1. Interpret the curves

The given equations are:

$y^2 = 8x$ $y^2 = 16(3-x)$

Rewrite the second one:

$y^2 = 48 - 16x$

or

$x = 3 - \frac{y^2}{16}$

From the first curve,

$x = \frac{y^2}{8}$

So for a given $y$, the region is between:

• left curve: $x = \frac{y^2}{8}$
• right curve: $x = 3 - \frac{y^2}{16}$

2. Find points of intersection

At intersection,

$\frac{y^2}{8} = 3 - \frac{y^2}{16}$

Multiply by $16$:

$2y^2 = 48 - y^2$

$3y^2 = 48$

$y^2 = 16$

$y = \pm 4$

Now find $x$:

$x = \frac{y^2}{8} = \frac{16}{8} = 2$

So the curves intersect at $(2,4)$ and $(2,-4)$.

3. Set up the area integral

Area between curves using horizontal strips:

$A = \int_{-4}^{4} \left[\left(3 - \frac{y^2}{16}\right) - \left(\frac{y^2}{8}\right)\right] dy$

Simplify:

$A = \int_{-4}^{4} \left(3 - \frac{y^2}{16} - \frac{y^2}{8}\right) dy$

$A = \int_{-4}^{4} \left(3 - \frac{3y^2}{16}\right) dy$

4. Evaluate the integral

Since the integrand is even,

$A = 2\int_{0}^{4} \left(3 - \frac{3y^2}{16}\right) dy$

$A = 2\left[3y - \frac{3}{16}\cdot \frac{y^3}{3}\right]_0^4$

$A = 2\left[3y - \frac{y^3}{16}\right]_0^4$

At $y=4$:

$3(4) - \frac{4^3}{16} = 12 - \frac{64}{16} = 12 - 4 = 8$

So,

$A = 2(8) = 16$

5. Compare with options

The area is

$\boxed{16}$

So the correct option is C.

6. Comparison with stored correct answer

Stored correct answer: C

Derived answer: C

They match.