xy coordinates to polar coordinates for double integral. hepl please!

[Andrew123]

1. The problem statement, all variables and given/known data
ok change the region R = { (x,y) | 1 <= X^2 + y^2 <= 4 , 0 <= y <= x } to polar region and perform the double integral over region R of z=arctan(y/x)dA


2. Relevant equations
r^2 = x^2 + y^2, x = r*sin(@), y = r * cos (@)


3. The attempt at a solution

i got R = { (rcos(@), rsin(@) | 1 <= r <= 2 , 0 <= @ <= pi/4 }

and 3/8 * pi ^2 answer in back of book is 3/64 * pi ^2


thankyou for your time!

[gabbagabbahey]

1. The problem statement, all variables and given/known data
ok change the region R = { (x,y) | 1 <= X^2 + y^2 <= 4 , 0 <= y <= x } to polar region and perform the double integral over region R of z=arctan(y/x)dA


2. Relevant equations
r^2 = x^2 + y^2, x = r*sin(@), y = r * cos (@)


3. The attempt at a solution

i got R = { (rcos(@), rsin(@) | 1 <= r <= 2 , 0 <= @ <= pi/4 }

and 3/8 * pi ^2 answer in back of book is 3/64 * pi ^2


thankyou for your time!

You've correctly converted to polar coordinates and found the limits of integration, but you somehow made a mistake evaluating the integral...Did you by chance forget that you are integrating the function \tan^{-1}\left(\frac{y}{x}\right)=\theta over this region, andf just find the area of the region instead?:wink:

[Andrew123]

thankyou veeery much!

[Mark44]

2. Relevant equations[/b]
r^2 = x^2 + y^2, x = r*sin(@), y = r * cos (@)


Not sure this made a difference in your answer, but the equations for x and y above are wrong. They should be
x = r*cos(theta)
y = r*sin(theta)