Find limits for integral

**Knut-o** · March 31st 2011, 04:46 AM

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

**Ackbeet** · March 31st 2011, 04:54 AM

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!

**HallsofIvy** · March 31st 2011, 05:40 AM

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

**Knut-o** · March 31st 2011, 05:53 AM

Ok, sorry. The integral is:

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

**Ackbeet** · March 31st 2011, 05:55 AM

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

**Knut-o** · March 31st 2011, 06:04 AM

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

**Ackbeet** · March 31st 2011, 06:07 AM

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?

**Knut-o** · March 31st 2011, 06:16 AM

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 $-\pi/2 \leq \theta \leq \pi/2$

**Ackbeet** · March 31st 2011, 06:26 AM

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.

**Knut-o** · March 31st 2011, 06:29 AM

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?

**Ackbeet** · March 31st 2011, 07:17 AM

Assuming a radius of 1, you are correct.

**Knut-o** · March 31st 2011, 07:30 AM

ok, thank you for your time!

**Ackbeet** · March 31st 2011, 07:33 AM

You're welcome! Have a good one.

Thread: Find limits for integral

LinkBack

Thread Tools

Display

Find limits for integral

Similar Math Help Forum Discussions

How to find these limits?

Volume integral, how do I find the limits for my integral?

Given a function, find points, find limits, etc.

Find limits

Using limits to find other limits

Search Tags