Area between $y = \sqrt{x}$ and $y = 4 - 0.5x$

Question

I have two functions: $x = y^2$ and $x + 2y = 8$ and need to find the area between them, first integrating with respect to $y$, then with respect to $x$. I was able to integrate with respect to $y$ fine and came out with an area of $36$ units.

However, when I integrated with respect to $x$ I got $\frac{148}{3} \approx 49.3$. The $x$ values for the intersection between the graphs of $y = \sqrt{x}$ and $y = 4 - 0.5x$ were $x = 16$ and $x = 4$.

I think the problem might be I am getting confused as to what function is the upper and what is the lower because $\sqrt{x}$ does not extend below the $x$-axis where the second intersection between the two graphs would be.

Any pointers on the correct way to find the area with respect to $x$?

Answer 1

Hint: Sketch the graphs. Maybe you can break down the area between the graphs into areas that you know how to calculate?

Answer 2

HINT: Draw the graph of the curve $y=\sqrt x$ & the given line: $y=4-0.5 x$. The area bounded is divided into two parts which are determined separately hence we get the total area bounded $$=\left|2\int_{0}^{4}\sqrt{x}dx\right|+\left|\int_{4}^{16}(4-0.5x-\sqrt x)dx\right|$$