Thread:

1. Find limits for integral

why does the unit circle with center in (1,0) have period $-\frac{\pi}{2} , \frac{\pi}{2}$ and not $0, 2\pi$ ?

2. 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!

3. As Ackbeet said, your question makes no sense. Functions have periods, not geometric objects.

4. Ok, sorry. The integral is:

$\int \int_R (x^2+y^2)^{(3/2)} dxdy$ R is the circle $(x-1)^2+y^2 \leq 1$

5. Are you asking why the limits in the integral are $-\pi/2$ to $\pi/2?$ I'm still not completely understanding the question.

6. 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

7. 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?

8. im just suppose to solve it using polar cordinates. thats all. And im suppose to integrate over the circle $(x-1)^2+y^2\leq 1$ so i guess they mean the whole circle. the only thing is that it has its center at (1,0) and not (0,0).
$(r\cos(\theta)-1)^2 + r^2\sin^2(\theta) \leq 1 \Rightarrow r \leq 2\cos(\theta)$, so thats for r, but I cant see how they get to $-\pi/2 \leq \theta \leq \pi/2$

9. 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.

10. 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?

11. Assuming a radius of 1, you are correct.

12. ok, thank you for your time!

13. You're welcome! Have a good one.
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