1. ## Another double integral

Evaluate

$\displaystyle \int_0^1\int_0^{\pi/2}\frac{dx\,dy}{\sqrt{1-\sqrt y\cdot\sin^2x}}$

2. ## Reverse order

If we attempt to do the x integral first, we get an elliptic integral, so we reverse the order of integration:
$\displaystyle \int_0^{\pi/2}\int_0^1\frac{dy\,dx}{\sqrt{1-\sqrt y\cdot\sin^2x}}$
Now, integrating with respect to y, we make the change of variables $\displaystyle u=\sqrt{y}\to{dy}=2u\,du$
$\displaystyle \int_0^{\pi/2}\int_0^1\frac{2u\,du\,dx}{\sqrt{1-u\cdot\sin^2x}}$
$\displaystyle -\frac{4}{9}\int_0^{\pi/2}\left[\frac{(2+u\sin^2{x})\sqrt{1-u\sin^2{x}}}{\sin^4{x}}\right]_{u=0}^{1}\,dx$ (via a table of integrals)
$\displaystyle -\frac{4}{9}\int_0^{\pi/2}\frac{(2+\sin^2{x})\cos{x}-2}{\sin^4{x}}\,dx$
Using the substitution $\displaystyle v=\sin{x}\to{dx}=\frac{dv}{\sqrt{1-v^2}}$
$\displaystyle -\frac{4}{9}\int_0^{1}\frac{1}{v^2}+\frac{2}{v^4}-\frac{2}{v^4\sqrt{1-v^2}}\,dv$
This is an improper integral, with the integrand undefined at both limits. The limit as $\displaystyle v\to0$ does exist, but the limit as $\displaystyle v\to1$ is infinite. A little work, however, could be used to do this integral (if it exists)

--Kevin C.

The integral is finite: the antiderivative of the integrand is defined at v=1, and the limit as $\displaystyle v\to0$ also exists and is finite.

--Kevin C.

4. Originally Posted by liyi
Evaluate

$\displaystyle \int_0^1\int_0^{\pi/2}\frac{dx\,dy}{\sqrt{1-\sqrt y\cdot\sin^2x}}$
Make the substitutions $\displaystyle x = \theta$ & $\displaystyle y = r^4$ and apply Fubini

$\displaystyle \int_0^{\pi /2} {\int_0^1 {\frac{{4r^3 }} {{\sqrt {1 - r^2 \sin \theta } }}\,dr\,d\theta } } .$

This is a quarter of circle with radius 1. Cartesian transform

$\displaystyle 4\int_0^1 {\int_0^{\sqrt {1 - y^2 } } {\frac{{x^2 + y^2 }} {{\sqrt {1 - y^2 } }}\,dx} \,dy},$ and you have a routine integral.