Change of Basis

**ragnar** · October 30th 2010, 01:24 PM

So I've done a few change of bases and they're hard but I'm getting the hang of them. Now I have this: $\displaystyle \int \int_{R} \sin(9x^{2} + 4y^{2}) \,\, dA$ where $R$ is the region in the first quadrant bounded by the ellipse $9x^{2} + 4y^{2}$ . It can't be any coincidence that the function I'm integrating is almost verbatim the equation of the region I'm integrating over. Still, in all previous examples I've been able to find out what region I've been applying my transformation to by tracing outlines. I can't do that here. Unless I'm supposed to do this by trig functions... Which is a bit daunting. $x = \frac{1}{3}\cos \theta, y = \frac{1}{2} \sin \theta$ ? But then how do I do traces? Set $r = 0$ and get just the point, then set $r = x^{2} + y^{2}$ ? This all seems like shaky ground to me.

**Opalg** · October 31st 2010, 01:11 AM

Originally Posted by ragnar

So I've done a few change of bases and they're hard but I'm getting the hang of them. Now I have this: $\displaystyle \int \int_{R} \sin(9x^{2} + 4y^{2}) \,\, dA$ where $R$ is the region in the first quadrant bounded by the ellipse $9x^{2} + 4y^{2}$ . It can't be any coincidence that the function I'm integrating is almost verbatim the equation of the region I'm integrating over. Still, in all previous examples I've been able to find out what region I've been applying my transformation to by tracing outlines. I can't do that here. Unless I'm supposed to do this by trig functions... Which is a bit daunting. $x = \frac{1}{3}\cos \theta, y = \frac{1}{2} \sin \theta$ ? But then how do I do traces? Set $r = 0$ and get just the point, then set $r = x^{2} + y^{2}$ ? This all seems like shaky ground to me.

The equation for the ellipse is incomplete. Should it be $9x^{2} + 4y^{2}= 1$ ?

For the substitution, take $x = \frac{1}{3}r\cos \theta$ and $y = \frac{1}{2} r\sin \theta$ . Then $dA = \frac16rdrd\theta$ , and $\displaystyle \iint_{R} \sin(9x^{2} + 4y^{2}) \,dA = \int_0^{\pi/2}\!\!\!\int_0^1\!\!\sin r^2\cdot\tfrac16rdrd\theta$ .

**HallsofIvy** · October 31st 2010, 04:35 AM

You are changing "coordinates", not "bases". "Bases" have to do with linear algebra. Also I am not sure what you mean by "tracing outlines".

To get Opalg's " $dA= \frac{1}{6}rdrd\theta$ ", write the "position vector" of a point as $\vec{R}= x\vec{i}+ y\vec{j}= \frac{1}{3}rcos(\theta)\vec{i}+ \frac{1}{2}rsin(\theta)\vec{j}$ . Then the two derivatives $\vec{R}_r= \frac{1}{3}cos(\theta)\vec{i}+ \frac{1}{2}sin(\theta)\vec{j}$ and $\vec{R}_\theta= -\frac{1}{3}rsin(\theta)\vec{i}+ \frac{1}{2}rcos(\theta)\vec{j}$ are in the tangent plane and their cross product, $\frac{1}{6}r\vec{k}$ is perpendicular to the plane and its length gives the "differential of area", $\frac{1}{6}r drd\theta$ .

**ragnar** · November 1st 2010, 02:08 PM

So thank you both for your responses. The only thing I don't totally get is why my limits of integration are what they are. I sort of get that I'm taking the coordinates from the rectangular to the polar, so that's a transformation. I guess because I'm using $\frac{1}{3} \cos x$ , it squishes the shape of the ellipse by just as much as $9 \cos x$ stretched it, thus it's squished back into a perfect circle? Thus I need only integrate over the quarter of the unit circle in the first quadrant... right?

**HallsofIvy** · November 2nd 2010, 05:16 AM

Essentially, yes. More specifically, $\theta$ is the angle a line from the origin to the point (x, y) makes with the x-axis. To cover the first quadrant, $\theta$ goes from 0 to $\pi/2$ .

Since $x= \frac{1}{3}rcos(\theta)$ and $y= \frac{1}{2}r sin(\theta)$ , $3x= r cos(\theta)$ and $2y= r sin(\theta)$ so that $9x^2+ 4y^2= r^2$ which equals 1 on the ellipse. Since r must cover all of the ellipse at any one angle, r must go from the center of the ellipse, (0, 0), where r= 0, to the boundary, the ellipse itself, where $r^2= 1$ and, since r is non-negative, r= 1.

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