Xy coordinates to polar coordinates for double integral. hepl please

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Andrew123
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Homework Statement


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


Homework Equations


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


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!
 
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Andrew123 said:

Homework Statement


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


Homework Equations


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


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 [itex]\tan^{-1}\left(\frac{y}{x}\right)=\theta[/itex] over this region, andf just find the area of the region instead?:wink:
 
thankyou veeery much!
 
Andrew123;2056564[h2 said:
Homework Equations[/h2]
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)