Question

@ganeshie8

Answer 1

\[\int\limits \int\limits_R \sin(9x^2+4y^2)dA\]
R: Region in 1st quadrant bounded by the ellipse 9x^2+4y^2=1, using transformations.

Answer 2

try u = 3x, v = 2y

Answer 3

please we have to change our coordinate from cartesian to these coordinates:
\[\begin{gathered}
x = \frac{\rho }{3}\cos \theta \hfill \\
y = \frac{\rho }{2}\sin \theta \hfill \\
\end{gathered} \]

Answer 4

Yeah, that should do the trick haha

Answer 5

then our integral becomes:
\[\int {\sin \left( {{\rho ^2}} \right)} \;\rho \;d\rho \;d\theta \]

Answer 6

looks changing to polar is equally fine too ^

Answer 7

oops.. I have made an error:
\[\int {\sin \left( {{\rho ^2}} \right)} \;\frac{\rho }{6}\;d\rho d\theta \]

Answer 8

Cool, thanks :)

Answer 9

thanks! :)