hmm 3 years ago Integrate

1. hmm

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

2. hmm

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

3. amistre64

polar integration eh ....

4. amistre64

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

5. hmm

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

6. amistre64
7. amistre64

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

8. amistre64

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

9. hmm

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

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

11. hmm

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

bound is 0 and pie/4

13. hmm

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

14. hmm

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

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

16. amistre64

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

17. hmm

The answer is pi/2 - 1

18. amistre64

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

19. shaan_iitk

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

20. hmm

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

21. amistre64

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

22. hmm

How did you get that?

23. amistre64

me?

24. hmm

Yep

25. amistre64

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

26. amistre64

|dw:1327854924962:dw|

27. hmm

How do I prove that in equations?

28. amistre64

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

29. amistre64

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

30. amistre64

but state it in sin cos stuff maybe lol

31. hmm

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

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

33. amistre64

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

34. hmm

Nope, still doesn't work.

35. amistre64

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

36. hmm

Yeah, I don't polar stuff either.

37. 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 ....