- #1
DryRun
Gold Member
- 838
- 4
Homework Statement
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.
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.