(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.

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Find area of region (double integral)

**Physics Forums | Science Articles, Homework Help, Discussion**