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anonymous

  • 5 years ago

Would the radius be from 1 to 2? Thanks!

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  1. anonymous
    • 5 years ago
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  2. anonymous
    • 5 years ago
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    it is 1

  3. anonymous
    • 5 years ago
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    So, wat would the interval be? if I were to integrate it.

  4. anonymous
    • 5 years ago
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    for the shaded region?

  5. anonymous
    • 5 years ago
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    Yea,

  6. anonymous
    • 5 years ago
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    x is the point of intersection of both the circles, y is from 0 to 1

  7. anonymous
    • 5 years ago
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    If you're talking radius, the radius is from the first circle to the second circle.

  8. anonymous
    • 5 years ago
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    Sigh, sorry, I'm being vague again -.- I mean I'm using polar system, so how would the iterated integral look like?

  9. anonymous
    • 5 years ago
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    For the shade region.

  10. anonymous
    • 5 years ago
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    What are you having problem with?

  11. anonymous
    • 5 years ago
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    radius from x^2+y^2-1 to x^2+y^2-2y angle is you have drawn the angles in the v absolute value shape from right to left. whatever angle you have there

  12. anonymous
    • 5 years ago
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    the boundaries for the radius should be between 1< r< 2y where y= r sin (theta) 1<r< rsin (thetat) this make sense because the boundary for the r the farthest is not constant so there should be a variable. Furthermore I think you should double check your boundaries for theta.

  13. anonymous
    • 5 years ago
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    I like doing double and triple integrals so if you have more questions feel free to ask them :)

  14. anonymous
    • 5 years ago
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    Ok, thx! I will lol. I'll examine the radius , but I think my theta is correct b/c I set the x^2+y^2=1 and x^2+(y-1)^2 = 1 to each other and solved to find the point of interception. Then, I used x = rcostheta, where x is one of the point of interception, r =1, and solved for theta. Right approach?

  15. anonymous
    • 5 years ago
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    oh, i mean, when i solved for the point of inteception, in terms of y, i got 1/2, which I then plugged into the equation to get x = +/- sqrtof3/2 , which I plugged into x=rcostheta

  16. anonymous
    • 5 years ago
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    well it says the first quadrant so I assume its (angle of intersection)< Theta < Pi / 2

  17. anonymous
    • 5 years ago
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    im not challenging your point of intersection angle but aren't you going beyond the first quadrant into your second?

  18. anonymous
    • 5 years ago
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    Oh, pellet, I totally forgot that its in the 1st quandratn LOLLOL

  19. anonymous
    • 5 years ago
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    Thanks , omg, faillll

  20. anonymous
    • 5 years ago
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    Hey, wouldn't the radius be sqrt(2y)? sinec its x^2 + y^2 = (sqrt(2y))^2 but I also solved for y, which is y =1/2, so if i plug that in, i get 1, weird.

  21. anonymous
    • 5 years ago
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    when going to polar coordinates you have x^2 + y^2 = r^2 r^2 = 2* y \[r= \sqrt{2y}\] but we also know y = r sin theta so I got it wrong sorry it should be \[\sqrt{2 * r \sin (\theta)}\]

  22. anonymous
    • 5 years ago
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    \[1<r< \sqrt{2* r \sin(\theta)}\]

  23. anonymous
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    lol, ok, thanks a lot.

  24. anonymous
    • 5 years ago
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    i got 1 because x^2 + y^2 = 1 so r^2 = 1 r=1 its really easy thats why I didnt show that one

  25. anonymous
    • 5 years ago
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    but if i get 1<r<2∗rsin(θ)√

  26. anonymous
    • 5 years ago
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    when i integrate

  27. anonymous
    • 5 years ago
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    using polar system

  28. anonymous
    • 5 years ago
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    i won't get rid of a variable, b/c there's two in the interval

  29. anonymous
    • 5 years ago
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    yeah I just saw that

  30. anonymous
    • 5 years ago
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    ok i said r^2 = 2 y where y = r sin (theta) r^2 = 2 * r sin (theta) divide r to both side and you have r= 2 sin (theta) idk thats the only way I can see this happening. try it out and see if it works.

  31. anonymous
    • 5 years ago
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    lol, ok. Thanks. Looks right. i'll get back to you when i'm done.

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