Hard Double Integral Homework: Solve & Understand

[CollinsArg]

Homework Statement

I'm given the integral show in the adjunct picture, in the same one there is my attempt at a solution.

Homework Equations

x = r.cos(Θ)
y = r.sin(Θ)
dA = r.dr.dΘ

The Attempt at a Solution

[/B]
I tried to do it in polar coordinates, so I substituted x=r.cos(Θ) y=r.sin(Θ) in the function, I know the answer for the first integral will be ln (1+r^2)/2, but I don't know what should I do with that theta as an exponent of two, inside of a function with a squared root xP. Thank you

 

[Mark44]

Homework Statement

I'm given the integral show in the adjunct picture, in the same one there is my attempt at a solution.

Homework Equations

x = r.cos(Θ)
y = r.sin(Θ)
dA = r.dr.dΘ

The Attempt at a Solution

View attachment 214717 [/B]
I tried to do it in polar coordinates, so I substituted x=r.cos(Θ) y=r.sin(Θ) in the function, I know the answer for the first integral will be ln (1+r^2)/2, but I don't know what should I do with that theta as an exponent of two, inside of a function with a squared root xP. Thank you

There's a mistake in your work. One of your steps is $\frac 1 2 \int_0^{\pi/4} \ln(1 + r^2) |_{r = \sqrt{2^{1 - \theta} - 1}}^1 d\theta$. When you evaluate the integrand at the two endpoints, remember that you have $r^2$ in the integrand, not r. That makes things a lot simpler.

 

[phyzguy]

You're on the right track. You forgot to square the lower limit when you plugged it in for r. What is $\log(1+r^2)$ when you plug in $r=\sqrt{2^{1-\theta}-1}$?

 

[CollinsArg]

Thank you so much!