# Thread: multintegral question with change of variable

1. ## 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.

2. Originally Posted by lllll
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 $2y = r \sin \theta$.

3. 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.