(adsbygoogle = window.adsbygoogle || []).push({}); The problem statement, all variables and given/known data

Find area of region bounded by

[tex]x^2 + y^2 = 4[/tex]above y=1

The attempt at a solution

OK, so i drew the graph.

http://s1.ipicture.ru/uploads/20111226/oR99IdxJ.jpg

The red part of the graph is the area that i need to find.

Put y = 1 in the equation of circle;

x^2 = 3

x = -√3 and +√3

Now, from what i understood from struggling with this problem, i must always describe the region area in the following way:

For x fixed, x varies between -√3 to √3

Then, y varies between 1 and √(4 - x^2)

Since this problem is found at the end of the section, "Double Integrals - Polar Coordinates", i'm going to transform to polar coordinates, instead of solving it directly:

x^2 +y^2 = 4

r^2 = 4

r = -2 or +2

Considering r= +2, since the graph is symmetric about the y-axis, i will evaluate the area of the positive half found in the first quadrant and then later when i get the answer, i will multiply the latter by 2 to get the final answer.

θ varies between 0 and ∏/2

So, rewriting the double integrals:

[tex]\int\int dydx = \int\int rdrd\theta[/tex]

And using the following limits:

For r:

y = 1

rsinθ=1

So,r = 1/sinθ

(1/sinθ)≤r≤2 and 0≤θ≤(∏/2)

For the inner integral, i get the answer:

2 - 1/[2sin^2(θ)]

which i integrate again w.r.t.θ

2θ + (1/2)cotθ

Applying the limits, i get:

∏

which is incorrect. I don't know where i made the mistake/s.

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# Homework Help: Find area of region (double integral)

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