Question

Double Integration of (e^-y)/y dxdy by changing into polarform

Answer 1

Answer:

Step-by-step explanation:

I have worked this out to I=π4.

My issue is with the following...I also tried the change of variables as:

x=cos(t) ,y=sin(t)

taking t in the range of: t∈[π/4,π/2]. These were taken this way since for those values of t: x,y≥0.

Differentiating we get:

dx=−sin(t)dtdy=cos(t)dt

However this does not work out to I=π4, which is the correct answer. Any reasons why? Is the range of t incorrect? Or is the change of variable incorrect? If so, why?

My working:

∬Re−(x2+y2)dxdy

=∫π/2π/4∫π/2π/4e−(cos(t)2+sin(t)2)(−sin(t)dt)(cos(t)dt)

=−e−1∫π/2π/4∫π/2π/4sin(t)cos(t)dtdt

=−12e∫π/2π/4∫π/2π/4sin(2t)dtdt

=−eπ16