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

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

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

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

    so my value associated to the Jacobian transformation would be:

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

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

    so putting everything together yields:

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

    now this is the part that I get stuck on since $\displaystyle \frac{\sin(u+v)}{\sqrt{u\cdot v}}$ 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 $\displaystyle \iint_R \sin(9x^2+4y^2) \ dA$, bounded by the ellipse $\displaystyle 9x^2+4y^2=1$ in the first quadrant.

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

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

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

    so my value associated to the Jacobian transformation would be:

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

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

    so putting everything together yields:

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

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

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

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