Find area of region (double integral)

[DryRun]

Homework Statement
Find area of region bounded by
x^2 + y^2 = 4above 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:
\int\int dydx = \int\int rdrd\theta
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.

 

[SammyS]

Homework Statement
Find area of region bounded by
x^2 + y^2 = 4above 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:
\int\int dydx = \int\int rdrd\theta
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.

Your integration limits for θ are incorrect. Look at your figure.

 

[DryRun]

OK, so now I'm using the limits:
(1/sinθ)≤r≤2 and (∏/6)≤θ≤(∏/2)

I get the answer: 2∏/3 - √3/2

To get the whole area, by symmetry, multiply by 2:

The final answer: 4∏/3 - √3 (correct!)

However, i have some doubts about the method:

1. Could i also have used y as fixed, instead of x, to find the area? Just wondering. As in evaluating the area (not through polar coordinates) I've used both methods.

2. Are these 2 expressions equivalent?
dydx = rdrd\theta
dxdy = rdrd\theta
In my notes, they are not used in any specific order, so I'm wondering, however i checked in a calculus book and it states:
dxdy = rdrd\theta

 

[SammyS]

Yes, dxdy = dydx .

But as you've learned recently, the limits of integration need to be changed to make sure that you integrate over the correct region in both cases.

BTW: rdrdθ = rdθdr .

 

[DryRun]

OK, thanks SammyS.