Integrate

- anonymous

Integrate

- Stacey Warren - Expert brainly.com

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- anonymous

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

- anonymous

Find the area shared by the polar equations \[r = 2 \cos \theta\] \[r = 2 \sin \theta\] |dw:1327853300525:dw|

- amistre64

polar integration eh ....

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## More answers

- amistre64

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

- anonymous

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

- amistre64

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

- amistre64

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

- amistre64

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

- anonymous

The 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

negative just means you got the bottom over the top i believe; just drop the sign and it should be good

- anonymous

The 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

bound is 0 and pie/4

- anonymous

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

- anonymous

Oh, 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

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

- amistre64

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

##### 1 Attachment

- anonymous

The answer is pi/2 - 1

- amistre64

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

- anonymous

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

- anonymous

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

- amistre64

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

- anonymous

How did you get that?

- amistre64

me?

- anonymous

Yep

- amistre64

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

- amistre64

|dw:1327854924962:dw|

- anonymous

How do I prove that in equations?

- amistre64

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

- amistre64

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

- amistre64

but state it in sin cos stuff maybe lol

- anonymous

I tried that but I must have made a mistake in the integration because I got \[(\sin2 \theta )/2 -3 \theta\] at the end.

- anonymous

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

- amistre64

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

- anonymous

Nope, still doesn't work.

- amistre64

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

- anonymous

Yeah, I don't polar stuff either.

- amistre64

ill 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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