Area between $ 2 y = 4 \sqrt{x}$, $y = 4$, and $2 y + 4 x = 8 $ Sketch the region enclosed by the curves given below. Then find the area of the region.
$ 2 y = 4 \sqrt{x}$, $y = 4$, and $2 y + 4 x = 8 $
Attempt at solution:
I guess I'm supposed to divide the areas into several parts, and then sum up the areas of those parts.
Wolfram alpha shows the sketched area... and I don't think those are the correct sketched areas, because the answer 14 isn't a correct answer.
So can someone tell me which sketched areas am I even looking at?

Simplify your boundary equations:
$$y = 2 \sqrt{x}$$
$$y = 4$$
$$y = -2x +4$$
Sketch the area. You ought to try hand-sketching it to verify.

Split into two double integrals. 
$$Area = \int_{x=0}^1\int_{y=lower curve}^{higher curve} dydx+ \int_{x=1}^4\int_{y=lower curve}^{higher curve}dydx$$.
UPDATE/EDIT: You ought to have solved it by now. For reference, here's the full integration and solution.
$$= \int_{x=0}^1\int_{y=-2x+4}^{4} dydx+ \int_{x=1}^4\int_{y=2\sqrt{x}}^{4}dydx$$
$$= \int_{x=0}^1 4 - (-2x+4) dx+ \int_{x=1}^4 4- 2\sqrt{x}\ dx$$
$$= 2\int_{x=0}^1 x\ dx+ \int_{x=1}^4 4- 2\sqrt{x}\ dx$$
$$= x^2 \bigg|_0^1 + 4x \bigg|_1^4 - \frac{4}{3}x^{\frac{3}{2}} \bigg|_1^4 $$
$$= 11/3 $$
