# define Region R limits =urgent please

• Nov 17th 2007, 10:49 PM
kittycat
define Region R limits =urgent please
given :
region R is x^2+y^2=2ax

we need to define the limit of R in terms of (r, theta) .

I got the answer from the book is 0< or = r < or = 2acostheta. :confused:I don't understand how this limit from. Could you please explain to me? Thank you very much.

For the theta limit is from -pi/2 to pi/2 . That is okay. I got this part.
• Nov 17th 2007, 10:58 PM
Jhevon
Quote:

Originally Posted by kittycat
given :
region R is x^2+y^2=2ax

we need to define the limit of R in terms of (r, theta) .

I got the answer from the book is 0< or = r < or = 2acostheta. :confused:I don't understand how this limit from. Could you please explain to me? Thank you very much.

For the theta limit is from -pi/2 to pi/2 . That is okay. I got this part.

no big deal here, they just switched to polar coordinates

recall that in polar coordinates, $\displaystyle x^2 + y^2 = r^2$ and $\displaystyle x = r \cos \theta$

so, $\displaystyle x^2 + y^2 = 2ax$

$\displaystyle \Rightarrow r^2 = 2a r \cos \theta$

if $\displaystyle r \ne 0$ we can divide by it.

$\displaystyle \Rightarrow r = 2a \cos \theta$
• Nov 17th 2007, 11:07 PM
kittycat
Hi Jhevon,
x^2 + y^2 = 2ax
then (x-a)^2 + y^2 = a^2

so we notice that this is a circle of radius a centered at (a, 0).

why can't we say that r's limits is from 0 < or = r < or = 2a?

Please teach me . thank you very much.
• Nov 17th 2007, 11:11 PM
Jhevon
Quote:

Originally Posted by kittycat
Hi Jhevon,
x^2 + y^2 = 2ax
then (x-a)^2 + y^2 = a^2

so we notice that this is a circle of radius a centered at (a, 0).

why can't we say that r's limits is from 0 < or = r < or = 2a?

Please teach me . thank you very much.

because, we measure r from (0,0) not (a,0). we have to account for the circle's center shifting from the origin, we do that as i did above