Finding the area of a region? I sketched out the graph to get this figure but I can't seem to find the area of the shaded region... would one Y = 4 and the other Y = 8? 
I, in all honesty, am so flabbergasted with this question, any guidance will be much appreciated. I got 28/3 for the first integral (0 to 4) and 292/3 for the second integral (4 to 16)... I feel like I'm doing this wrong.
Okay, I'm going to be honest, can someone please solve this for me? I got 320/3 as my final answer and yet it's still wrong, I'm just about given up at this point, it's literally my last question of this semester.
 A: HINT Find the point of intersection, set the limits $a = 0$ and $b =$ the point you found, set up the integral with those limits and the function $f(x) - g(x)$, and set up another integral from $b$ to $16$ with $g(x) - f(x)$.
A: You need to set up two double integrals.
The first one:
$\int_0^4 \int_{.5x}^ \sqrt{x} dydx$
The second one:
$\int_4^{16} \int_\sqrt{x}^{.5x} dydx$
The computation is straightforward.
A: The area between two curves given by functions $f(x)$ and $g(x)$ is given by,
$$ \int_a^b \vert f(x)-g(x) \vert dx = \int_a^b \textbf{Top Function} - \textbf{Bottom Function }\mathrm{d}x$$
A common mistake students make when trying to compute this is that they don't change the order of the subtraction when the "Top Function" changes from $f(x)$ to $g(x)$. This happens whenever the two curves cross eachother. In your case they seem to cross at $x=4$ so that you need to evaluate an integral from $0$ to $4$ and then another one from $4$ to $16$ always making sure to have to top function come first in the subtraction.
A: We want,
$$\left(\int_{0}^{4} \sqrt{x}\,\,dx-\int_{0}^{4}\dfrac12x\,\,dx\right)+\left(\int_{4}^{16}\dfrac12x\,\,dx-\int_{4}^{16}\sqrt{x}\,\,dx\right)$$
where the first term two terms give the area from $0$ to $4$ and the second term gives it from $4$ to $16$.
The expression becomes: $\left(\dfrac23x^{3/2}-\dfrac14x^2\right)\left.\right|_0^4+\left(\dfrac14x^2-\dfrac23x^{3/2}\right)\left.\right|_4^{16}$
