It is given quarter circle in first quadrant $G=\left\{ \left(x,y\right) \in \mathbb R ^{2}: x\geq 0, y\geq 0, 9\leq x^{2}+y^{2}\leq 49 \right\} $ and $g(x,y)=x^2y^2+4y$.
Calculate integral over $G$ (use polar coordinates): $\int_{G}^{} \! g(x,y)d(x,y) \, $
Idea:
$x=rcos\theta $
$y=rsin\theta$
$\int_{0}^{\frac{\pi}{2} } \! \int_{3}^{7} \! r(r^2cos(\theta)^2 r^2 sin(\theta)^2+4rsin(\theta)) \, dr \, d\theta $
Am I on the good way?
Let continue your good start
$$I=\int_3^7r^5dr\int_0^{\frac{\pi}{2}}\frac{1-\cos(4\theta)}{8}d\theta$$
$$+4\int_3^7r^2dr\int_0^{\frac{\pi}{2}}\sin(\theta)d\theta$$
$$=\frac{\pi}{96}(7^6-3^6)+\frac{4}{3}(7^3-3^3)$$ Yes you can finish it.