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\[\int\limits_{0}^{2\pi} \sqrt{2 \cos \theta + 2} d \theta\]

polar integration eh ....

A = 1/2 .... something or another, id have to refresh my memory

I asked the same question earlier but nobody could answer me and I felt pretty smart for that. :P

http://tutorial.math.lamar.edu/Classes/CalcII/PolarArea.aspx

im guessing that since this is 2 parts that it might be a double integral tho, sound about right?

|dw:1327854057363:dw|
or maybe even this rendition?

bound is 0 and pie/4

Nope, I got - 11.57 but the answer is 8.

I still get a negative number for the polar area with the new upper and lower bound.

if we move the circles around we might get an easier read? or at least I do

The answer is pi/2 - 1

this is just the quarter of a circle - the cresent part then

it intersects at tan(thetha) = 1 which means at pie/4 and 3*pie/4 ..

Ah, I don't understand what you just said. :O

(pi/4)-(1-pi/4) = 2pi/4 - 1
pi/2 - 1

How did you get that?

me?

Yep

|dw:1327854888415:dw|you end up with a 1x1 square with an area of 1;

|dw:1327854924962:dw|

How do I prove that in equations?

id move the circles around to equavelent positions and do:\[\int_{0}^{pi/2}\frac{1}{2}(R^2-r^2)dr\]

whre R = (x^2+y^2=1)
and r = ((x-1)^2+(y-1)^2=1)

but state it in sin cos stuff maybe lol

I integrated \[ (2\cos \theta) ^2 - (2 \sin \theta )^2\] /2

keep the same equations perhaps and chnge integraion to -pi/2 to 0

Nope, still doesn't work.

yeah, polars im still a bit rusty on, got the concept but terrible at polarizing it

Yeah, I don't polar stuff either.