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anonymous
 4 years ago
Integrate
anonymous
 4 years ago
Integrate

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{0}^{2\pi} \sqrt{2 \cos \theta + 2} d \theta\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Find the area shared by the polar equations \[r = 2 \cos \theta\] \[r = 2 \sin \theta\] dw:1327853300525:dw

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0polar integration eh ....

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0A = 1/2 .... something or another, id have to refresh my memory

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I asked the same question earlier but nobody could answer me and I felt pretty smart for that. :P

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0http://tutorial.math.lamar.edu/Classes/CalcII/PolarArea.aspx

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0im guessing that since this is 2 parts that it might be a double integral tho, sound about right?

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1327854057363:dw or maybe even this rendition?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0The first question I did a double integral and got \[\theta \sqrt{2 \cos \theta + 2} \theta \sin \theta  \cos \theta  \theta^2/2\] from 2 pi to zero. Since this was originally \[\int\limits_{0}^{2\pi} \sqrt{(1 + \cos^2 \theta) + \sin^2 \theta}\] for the arc length of the cardioid. It's not supposed to be negative, which I got.

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0negative just means you got the bottom over the top i believe; just drop the sign and it should be good

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0The second questions about the polar area is the area between two circles whihc I drew on the top. I used the calculator to find the bound to be 0 and pi, but it might be wrong.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Nope, I got  11.57 but the answer is 8.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Oh, I mgiht have gotten the upper and lower bound wrong. But I was trying to find the area without the shared part first then use that to be subtracted from the area of the entire circle but I am wondering if there's a better idea. I'll try the pi over 4 bound.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I still get a negative number for the polar area with the new upper and lower bound.

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0if we move the circles around we might get an easier read? or at least I do

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0The answer is pi/2  1

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0this is just the quarter of a circle  the cresent part then

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0it intersects at tan(thetha) = 1 which means at pie/4 and 3*pie/4 ..

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Ah, I don't understand what you just said. :O

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0(pi/4)(1pi/4) = 2pi/4  1 pi/2  1

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0How did you get that?

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1327854888415:dwyou end up with a 1x1 square with an area of 1;

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1327854924962:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0How do I prove that in equations?

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0id move the circles around to equavelent positions and do:\[\int_{0}^{pi/2}\frac{1}{2}(R^2r^2)dr\]

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0whre R = (x^2+y^2=1) and r = ((x1)^2+(y1)^2=1)

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0but state it in sin cos stuff maybe lol

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I tried that but I must have made a mistake in the integration because I got \[(\sin2 \theta )/2 3 \theta\] at the end.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I integrated \[ (2\cos \theta) ^2  (2 \sin \theta )^2\] /2

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0keep the same equations perhaps and chnge integraion to pi/2 to 0

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Nope, still doesn't work.

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0yeah, polars im still a bit rusty on, got the concept but terrible at polarizing it

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yeah, I don't polar stuff either.

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0ill reread my book at the house and know this stuff better tomorrow :) i cant seem to make sense of what I find online ....
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