## hmm Group Title Integrate 2 years ago 2 years ago

1. hmm Group Title

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

2. hmm Group Title

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

3. amistre64 Group Title

polar integration eh ....

4. amistre64 Group Title

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

5. hmm Group Title

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

6. amistre64 Group Title
7. amistre64 Group Title

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

8. amistre64 Group Title

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

9. hmm Group Title

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.

10. amistre64 Group Title

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

11. hmm Group Title

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.

12. shaan_iitk Group Title

bound is 0 and pie/4

13. hmm Group Title

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

14. hmm Group Title

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.

15. hmm Group Title

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

16. amistre64 Group Title

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

17. hmm Group Title

The answer is pi/2 - 1

18. amistre64 Group Title

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

19. shaan_iitk Group Title

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

20. hmm Group Title

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

21. amistre64 Group Title

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

22. hmm Group Title

How did you get that?

23. amistre64 Group Title

me?

24. hmm Group Title

Yep

25. amistre64 Group Title

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

26. amistre64 Group Title

|dw:1327854924962:dw|

27. hmm Group Title

How do I prove that in equations?

28. amistre64 Group Title

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

29. amistre64 Group Title

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

30. amistre64 Group Title

but state it in sin cos stuff maybe lol

31. hmm Group Title

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

32. hmm Group Title

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

33. amistre64 Group Title

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

34. hmm Group Title

Nope, still doesn't work.

35. amistre64 Group Title

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

36. hmm Group Title

Yeah, I don't polar stuff either.

37. amistre64 Group Title

ill reread my book at the house and know this stuff better tomorrow :) i cant seem to make sense of what I find online ....