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Math Help - multintegral question with change of variable

  1. #1
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    multintegral question with change of variable

    I need a little help with the following question:

    find the region \iint_R \sin(9x^2+4y^2) \ dA, bounded by the ellipse 9x^2+4y^2=1 in the first quadrant.

    I drew out the bounds which gave me a function where 0 \leq x \leq \frac{1}{3} and 0 \leq y \leq \frac{1}{2} .

    now if I let u=9x^2 and v= 4y^2 I would get:

    x=\frac{1}{3}\sqrt{u} and y =\frac{1}{2}\sqrt{v}

    so my value associated to the Jacobian transformation would be:

    \frac{1}{24}\cdot \frac{1}{\sqrt{u\cdot v}}

    the bounds associated with both u and v are 0\leq u \leq 1 and 0\leq y \leq 1-u

    so putting everything together yields:

    \int_{u=0}^{u=1} \int_{v=0}^{v=1-u} \sin(u+v) \cdot \bigg{|}\frac{1}{24}\cdot \frac{1}{\sqrt{u\cdot v}} \bigg{|} \ dv \ du

    now this is the part that I get stuck on since \frac{\sin(u+v)}{\sqrt{u\cdot v}} doesn't integrate nicely, and checking out my work on maple gives me a bizarre result.
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  2. #2
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    Quote Originally Posted by lllll View Post
    I need a little help with the following question:

    find the region \iint_R \sin(9x^2+4y^2) \ dA, bounded by the ellipse 9x^2+4y^2=1 in the first quadrant.

    I drew out the bounds which gave me a function where 0 \leq x \leq \frac{1}{3} and 0 \leq y \leq \frac{1}{2} .

    now if I let u=9x^2 and v= 4y^2 I would get:

    x=\frac{1}{3}\sqrt{u} and y =\frac{1}{2}\sqrt{v}

    so my value associated to the Jacobian transformation would be:

    \frac{1}{24}\cdot \frac{1}{\sqrt{u\cdot v}}

    the bounds associated with both u and v are 0\leq u \leq 1 and 0\leq y \leq 1-u

    so putting everything together yields:

    \int_{u=0}^{u=1} \int_{v=0}^{v=1-u} \sin(u+v) \cdot \bigg{|}\frac{1}{24}\cdot \frac{1}{\sqrt{u\cdot v}} \bigg{|} \ dv \ du

    now this is the part that I get stuck on since \frac{\sin(u+v)}{\sqrt{u\cdot v}} doesn't integrate nicely, and checking out my work on maple gives me a bizarre result.
    You should make the change of variable 3x = r \cos \theta and 2y = r \sin \theta.
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  3. #3
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    This method is essentially what mr fantastic posted above.

    We have that R=\left\{ (x,y)\in {{\mathbb{R}}^{2}}|9{{x}^{2}}+4{{y}^{2}}\le 1 \right\}.

    Now make the subsitutions (x,y)=\left( \frac{u}{3},\frac{v}{2} \right) so that dx\,dy=\frac{du\,dv}{6}, so R will be transfered to the unit circle u^2+v^2\le1. Now just do the polar transformation and you'll end up with the classic bounds 0\le r\le1 and 0\le\varphi\le2\pi and we're done.
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