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anonymous
 5 years ago
Would the radius be from 1 to 2? Thanks!
anonymous
 5 years ago
Would the radius be from 1 to 2? Thanks!

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So, wat would the interval be? if I were to integrate it.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0for the shaded region?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0x is the point of intersection of both the circles, y is from 0 to 1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If you're talking radius, the radius is from the first circle to the second circle.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Sigh, sorry, I'm being vague again . I mean I'm using polar system, so how would the iterated integral look like?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0For the shade region.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0What are you having problem with?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0radius from x^2+y^21 to x^2+y^22y angle is you have drawn the angles in the v absolute value shape from right to left. whatever angle you have there

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I like doing double and triple integrals so if you have more questions feel free to ask them :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ok, 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+(y1)^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?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh, 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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well it says the first quadrant so I assume its (angle of intersection)< Theta < Pi / 2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0im not challenging your point of intersection angle but aren't you going beyond the first quadrant into your second?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Oh, pellet, I totally forgot that its in the 1st quandratn LOLLOL

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Thanks , omg, faillll

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Hey, 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0when 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 \[\sqrt{2 * r \sin (\theta)}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[1<r< \sqrt{2* r \sin(\theta)}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lol, ok, thanks a lot.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i got 1 because x^2 + y^2 = 1 so r^2 = 1 r=1 its really easy thats why I didnt show that one

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but if i get 1<r<2∗rsin(θ)√

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i won't get rid of a variable, b/c there's two in the interval

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lol, ok. Thanks. Looks right. i'll get back to you when i'm done.
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