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Math Help - change of variables formula for integrals

  1. #1
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    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!
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    Quote Originally Posted by sfitz View Post
    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
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  3. #3
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    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!
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