Double Integral Polar Coordinates to find area of regionby Nymphetamine Tags: area, double integral, polar coordinates 

#1
Dec211, 07:27 PM

P: 4

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 x^{2}+(y1)^{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=x^{2}+(y1)^{2}=1 R=r^{2}cos^{2}θ+(rsinθ1)^{2}=1 R=r^{2}cos^{2}θ+r^{2}sin^{2}θ2rsinθ+1=1 R=r^{2}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. 



#2
Dec211, 08:21 PM

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



#3
Dec211, 09:03 PM

P: 4

Calculating it out: ∫(0 to ∏)∫(0 to 2sinθ) rdrdθ First integral: ∫(0 to 2sinθ) rdr > r^{2}/2 (0 to 2sinθ) = 2sin^{2}θ Second integral: ∫(0 to ∏) 2sin^{2}θ dθ > θ(sinθ)(cosθ) (0 to ∏) = [(∏0)(00) = ∏ but using the first integral again and doing the second as: ∫(3∏/2 to ∏/2) 2sin^{2}θ 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?) 



#4
Dec211, 09:10 PM

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PF Gold
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Double Integral Polar Coordinates to find area of region
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. 



#5
Dec311, 12:45 AM

P: 4

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. 



#6
Dec311, 12:17 PM

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PF Gold
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#7
Dec311, 07:02 PM

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