double integral evaluation Evaluate the integral $\int\int_R(x^2-2y^2) \, dA$ where $R$ is the first quadrant region between the circles of radius $4$ and radius $5$.
I am having troubles with finding the limits of the integral. Thank you!

Just use polar coordinates.
$x^2=r^2\cos^2{\theta};y^2=r^2\sin^2{\theta}$
$x^2-2y^2=r^2(\cos^2{\theta}-2\sin^2{\theta})$
$x^2-2y^2=r^2(1-\sin^2{\theta}-2\sin^2{\theta})=r^2(1-3\sin^2{\theta})$
$x^2-2y^2=r^2(1-\frac{3}{2}+\frac{3}{2}cos2{\theta})=r^2(-\frac{1}{2}+\frac{3}{2}cos2{\theta}))$
$\iint_Rx^2-2y^2\,dx\,dy=-\frac{1}{2}\int_{0}^{\pi/2}\int_{4}^5r^3\,dr\,d\theta +\frac{3}{2}\int_{0}^{\pi/2}\int_{4}^5r^3\cos2{\theta}\,dr\,d\theta$
Note that "first cuadrant" means $\theta$  goes from $0$ to $\pi/2$
