Calculus Question: Polar Graphs and Area

[StudioAudience]

The question is if you have the equation r^2 = cos (n*theta), where n is some positive integer, how would you go about finding a general formula to calculate the area?

I'm stumped, and have only got the following so far:

The graph of this is going to be some sort of petal shaped thing, with the values of r^2 being equal to a negative number being discarded for obvious reasons.

If you have a value such as n = 2 and theta = 0, however, you get r^2 =1, which means r could be +/- 1. Does this mean I would plot a value of r at (0, 1) and (pi, 1)? The last coordinate is gotten using the fact that (-r, theta) = (r, pi+theta).

I know the formula for getting an area section is by integrating from a to b w/ the equation r^2*(dtheta), so if I could somehow figure out how many petals an integer n would yield, I would be closer, I think.

Any tips/hints would be appreciated!

 

[saymyname]

I don't mean to thread crap, but I'm curious if anyone here remembers this stuff still if it's been over 5 years since they did it and used it? I'm sure I could review it and learn it real quick, but seriously, who can answer this who hasn't just recently studied it?

 

[chuckywang]

You have to think about what you're integrating over...ie what values of theta. Also, make sure you remember dx dy = r dr dtheta

 

[fitzov]

math is fun

 

[Amplifier]

not to interupt but im wondering do you think it would be worth it to get another 7800gt or hold out for the 8 series*


*no radeon cards those always sux

 

[cleanerupper]

get teh radeon xtxxxxtxxx 2800xtx

 

[esun]

Wait, as I understand it, the area is r^2 dtheta, or cos (n * theta) dtheta (let t = theta for convenience).

Hence, you just need the integral of cos (n * t) dt, which is (1 / n) sin (n * t), correct? Evaluated from a to b would be:

(1 / a) sin (a * t) - (1 / b) sin (b * t)

I don't know if this is right since I haven't done polar stuff in ages.