what would bet the limits of integration for the polar curve r=1-sin(theta)

Question

tkhunny · Answer

Well, that certainly would depend on what it is you are doing.

$\theta = 0 \implies r = 1$
$\theta = \pi/6 \implies r = 1/2$
$\theta = \pi/4 \implies r = (2-\sqrt{2})/2$
$\theta = \pi/3 \implies r = (2-\sqrt{3})/2$
$\theta = \pi/2 \implies r = 0$

After that, you may wish to realize the symmetry and just multiply by 2.

anonymous · Answer

I'm just trying to find the area of the curve. I know the integral is 1/2 integral (1-sintheta)^2 but I don't know the how to bound it. My teacher didn't really explain..

tkhunny · Answer

Okay, now look at the information I gave you and see where that leads.  You want to trace it all just once.

anonymous · Answer

would it be from o to pi/2 times 2? i still have no idea..because there aren't two values for theta between 0 and 2pi that make sin 1...

tkhunny · Answer

Generally, you want to trace it once in one logical piece.  It usually helps to start somewhere meaningful (The Origin is always good if that's possible) and end up in the same place.  You must be careful not to tract things twice and you must watch out for negative values of 'r'.

Try either -pi/2 to pi/2
or 2 (0 to pi/2)

If it's symmetrical, either should work.

anonymous · Answer

I have the answer key..and the answer is 3pi/2. I tried both of those and the answer doesn't match..

tkhunny · Answer

Well, you didn't actually answer my question about the symmetry.  From 0 to pi/2 is a tiny little piece.  Try -pi/2 to 3pi/2 or 2(-pi/2 to pi/2).  Unfortunately, you will find that 3pi/2 is NOT the correct result.  If you want only half of it, that is fine.  Reread the problem statement and make sure we are working the right problem.