# Another double integral

• Jan 19th 2008, 03:15 PM
liyi
Another double integral
Evaluate

$\int_0^1\int_0^{\pi/2}\frac{dx\,dy}{\sqrt{1-\sqrt y\cdot\sin^2x}}$
• Jan 19th 2008, 06:11 PM
TwistedOne151
Reverse order
If we attempt to do the x integral first, we get an elliptic integral, so we reverse the order of integration:
$\int_0^{\pi/2}\int_0^1\frac{dy\,dx}{\sqrt{1-\sqrt y\cdot\sin^2x}}$
Now, integrating with respect to y, we make the change of variables $u=\sqrt{y}\to{dy}=2u\,du$
$\int_0^{\pi/2}\int_0^1\frac{2u\,du\,dx}{\sqrt{1-u\cdot\sin^2x}}$
$-\frac{4}{9}\int_0^{\pi/2}\left[\frac{(2+u\sin^2{x})\sqrt{1-u\sin^2{x}}}{\sin^4{x}}\right]_{u=0}^{1}\,dx$ (via a table of integrals)
$-\frac{4}{9}\int_0^{\pi/2}\frac{(2+\sin^2{x})\cos{x}-2}{\sin^4{x}}\,dx$
Using the substitution $v=\sin{x}\to{dx}=\frac{dv}{\sqrt{1-v^2}}$
$-\frac{4}{9}\int_0^{1}\frac{1}{v^2}+\frac{2}{v^4}-\frac{2}{v^4\sqrt{1-v^2}}\,dv$
This is an improper integral, with the integrand undefined at both limits. The limit as $v\to0$ does exist, but the limit as $v\to1$ is infinite. A little work, however, could be used to do this integral (if it exists)

--Kevin C.
• Jan 19th 2008, 06:18 PM
TwistedOne151
The integral is finite: the antiderivative of the integrand is defined at v=1, and the limit as $v\to0$ also exists and is finite.

--Kevin C.
• Jan 19th 2008, 06:24 PM
Krizalid
Quote:

Originally Posted by liyi
Evaluate

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

Make the substitutions $x = \theta$ & $y = r^4$ and apply Fubini

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

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

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