- anonymous

Would the radius be from 1 to 2? Thanks!

- jamiebookeater

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- anonymous

##### 1 Attachment

- anonymous

it is 1

- anonymous

So, wat would the interval be? if I were to integrate it.

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## More answers

- anonymous

for the shaded region?

- anonymous

Yea,

- anonymous

x is the point of intersection of both the circles, y is from 0 to 1

- anonymous

If you're talking radius, the radius is from the first circle to the second circle.

- anonymous

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

- anonymous

For the shade region.

- anonymous

What are you having problem with?

- anonymous

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

- anonymous

the boundaries for the radius should be between 1< r< 2y
where y= r sin (theta)
1

- anonymous

I like doing double and triple integrals so if you have more questions feel free to ask them :)

- anonymous

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?

- anonymous

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

- anonymous

well it says the first quadrant so I assume its
(angle of intersection)< Theta < Pi / 2

- anonymous

im not challenging your point of intersection angle but aren't you going beyond the first quadrant into your second?

- anonymous

Oh, pellet, I totally forgot that its in the 1st quandratn LOLLOL

- anonymous

Thanks , omg, faillll

- anonymous

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.

- anonymous

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
\[\sqrt{2 * r \sin (\theta)}\]

- anonymous

\[1

- anonymous

lol, ok, thanks a lot.

- anonymous

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

- anonymous

but if i get 1

- anonymous

when i integrate

- anonymous

using polar system

- anonymous

i won't get rid of a variable, b/c there's two in the interval

- anonymous

yeah I just saw that

- anonymous

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.

- anonymous

lol, ok. Thanks. Looks right. i'll get back to you when i'm done.

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