∫∫ (x+y)^2/(X^2+Y^2) dxdy

The region bounded by the positive x and y axes and the line y = 1-x

Question

∫∫ (x+y)^2/(X^2+Y^2) dxdy

The region bounded by the positive x and y axes and the line y = 1-x

zarkon · Answer

expand $(x+y)^2$ then simplify...then integrate

anonymous · Answer

i know how to intefrate this i just dont know how to convert the limits into polar form?

anonymous · Answer

i know that im left with sin2theta but the integrals of x: 0 to 1 and 0 to 1-y, how can i convert the limits into polar form?

anonymous · Answer

Can anyone confirm that im on the right lines ?

zarkon · Answer

$$y = 1-x$$

$$r\sin(\theta)=1-r\cos(\theta)$$

$$r\sin(\theta)+r\cos(\theta)=1$$

$$r(\sin(\theta)+\cos(\theta))=1$$

$$r=\frac{1}{\sin(\theta)+\cos(\theta)}$$

anonymous · Answer

Are the limits r=1, r=0 and theta=pi/4, to theta=0 ?

anonymous · Answer

so what are the limits ? are mine correct?

zarkon · Answer

those are not correct

anonymous · Answer

i sketched a graph and theses are the one that i see.. what im i doing wrong?

zarkon · Answer

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zarkon · Answer

why do you have r=1 and $\theta=\pi/4$

zarkon · Answer

that makes 1/8th of a circle

anonymous · Answer

polar coordinates have to be with respect to rdrdtheta

zarkon · Answer

yes...I'm questioning the choice of 1 and pi/4

zarkon · Answer

you are integrating over a triangle in the 1st quadrant

zarkon · Answer

not a circle

anonymous · Answer

i think r =1 to r=0 is correct but not sure about theta

zarkon · Answer

it is not

anonymous · Answer

hmmm let me see...

zarkon · Answer

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