Where where f is a continous function on $[0,1]$ and The region R is bounded by the ellipse $5x^2 + 3y^2 = 15$
I'm confused how to approach this, I just recently started with double integration and i have only completed the topic Change of order of integration and now i moved onto Change of variable. So far i am familiar with question in change of variable where the transformation of the plane is given clearly for example $$T:ℝ^2\rightarrow \ ℝ^2,\ T\left(x,y\right)=\left(u+av,v\right)$$
In such cases like this transforming from xy-plane to uv-plane is fairly easy for me as the instructions are given clearly, but the question at hand is a bit confusing since its a little unclear to me. The only progress i made is draw the region R which is attached below:
Any help to walk me through the derivation is greatly appreciated
I think i figured out how to solve this, I'm simply posting the answer in case someone else is also stuck in the same question:
First we can use polar coordinate system to solve this, our region R is enclosed in the ellipse given by $5x^2 +3y^2 =15$ rewriting it as $\frac{x^2}{3}+\frac{y^2}{5}=1$, take the substitution as : $$x=r\sqrt{\ 3}\cos \theta, y= r\sqrt{\ 5}\sin \theta $$ putting it into the ellipse equation we get: $r=1$ and taking the Jacabian we get $J\left(r,\theta \right) = \left|\begin{array} {} \frac {dx}{dr} & \frac {dx}{d\theta}\\ \frac {dy}{dr} & \frac {dy}{d\theta}\end{array}\right|=r\sqrt{\ 15}$
$\int _0^1\ \int _0^{2\pi}f\left(\sqrt{\ r^2}\ \right)r\sqrt{\ 15}d\theta dr$
$2\pi \sqrt{\ 15}\int _0^1\ f\left(\sqrt{\ r^2}\ \right)rd\theta dr$