Question

Express a circle with radius =1 centered at (0,1) as a double integral? I know it's with polar coordinates, but I'm not sure I'm doing it right.

Answer 1

im not sure your doing it right either, what do you got so far?

Answer 2

x = r cos(t-h) , y = r sin(t-k) , given that the center is (h,k)

Answer 3

opps, x-h = r cos(t), y-k = r sin(t)

Answer 4

and r = sqrt(x^2+y^2) might be useful to define a polar function with

Answer 5

lol, that bombed on the wolf

Answer 6

r = 2sin(t)

works, but i cant seem to recall how to get there from here

Answer 7

yeah i got r=2sin(t) as well, then set r=1 and r=2sin(t) equal to each other and got t=pi/6. I'm stuck from here. I'm not sure how to write the integral and which values to use for bounds..

Answer 8

im not sure i follow the instructions that well .... integrals are usually not used for "expressing" functions.

Do they mean that you want to integrate some doubler up into the equation for a circle r = 2sin(t) ?

Answer 9

...and does it have to be with polars?

Answer 10

it just wants us to write the integral in polar coordinates, but we don't have to solve it.

Answer 11

t = 0 to pi is fine for creating the circle; a r = 0 to 2sin(t)

\[\int_{0}^{pi}\int_{0}^{2sin(t)}~drdt\]??

Answer 12

could you please explain why the bound is 0 to 2sin(t)?

Answer 13

usually intergrals are written to find some property ..... area of shape, length of a perimeter, etc ....

the question makes no sense as stated: Express a circle