Efficiently Solve Double Integral of 2/(2-x^2+y^2) with Trig Substitution

[cragar]

Homework Statement

double integral of 2/(2-x^2+y^2) x's from -y to y and y's from 0 to sqrt(2)/2

The Attempt at a Solution

okay so i first started by using a trig substitution
and can i call a=(2+y^2) my a to simplify things so i get
2/(a-x^2) x=sqrt(a)sin(t)
dx=sqrt(a)cos(t)dt
then we get 2/(sqrt(a)) ln|sec(t)+tan(t)| evaluated from -y to y
then i get 2ln|(y+sqrt(a))(sqrt(a)-y))| after i simplified then I’m not sure what to do here .

 

[lanedance]

have you considered a double variable change?

based on the shape of the intergation region, i looked at a basis rotated by pi/4 & scaled to simplifythe integral, which ithink simplified things a fair bit...

that said it still a bit messy & i didn't follow it all the way through...

 

[cragar]

this original integral was rotated by pi/4 , do you know where i could look at an example of this.

 

[lanedance]

do you man you had already rotated teh coordinates?

not really sure about an example, but by the rotation i meant try a subsititution something like
y = u+v
x = u-v

this give the roatated basis frame. You just need to compute a Jacobian (which will eb a constat number in thsi case) & work out the limits

for the region of integration, i had it as a (90,45,45) triangle with hypotenuse at y = 1/sqrt(2) and short sides along y=x, and y=-x

in the variable change, the y=x and y=-x beome the varibale axis (u&v) and with a bit of work you should be able to read off the limts

though as mentioned i haven't tried this through fully, just seems like its worth an attempt