Double Integrals using Polar Coordinates

[ThiagoG]

Homework Statement

∫∫_Rarctan(y/x) dA, where R={(x,y) | 1\leqx²+y²\leq4, 0\leqy\leqx

Homework Equations

x=rcos(θ)
y=rsin(θ)
x²+y²=r²

The Attempt at a Solution

I know that the range of r is 1 to 2 but I can't figure out how to change the second part into θ. If I change y and x to rsin(θ) and rcos(θ), respectively, I'm not sure what to do after this point. How would I get the theta by iteslf?

 

[slider142]

The limits or intervals of integration are sometimes easier to see if you look directly at the geometry of the region, as opposed to its algebraic representation in terms of coordinate equations.
Can you sketch the region of points that solves the pair of inequalities represented by 0 ≤ y ≤ x ? How would you describe the region you sketched in terms of r and θ ?
Remember the intuitive descriptions of r and θ that lead to the definition of the equations you cited. θ is meant to denote the counterclockwise angle between the point's radius (or position vector) and the positive x-axis. r is just meant to be the distance between the point and the origin, or the length of the point's position vector.
In the region described by those two inequalities, what is the minimum possible value of θ ? The maximum ? Is r restricted in any way ?

 

[LCKurtz]

Homework Statement

∫∫_Rarctan(y/x) dA, where R={(x,y) | 1\leqx²+y²\leq4, 0\leqy\leqx

Homework Equations

x=rcos(θ)
y=rsin(θ)
x²+y²=r²

The Attempt at a Solution

I know that the range of r is 1 to 2 but I can't figure out how to change the second part into θ. If I change y and x to rsin(θ) and rcos(θ), respectively, I'm not sure what to do after this point. How would I get the theta by iteslf?

Draw a picture of the region. Think of a windshield wiper cleaning the region. What values of $\theta$ would it sweep through?

 

[pasmith]

Homework Statement

∫∫_Rarctan(y/x) dA, where R={(x,y) | 1\leqx²+y²\leq4, 0\leqy\leqx

Homework Equations

x=rcos(θ)
y=rsin(θ)
x²+y²=r²

The Attempt at a Solution

I know that the range of r is 1 to 2 but I can't figure out how to change the second part into θ. If I change y and x to rsin(θ) and rcos(θ), respectively, I'm not sure what to do after this point. How would I get the theta by iteslf?

Look at \frac yx = \tan \theta.

 

[ThiagoG]

Draw a picture of the region. Think of a windshield wiper cleaning the region. What values of $\theta$ would it sweep through?

So would it be from 0 to ∏/4?

 

[ThiagoG]

The limits or intervals of integration are sometimes easier to see if you look directly at the geometry of the region, as opposed to its algebraic representation in terms of coordinate equations.
Can you sketch the region of points that solves the pair of inequalities represented by 0 ≤ y ≤ x ? How would you describe the region you sketched in terms of r and θ ?
Remember the intuitive descriptions of r and θ that lead to the definition of the equations you cited. θ is meant to denote the counterclockwise angle between the point's radius (or position vector) and the positive x-axis. r is just meant to be the distance between the point and the origin, or the length of the point's position vector.
In the region described by those two inequalities, what is the minimum possible value of θ ? The maximum ? Is r restricted in any way ?

After I drew it, I think it's ∏/4 because y=x bisects the first quadrant which would mean it is at ∏/4. So it ranges from 0 to ∏/4

 

[slider142]

After I drew it, I think it's ∏/4 because y=x bisects the first quadrant which would mean it is at ∏/4. So it ranges from 0 to ∏/4

That's it exactly. :) The limits you have for r are perfect as well.