Area of the region enclosed by three curves

Question

find the area between:

$$\begin{array}{} y=4\\ y=2\sqrt{x}\\ y=3-x \end{array}$$

I have found the intersection points between the curves to find the interval but even while doing it with respect to x or to y, my answer is always wrong. please help

Answer 1

HINT: Draw a diagram indicating the required area bounded by $y=4$, $y=2\sqrt x$ & $y=3-x$. Divide the bounded region into two smaller regions, then one should get area bounded $$=\frac{1}{2}(2\times 2)+\left(3\times 4-\int_{1}^{4}2\sqrt x\ dx\right) $$

Answer 2

Hints

1. Take a look at this picture

2. You should compute

$$\text{Area}=\color{blue}{\text{Blue}}+\color{green}{\text{Green}}=\int_{-1}^{1} (4-(3-x)) dx + \int_{1}^{4} (4-2\sqrt x) dx$$

3. The final answer is $\frac{14}{3}$