## anonymous 5 years ago Can someone help me solve this iterated integral? Tanks.

1. anonymous

Here's the equation.

2. anonymous

the one marked red?

3. anonymous

Yes, xD

4. anonymous

use polar

5. anonymous

substitutions

6. anonymous

I have a test in 4days time on this stuff, double/triple integrals etc

7. anonymous

Oh, okay, thanks. Good luck to you! I'll get back here if I can't do it lol.

8. anonymous

let A be the angle

9. anonymous

x=rcos(A) y= rsin(A)

10. anonymous

the jacobian of the substitution is "f"

11. anonymous

edit "r"

12. anonymous

how would I get r? is it = 1? I figured out theta, which is from 0 to pi/4

13. anonymous

14. anonymous

not the best picture lol

15. anonymous

lol, i'll upload mine

16. anonymous

the angle goes from 90degrees to 45degrees actually

17. anonymous

but the difference is still the same

18. anonymous

and our r goes from 0<=r<=1

19. anonymous

because we are inside the disc x^2 +y^2 <1 , so [rcos(A)]^2 +[ rsin(A)]^2 <1 , r^2<1 but r is also positive

20. anonymous

This is what I came up with, is it correct?

21. anonymous

Elec, your drawing is correct, but the sector described is actually the portion you have marked 45 degrees. Good job

22. anonymous

I got it based on the 2 iterated integrals below in pic. ok, thx, i think i can do it now, 0<=r<=1

23. anonymous

wait, the region I drew was correct I am fairly sure

24. anonymous

I know it doesnt impact the final answer , but the integral has x as the lower limit, and the semicircle as the upper limit

25. anonymous

is it 2y+2x

26. anonymous

$= \int\limits_{\frac{\pi}{4}}^{\frac{\pi}{2}} d \theta \int\limits_{0}^{1} r^2 dr$

27. anonymous

= $\frac{\pi}{4} \times \frac{1}{3} = \frac{\pi}{12}$

28. anonymous

What about the region I drew? Is that acceptable?

29. anonymous

Oh yeah, you're right, I was looking at it cross ways. Choose some x, and the y goes from x to the semicircle.

30. anonymous

Revan, elec's drawing is right. Also you have a sq rt 2/2 in there, that I think you got from your other problem.

31. anonymous

But isn't it, 0<=y<=x?

32. anonymous

no!

33. anonymous

the region you drew was y<=x and y<=sqrt(1-x^2)

34. anonymous

the intersection of those region , thats what your picture shows

35. anonymous

even though they would both give the same answer it is very likely you would lost a mark or two , because your drawing and the limits of the angle dont reflect the correct region

36. anonymous

Out of curiosity, revan, where did you get sq rt 2/2 from?

37. anonymous

http://openstudy.com/updates/attachments/4dcf65a1dfd28b0b901b0dfd-revangale-1305438547251-photo1.jpg from the two integrals at the bot of the pic. I have to sketch Region of the sum of the two and then replace the sum by an equivalent single iterated integral and evalute

38. anonymous

Part of confusion might be: that's not the question you posted here for this post.

39. anonymous

Lol, sorry >.< The question I posted is the single iterated integral. I just needed help solving it.

40. anonymous

yes, the way you did was very wrong lols

41. anonymous

:D thats why i posted lol

42. anonymous

$\int\limits(ax+b)^n = \frac{(ax+b)^{n+1} } {a(n+1)} +C$

43. anonymous

only works for linear functions

44. anonymous

raised to a power

45. anonymous

Now elec, is writing wrong thing

46. anonymous

You said to do poloar coordinates. So, I think I can solve it now. I'll post it when I'm done.

47. anonymous

Btw, y is r from 0 to 1? o.0

48. anonymous

nvm . ..

49. anonymous

Revan, if given this from scratch, you choose polar, but if given the integrals already compiled it is very easy to read.

50. anonymous

Yea, but I cant' solve it w/o using polar, but either system, the answer will be the same?

51. anonymous

What do you mean by 'solve'

52. anonymous

or find volume? that's what you mean?

53. anonymous

http://openstudy.com/updates/attachments/4dcf65a1dfd28b0b901b0dfd-revangale-1305438547251-photo1.jpg I'm supposed to combine the sum of the two iterated integrals at the bottom of this pic into a single iterated integral and evalute it.

54. anonymous

So, I drew the graph of the two iterated integrals together, to get the overall Region, and then use the overall Region to set up a single iterated integral. But, I couldn't solve it, unless it was in polar system. Is this the right approach?

55. anonymous

Well, just looking at the picture, it is a sector of a circle and it's a polar type thing. So sure. It is just a kind of trick question, your teacher wants to know if you understand what's going on

56. anonymous

Lol, okay, thx., yea I didn't know wat was going on.

57. anonymous

BTW ele engineer is right in using transformation but going from rectangular coordinates to polar is something that should be easy we know da---> r dr dtheta where x= r cos theta y = r sin theta and x^2 + y^2 = r ^2 and if you actually draw it out its easier to know the boundaries.