Double Integration of (x-y)/(x^2+y^2) dy dx?

Question

the limit of dy is from 0 to sqrt(2x-x^2)
the limit of dx is from 0 to 2

Is this solvable without changing to polar ..double INT

I changed it to [r cos @ - r sin @)/ r^2] *r dr d@
but I am confused with the limit of iNT ??

is that right? can u help with the limit of INT

Thank you

++

I got (pi/2) -1

Answer 1

This can be done either way.
The boundary of integration is the circle
x^2 + y^2 - 2x = 0
or (x-1)^2 + y^2 = 1
Converting to polar coords give r = 2cos@
You initial integral converts to
(cos@ - sin@)*d@*dr
with @ going from 0 to pi/2
and r going from 0 to 2*cos@

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You can also use cartesian cords.

Since you have the limits of y in terms of x, start with integrating y, then move on to x.

To integrate wrt y, treat x as a constant.

int (x-y) / (x^2 + y^2) dy
= x / (x^2 + y^2) dy - y / (x^2 + y^2) dy
The first one will be an arc tan, the second a log.

Put in the limits of y and you get an expression in terms of x.
Now integrate this wrt x.

It's complicated but it should work.

Use the site below to help with some of the integrations.
I think you should end up getting 1 + pi/2.

When you do it using cartesian, you have to be careful evaluating the arc tan of +ve and -ve infinity.

Answer 2

Yes, it is right.

[r cos @ - r sin @)/ r^2] *r dr d@
=[cos @ - sin @ ] * dr d@

x=rcos@
0=rcos@

lower value
r1=0

upper value
2=rcos@

r2=2/cos@


y=rsin@
0=rsin@
sin@=0
@1=0


y=rsin@
sqrt(2*rcos@-r^2cos^2 @)=r sin@
take square

2rcos@=1
cos@=1/2r

@2=arc cos(1/2r)