anonymous
  • anonymous
Would the radius be from 1 to 2? Thanks!
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
1 Attachment
anonymous
  • anonymous
it is 1
anonymous
  • anonymous
So, wat would the interval be? if I were to integrate it.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
for the shaded region?
anonymous
  • anonymous
Yea,
anonymous
  • anonymous
x is the point of intersection of both the circles, y is from 0 to 1
anonymous
  • anonymous
If you're talking radius, the radius is from the first circle to the second circle.
anonymous
  • anonymous
Sigh, sorry, I'm being vague again -.- I mean I'm using polar system, so how would the iterated integral look like?
anonymous
  • anonymous
For the shade region.
anonymous
  • anonymous
What are you having problem with?
anonymous
  • 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
  • anonymous
the boundaries for the radius should be between 1< r< 2y where y= r sin (theta) 1
anonymous
  • anonymous
I like doing double and triple integrals so if you have more questions feel free to ask them :)
anonymous
  • 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
  • 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
  • anonymous
well it says the first quadrant so I assume its (angle of intersection)< Theta < Pi / 2
anonymous
  • anonymous
im not challenging your point of intersection angle but aren't you going beyond the first quadrant into your second?
anonymous
  • anonymous
Oh, pellet, I totally forgot that its in the 1st quandratn LOLLOL
anonymous
  • anonymous
Thanks , omg, faillll
anonymous
  • 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
  • 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
  • anonymous
\[1
anonymous
  • anonymous
lol, ok, thanks a lot.
anonymous
  • 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
  • anonymous
but if i get 1
anonymous
  • anonymous
when i integrate
anonymous
  • anonymous
using polar system
anonymous
  • anonymous
i won't get rid of a variable, b/c there's two in the interval
anonymous
  • anonymous
yeah I just saw that
anonymous
  • 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
  • anonymous
lol, ok. Thanks. Looks right. i'll get back to you when i'm done.

Looking for something else?

Not the answer you are looking for? Search for more explanations.