# change of variables formula for integrals

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• Apr 8th 2008, 12:32 AM
sfitz
change of variables formula for integrals
Does anyone have any hints on how to figure out a change of variables formula for integrals? There aren't many tips in the book, and I just can't figure this one out. I think the fact that $f(x,y)$ isn't linear is what's messing me up.

Here is the problem:

$\int\int_R\sin(9x^2+4y^2)dA$ where $R$ is the region in the first quadrant bounded by the ellipse $9x^2+4y^2=1$.

I guess I could do $u=9x^2+4y^2$, but what about $v$? What am I missing? Thank you!
• Apr 8th 2008, 03:26 AM
topsquark
Quote:

Originally Posted by sfitz
Does anyone have any hints on how to figure out a change of variables formula for integrals? There aren't many tips in the book, and I just can't figure this one out. I think the fact that $f(x,y)$ isn't linear is what's messing me up.

Here is the problem:

$\int\int_R\sin(9x^2+4y^2)dA$ where $R$ is the region in the first quadrant bounded by the ellipse $9x^2+4y^2=1$.

I guess I could do $u=9x^2+4y^2$, but what about $v$? What am I missing? Thank you!

Mind you, I don't actually know how to do this integral directly, but if we use your substitution your integral is
$\int_0^{1/3}\int_0^{(1/2)\sqrt{1 - 9x^2}} \sin(9x^2+4y^2)~dydx$

For the y integral, take $u = 9x^2 + 4y^2$ and recall that for this integration, x is simply a constant. So $du = 8y~dy$ which means your integral is
$= \int_0^{1/3}\int_{9x^2}^1 \sin(u)~\left [ 8 \cdot \left ( \frac{1}{2} \cdot \sqrt{u - 9x^2} \right ) \right ] ^{-1}~dudx$

Given the result I have a feeling this is not the best way to attack this problem.

I would suggest, perhaps adjusting your variables so that you are integrating over the unit circle rather than an ellipse by using
$u = 3x$
and
$v = 2y$

Then your integral becomes:
$\int_0^{1/3}\int_0^{(1/2)\sqrt{1 - 9x^2}} \sin(9x^2+4y^2)~dydx = \int_0^1\int_0^{\sqrt{1 - u^2}} sin(u + v) \cdot \left ( \frac{\partial (u, v)}{\partial (x, y)} \right ) ~dvdu$
where the quantity in ( ) is the Jacobian determinant. This is probably more along the lines of what you are looking for.

-Dan
• Apr 8th 2008, 02:02 PM
sfitz
OK, that made the integral very easy. So in this case, we're not trying to simplify the integrand, we're trying to simplify the region. Thank you!