Finding area of circle exterior to parabola using double integral. 
Find the area of the circle $x^2 + y^2 = 16$, which is exterior to the parabola $y^2 = 6x$, using integration.

This problem is given in my textbook which asks the question to be solved using integration. This is a calculus - 1 problem.
I'm able to calculate the area using a single integral but I'm just wondering if we could solve it using double integral more easily. I'm not well versed with double integration as I'm just learning Calculus-1.
Anyways, here's what I tried:
First of all I found the area of the circle to be $16\pi$ sq. units.
Now I've to subtract the area of parabola which is common to the area of circle.
For finding the common area in first quadrant, I made the following integral $$\int_{0}^{2\sqrt{3}}\int_{\frac{y^{2}}{6}}^{\sqrt{16-\ y^{2}}}dxdy$$
Now by symmetry the total common area is $$2\int_{0}^{2\sqrt{3}}\int_{\frac{y^{2}}{6}}^{\sqrt{16-\ y^{2}}}dxdy$$
I'm searching for a way so  that I could find the required area using a single "Double integral" itself. How should I select elemental strip? Vertical/Horizontal? Can it be solved in this manner?

 A: There are many ways you can set up the integral to find area. For example, you could set up in the order $dy ~ dx$ to find the area to the right of y-axis and then add the area of the semi-circle to the left of y-axis, which is $8 \pi$.
So the area of the desired region is,
$ \displaystyle 8 \pi + 2 \int_0^2\int_{\sqrt{6x}}^{\sqrt{16-x^2}} ~ dy ~ dx$
Or in polar coordinates,
$\displaystyle 8 \pi + 2 \int_{\pi/3}^{\pi/2} \int_{6 \cot \theta \csc \theta}^{4} r ~ dr ~ d\theta$
Of course, the above uses the formula for the area of the circle, which is $\pi r^2$.
If you want to truly set up one double integral to find the area of the region, one of the ways is to set up the integral in polar coordinates in the order $d\theta$ followed by $dr$ but it is more cumbersome to set up and definitely more difficult to evaluate. Just in case, here is the set up  -
For any value of $r \in (0, 4)$, the circular strip starts at the top half of the parabola and end at the lower half. Writing parabola in polar coordinates using $x = r \cos\theta, y = r \sin\theta$,
$y^2 = 6x \implies r (1 - \cos^2\theta) = 6 \cos\theta$
Solving quadratic in $\cos \theta$,
$ \displaystyle \cos\theta = \frac{-3 + \sqrt{9 + r^2}}{r} ~$ for the upper half of the parabola.
So the integral is,
$\displaystyle 2 \int_0^4 \int_{\arccos ((-3 + \sqrt{9 + r^2}~) /r)}^{\pi} r ~d\theta ~dr $
