Question

Evaluate the given integral by making an appropriate change of variables, where R is the region in the first quadrant bounded by the ellipse 64x2 + 81y2 = 1. $ L=\iint_{R} {\color{red}9} \sin ({\color{red}384} x^{2} + {\color{red}486} y^{2})\,dA $.

Answer 1

$\displaystyle\iint_R\sin(384x^2+486y^2)\,\mathrm dA$

Notice that Given that $R$ is an ellipse, consider a conversion to polar coordinates:

$\begin{cases}x(r,\theta)=\frac r8\cos\theta\\y(r,\theta)=\frac r9\sin\theta\end{cases}$

The Jacobian for this transformation is

$J=\begin{bmatrix}\frac18\cos\theta&-\frac r8\sin\theta\\\frac19\sin\theta&\frac r9\cos t\end{bmatrix}$

with determinant $\det J=\frac r{72}$

Then the integral in polar coordinates is

$\displaystyle\frac1{72}\int_0^{\pi/2}\int_0^1\sin(6r^2\cos^2t+6r^2\sin^2t)r\,\mathrm dr\,\mathrm d\theta=\int_0^{\pi/2}\int_0^1r\sin(6r^2)\,\mathrm dr\,\mathrm d\theta=\boxed{\frac{\pi\sin^23}{864}}$

where you can evaluate the remaining integral by substituting $s=6r^2$ and $\mathrm ds=12r\,\mathrm dr$.