anonymous
  • anonymous
Integrate
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
\[\int\limits_{0}^{2\pi} \sqrt{2 \cos \theta + 2} d \theta\]
anonymous
  • anonymous
Find the area shared by the polar equations \[r = 2 \cos \theta\] \[r = 2 \sin \theta\] |dw:1327853300525:dw|
amistre64
  • amistre64
polar integration eh ....

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

amistre64
  • amistre64
A = 1/2 .... something or another, id have to refresh my memory
anonymous
  • anonymous
I asked the same question earlier but nobody could answer me and I felt pretty smart for that. :P
amistre64
  • amistre64
http://tutorial.math.lamar.edu/Classes/CalcII/PolarArea.aspx
amistre64
  • amistre64
im guessing that since this is 2 parts that it might be a double integral tho, sound about right?
amistre64
  • amistre64
|dw:1327854057363:dw| or maybe even this rendition?
anonymous
  • 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
  • amistre64
negative just means you got the bottom over the top i believe; just drop the sign and it should be good
anonymous
  • 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
  • anonymous
bound is 0 and pie/4
anonymous
  • anonymous
Nope, I got - 11.57 but the answer is 8.
anonymous
  • 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
  • anonymous
I still get a negative number for the polar area with the new upper and lower bound.
amistre64
  • amistre64
if we move the circles around we might get an easier read? or at least I do
1 Attachment
anonymous
  • anonymous
The answer is pi/2 - 1
amistre64
  • amistre64
this is just the quarter of a circle - the cresent part then
anonymous
  • anonymous
it intersects at tan(thetha) = 1 which means at pie/4 and 3*pie/4 ..
anonymous
  • anonymous
Ah, I don't understand what you just said. :O
amistre64
  • amistre64
(pi/4)-(1-pi/4) = 2pi/4 - 1 pi/2 - 1
anonymous
  • anonymous
How did you get that?
amistre64
  • amistre64
me?
anonymous
  • anonymous
Yep
amistre64
  • amistre64
|dw:1327854888415:dw|you end up with a 1x1 square with an area of 1;
amistre64
  • amistre64
|dw:1327854924962:dw|
anonymous
  • anonymous
How do I prove that in equations?
amistre64
  • amistre64
id move the circles around to equavelent positions and do:\[\int_{0}^{pi/2}\frac{1}{2}(R^2-r^2)dr\]
amistre64
  • amistre64
whre R = (x^2+y^2=1) and r = ((x-1)^2+(y-1)^2=1)
amistre64
  • amistre64
but state it in sin cos stuff maybe lol
anonymous
  • 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
  • anonymous
I integrated \[ (2\cos \theta) ^2 - (2 \sin \theta )^2\] /2
amistre64
  • amistre64
keep the same equations perhaps and chnge integraion to -pi/2 to 0
anonymous
  • anonymous
Nope, still doesn't work.
amistre64
  • amistre64
yeah, polars im still a bit rusty on, got the concept but terrible at polarizing it
anonymous
  • anonymous
Yeah, I don't polar stuff either.
amistre64
  • 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 ....

Looking for something else?

Not the answer you are looking for? Search for more explanations.