# Evaluating integral using change of variables

• May 17th 2009, 07:02 PM
MrSplashypants1
Evaluating integral using change of variables
Hey guys, just having a bit of trouble getting started with this.

Have to use a change of variables to evaluate the integral
$\displaystyle \int\int_{D}sin(x+2y)sin(3x-y)dxdy$
where $\displaystyle D={ (x,y)\in\Re^2 | 0 \leq x+2y \leq (\Pi/2) and 0\leq 3x-y\leq (\Pi/2) }$

I assume we use polar coordinates, I'm just a bit lost with the inequalities in D
• May 17th 2009, 08:02 PM
NonCommAlg
Quote:

Originally Posted by MrSplashypants1
Hey guys, just having a bit of trouble getting started with this.

Have to use a change of variables to evaluate the integral
$\displaystyle \int\int_{D}sin(x+2y)sin(3x-y)dxdy$
where $\displaystyle D={ (x,y)\in\Re^2 | 0 \leq x+2y \leq (\Pi/2) and 0\leq 3x-y\leq (\Pi/2) }$

I assume we use polar coordinates, I'm just a bit lost with the inequalities in D

substitution is what you're supposed to do here. let $\displaystyle x+2y=u, \ 3x - y= v,$ find the Jacobian of this transformation, etc. you'll end up with a very simple integral.

by the way that's $\displaystyle \frac{\pi}{2}$ not $\displaystyle \frac{\Pi}{2}.$
• May 17th 2009, 11:05 PM
MrSplashypants1
Quote:

Originally Posted by NonCommAlg
substitution is what you're supposed to do here. let $\displaystyle x+2y=u, \ 3x - y= v,$ find the Jacobian of this transformation, etc. you'll end up with a very simple integral.

by the way that's $\displaystyle \frac{\pi}{2}$ not $\displaystyle \frac{\Pi}{2}.$

Alright, so we define
$\displaystyle g: D^* \longrightarrow D$
$\displaystyle (u,v) \longmapsto (x,y)$
where u=x+2y and v=3x-y
Then $\displaystyle D^*$ is the square region bounded by $\displaystyle 0 \leq u \leq (\pi/2) and 0\leq v \leq (\pi/2)$

And so the original integral becomes $\displaystyle \int_0^\frac{\pi}{2}\int_0^\frac{\pi}{2}sin(u)sin( v)|detJ_{g} (u,v)|dudv$

where the determinant = $\displaystyle \frac{1}{7}$

and so the integral comes out as just $\displaystyle \frac{1}{7}$ as well?
• May 18th 2009, 04:51 AM
NonCommAlg
Quote:

Originally Posted by MrSplashypants1
Alright, so we define
$\displaystyle g: D^* \longrightarrow D$
$\displaystyle (u,v) \longmapsto (x,y)$
where u=x+2y and v=3x-y
Then $\displaystyle D^*$ is the square region bounded by $\displaystyle 0 \leq u \leq (\pi/2) and 0\leq v \leq (\pi/2)$

And so the original integral becomes $\displaystyle \int_0^\frac{\pi}{2}\int_0^\frac{\pi}{2}sin(u)sin( v)|detJ_{g} (u,v)|dudv$

where the determinant = $\displaystyle \frac{1}{7}$

and so the integral comes out as just $\displaystyle \frac{1}{7}$ as well?

yes. $\displaystyle \int_0^{\frac{\pi}{2}} \int_0^{\frac{\pi}{2}} \sin u \sin v |\det J| \ du \ dv = \frac{1}{7} \left( \int_0^{\frac{\pi}{2}} \sin u \ du \right)^2=\frac{1}{7}.$
• May 20th 2009, 12:37 AM
MrSplashypants1
Thank you very much sir :)