Convert the double integral to polar coordinates

[tag16]

Homework Statement

Evaluate the double integral by converting to polar coordinates.
∫∫ arctan y/x dA; R is the sector in the first quadrant between the circles 1/4= x^2+y^2 and x^2+y^2=1 and the lines y=x/√3 and y=x.

Homework Equations

arctan y/x= θ

The Attempt at a Solution

D={(r,θ) l 1/4≤r≤1, ∏/6≤∏/4}

∫∫ θr drdθ

Which gives me 25∏^2/3072 which is incorrect according to my book. The correct answer is 15∏^2/2304.

 

[Dick]

Homework Statement

Evaluate the double integral by converting to polar coordinates.
∫∫ arctan y/x dA; R is the sector in the first quadrant between the circles 1/4= x^2+y^2 and x^2+y^2=1 and the lines y=x/√3 and y=x.

Homework Equations

arctan y/x= θ

The Attempt at a Solution

D={(r,θ) l 1/4≤r≤1, ∏/6≤∏/4}

∫∫ θr drdθ

Which gives me 25∏^2/3072 which is incorrect according to my book. The correct answer is 15∏^2/2304.

Your lower bound for r is wrong. Think about it again.

 

[tag16]

I'm not sure, I thought maybe 1/8 but that's not right.

 

[Dick]

I'm not sure, I thought maybe 1/8 but that's not right.

A circle around the origin has the equation x^2+y^2=r^2. If r^2=1/4, what's r?

 

[tag16]

I tried 1/2 when I was originally having difficulty with the problem and still got the wrong answer. I guess I must be missing something else.

 

[Dick]

I tried 1/2 when I was originally having difficulty with the problem and still got the wrong answer. I guess I must be missing something else.

Their given answer isn't in lowest terms for some reason. 15*pi/2304=5*pi/768. Is that part of the trouble?

 

[SammyS]

I tried 1/2 when I was originally having difficulty with the problem and still got the wrong answer. I guess I must be missing something else.

Try again, This gives the right answer using the rest of what you have.