why does the unit circle with center in (1,0) have period and not ?

Printable View

- March 31st 2011, 04:46 AMKnut-oFind limits for integral
why does the unit circle with center in (1,0) have period and not ?

- March 31st 2011, 04:54 AMAckbeet
The problem statement here makes no sense. Can you please either give the context of the problem, or better yet, simply state the original problem? Thanks!

- March 31st 2011, 05:40 AMHallsofIvy
As Ackbeet said, your question makes no sense.

**Functions**have periods, not geometric objects. - March 31st 2011, 05:53 AMKnut-o
Ok, sorry. The integral is:

R is the circle - March 31st 2011, 05:55 AMAckbeet
Are you asking why the limits in the integral are to I'm still not completely understanding the question.

- March 31st 2011, 06:04 AMKnut-o
im supposed to find the limits and the "blueprint" says (-pi/2 to pi/2) for theta and (0 to 2cos(theta) for r. I cant see why -pi/2 to pi/2 and not 0 to 2pi.. it is a circle

- March 31st 2011, 06:07 AMAckbeet
My guess is that the limits are using some sort of symmetry condition in the integral, whereby you're only integrating over half the area, or something along those lines. What substitutions do you want to use?

- March 31st 2011, 06:16 AMKnut-o
im just suppose to solve it using polar cordinates. thats all. And im suppose to integrate over the circle so i guess they mean the whole circle. the only thing is that it has its center at (1,0) and not (0,0).

, so thats for r, but I cant see how they get to - March 31st 2011, 06:26 AMAckbeet
Ignore my comment in post # 7. The reason the angle limits are what they are is that the entire circle is in the right half-plane, which is described by those limits, and not 0 to 2pi.

- March 31st 2011, 06:29 AMKnut-o
ahh.. ok . that made sense. so if the circle had been in (0,1) (the upper-pane) the limits would have been 0 to pi?

- March 31st 2011, 07:17 AMAckbeet
Assuming a radius of 1, you are correct.

- March 31st 2011, 07:30 AMKnut-o
ok, thank you for your time!

- March 31st 2011, 07:33 AMAckbeet
You're welcome! Have a good one.