{3}}\)

\(\tan(\theta) = \dfrac{1}{\sqrt{3}}\)

Everything is for an exam. That's why I intend to leave parts out to see if you are getting it. We're trying to learn calculus. If we have to review trigonometry, it will be harder to learn calculus.

Seeing it now?

Answer 8

oh u just set the values of r equal to each other...i know what u did now.

and sometimes u have to review trig in order to move forward with calc. reviewing isnt a bad thing.

Answer 9

Review is definitely not bad, but it does slow us down. Let it not be said that we shoudl always be in a hurry, either. Have to find a balance.

Okay, now we have the intersection point, \(\theta = \dfrac{\pi}{6}\). What shall we do with it. Have you the official formula for Area using Polar Coordinates?