What is the most efficient strategy to compute the area of the following region? $y=x^2-16$, $3y-16x=0$, $y=0$ for $y \geq 0$ Please keep in mind, I have included the original question as an image, for any references. So here is what I know: I know I am supposed to set the two equations equal to each other, and then integrate. But for the second equation, $3y-16x-16=0$ am I supposed to solve for $y$, and then set it equal to the first equation? If so, I get some weird numbers, and I am not sure if there is a more efficient way to do this. I know I integrate eventually. Also, what do I do with the $y=0$? Is that for the integration part?

Also, I'm new here, so I apologize in advance for any misconceptions I have, or for any formatting issues. Please notify me of any issues before choosing to downvote me.
You have been supplied 3 curves.

You should plot the 3 curves.  And you should algebraically find the points of intersection of the curves.

Then based on the shape of the curve, make a decision of whether it will be easier to integrate with respect to x or with respect to y.  Which curve will make your upper bound, your lower bound, and what will be your interval to integrate.

Finally you will integrate the result.

$y = x^2 - 16$ is a parabola.
$3y-16 x -16 = 0$ is a line
$y = 0$ is also a line.

$y = 0$ intercepts $y = x^2 - 16$ at $(-4,0),(4,0)$

$3y-16 x-16 = 0$ intercepts $y=0$ at (-1,0).

$y = x^2 - 16$ intercepts $3y-16 x -16= 0$

$y = \frac {16}{3} (x+1)\\
\frac {16}{3} (x+1) = x^2 - 16\\
x^2 - \frac {16}3 x - \frac {64}{3} = 0\\
x = \frac 83 + \frac {\sqrt {64+9\cdot 64}}{3} = \frac {8 + 8\sqrt {10}}{3}\\
y = \frac {176 + 128\sqrt {10}}{9}$

We only care about one of the points of intersection as $y&lt;0$ at the other.

The region is vaguely triangular.  If you integrate with respect to x, you are going to need to break the region in two, if you integrate with respect to y, you do not.

$\int_0^{\frac {176 + 128\sqrt {10}}{9}} \sqrt {y+16} - \frac {3}{16} y + 1 \ dy$

or

$\int_1^4 \frac {16}{3} (x+1)\ dx + \int_4^{\frac {8 +8 \sqrt{10}}{3}} \frac {16}{3}(x+1) - (x^2 - 16)\ dx$
A segment of a parabola is known to occupy two thirds of its bounding rectangle.

By solving a quadratic equation, the intersection of the parabola and the straight line is at coordinates $\left(-\dfrac83,-\dfrac{80}9\right)$.

We can decompose the given domain in a large parabolic segment on the right (of area $\dfrac23\cdot4\cdot16$) and a smaller on the left (of area $\dfrac23\cdot\dfrac83\cdot\dfrac{64}9)$, plus a trapezoid (of area $\dfrac12\left(\dfrac83+1\right)\dfrac{80}9$).



Total: $\dfrac{5800}{81}$.