# Area of the region enclosed by three curves

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

• You should post your answer so that we can point out the mistake, if any. – Justpassingby Dec 9 '15 at 19:22
• If you show us your work then we might be able to tell you where you’re going wrong. – amd Dec 9 '15 at 19:22
• the problem is not so clear have you made a picture? – Dr. Sonnhard Graubner Dec 9 '15 at 19:30
• It looks straight forward to me. Exactly WHAT did you do? What answer did you get? What answer do you think you should have gotten? – user247327 Dec 9 '15 at 19:32
• help yourself with wolframalpha.com/input/… – janmarqz Dec 9 '15 at 19:34

$$\text{Area}=\color{blue}{\text{Blue}}+\color{green}{\text{Green}}=\int_{-1}^{1} (4-(3-x)) dx + \int_{1}^{4} (4-2\sqrt x) dx$$
1. The final answer is $\frac{14}{3}$
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 $$=\frac{1}{2}(2\times 2)+\left(3\times 4-\int_{1}^{4}2\sqrt x\ dx\right)$$