Double Integral Polar Coordinates to find area of region

Nymphetamine

I'm studying for my final and tutors/my professor isn't available over the weekend. Could someone please spend a little time to help me? My problem is stated as:

Let R be the right half of the circle x²+(y-1)²=1. Use a double integral polar coordinates to find the area of the region R.

I could solve for this easily if the circle wasn't shifted up 1, could someone tell me what I should do in order to solve this problem?

Some of my guesses:
I think it should still be taken in the region 3∏/2 to ∏/2 as my outside integral (right half of the circle). I could be wrong, would adding 1 to those values fix it?

I don't know how to express r (radius) in my inside integral (unless it's still going to be 0 to 1). My professor attached the note of "You will need to convert the given circle into polar form."

Doing this, I got:

R=x²+(y-1)²=1
R=r²cos²θ+(rsinθ-1)²=1
R=r²cos²θ+r²sin²θ-2rsinθ+1=1
R=r²-2rsin+1=1

I'm not sure if I'm doing this right, but what's after that?

Any help is appreciated, please let me know if you require any additional information or if I'm making something unclear.

LCKurtz

The first thing to do is remove the R= from the left margin of all those equations. What is that all about? Once you have done that, your last equation can easily be solved for r to get r = 2sin(θ). (After cancelling the 1's, dividing out an r only potentially removes the single point r = 0). That is the correct polar equation of that circle with θ going from 0 to π.

So in your double integral r is going to go from r = 0 to r on the circle and θ from 0 to π.

Nymphetamine

Thank you for your help, I can see how the polar equation for θ being 3∏/2 going to ∏/2 is the same as saying 0 to ∏ in terms of both being a half circle, but when I was seeing if they were the same this is what I did:

Calculating it out:

∫(0 to ∏)∫(0 to 2sinθ) rdrdθ

First integral:

∫(0 to 2sinθ) rdr -> r²/2 (0 to 2sinθ) = 2sin²θ

Second integral:

∫(0 to ∏) 2sin²θ dθ -> θ-(sinθ)(cosθ) (0 to ∏)
= [(∏-0)-(0-0) = ∏

but using the first integral again and doing the second as:

∫(3∏/2 to ∏/2) 2sin²θ dθ -> θ-(sinθ)(cosθ) (3∏/2 to ∏/2)
= [((∏/2)-0)-((3∏/2)-0) = -∏

Did I do anything wrong? Does it matter if the final answer is positive or negative (or no because it all depends on what your reference for θ's polar equation is?)

LCKurtz

Of course it matters if the answer is negative. That means it is wrong for an area. The limits I gave you are for the full circle. You want only the right half of the circle. I suggest you plot that circle in polar coordinates so you understand the limits. θ wouldn't go from 0 to 2π in this problem even if you wanted the whole circle. And, of course, it should be easy to check your answer since it is half the area of a circle of radius 1.

[Edit] I didn't notice in my original post you wanted only half the circle. Sorry about that.

Nymphetamine

I called a friend and he mentioned how I can't do that due to limits and continued studying my notes when I found an example similar to my problem only the circle isn't shifted up one.

My professor broke up the problem into the polar regions of θ to be 0 to ∏/2 + another integral of region 3∏/2 to 2∏ which makes sense to me now. An alternative I wrote down was -∏/2 to ∏/2.

Using that alternative region and replacing it with my previous polar region for θ I get ∏ as my answer (∏/2 - (-∏/2)). I assume my final answer should be ∏/2 based off what you told me, but I can't figure out what I'm doing wrong.

LCKurtz

Using 3π/2 to 2π or -π/2 to π/2 makes no sense in this problem and I have no idea why your professor would break the region into two parts. I suggested earlier that you draw the curve in polar coordinates. Have you done that? It would have saved you a lot of time. If (when) you do, you will find that the whole circle is completed by the time θ goes from 0 to π and the right half is already completed by the time θ is π/2, and that will give you the correct answer.

Nymphetamine

Alright, I figured it out, this link helped me out as well.

Thanks for the help!