Currently between circles

The first term, $\int\int_R 3x\:dA = 0$ because the integrand is odd in $x$ and the region of integration is symmetric under $x \leftrightarrow -x$ .

The second term I evaluate using polar coordinates. $\int\int_R f(x,y)\:dx = \int_1^3 dr\:\int_0^\pi r\:d\theta\:f(r,\theta) \\ = \int_1^3 dr\:\int_0^\pi r\:4r^2\sin^2\theta \\ = 4 \left(\int_1^3 r^3\:dr\right)\left(\int_0^\pi \sin^2\theta\:d\theta\right) \\ = 4\cdot \left.\tfrac14r^4\right|_1^3\cdot \tfrac12\pi \\ = \boxed{40\pi}.$

(Over an interval whose length is a multiple of $\pi$ , the average value of $\sin^2\theta$ is equal to $\tfrac12$ . That makes it easy to evaluate the integral over $\theta$ .)

I did it in polar co-ordinates. As x^2 + y^2 varies from 1 to 9, r must vary from 1 to 3. Theta varies from 0 to pi as we are only dealing with y>/= 0. Therefore it becomes the double integral over the aforementioned limits of 3r^2 cos(theta) + 4r^3cos^2(theta) which comes out as 40 pi or 126 to 3 significant figures.