1. Interpret the region

We need the area of

$$\{(x,y): |x-y|\le y \le 4\sqrt{x}\}.$$

So the point $(x,y)$ must satisfy both:

• $|x-y|\le y$
• $y\le 4\sqrt{x}$

Since $4\sqrt{x}$ is real, we must have $x\ge 0$.

Also, from $|x-y|\le y$, we must have $y\ge 0$.

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2. Simplify the inequality $|x-y|\le y$

Because $y\ge 0$, we can write

$$-y \le x-y \le y.$$

Add $y$ throughout:

$$0 \le x \le 2y.$$

Equivalently,

$$y \ge \frac{x}{2}.$$

So the region is described by

$$\frac{x}{2} \le y \le 4\sqrt{x}, \quad x\ge 0.$$

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3. Find the intersection points

For the region to exist, we need

$$\frac{x}{2} \le 4\sqrt{x}.$$

Solve:

$$x \le 8\sqrt{x}.$$

Let $t=\sqrt{x}\ge 0$. Then $x=t^2$, so

$$t^2 \le 8t \implies t(t-8)\le 0.$$

Since $t\ge 0$, this gives

$$0\le t\le 8.$$

Hence

$$0\le x\le 64.$$

The curves intersect at:

• $x=0 \Rightarrow y=0$
• $x=64 \Rightarrow y=32$

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4. Set up the area integral

Area between the upper curve $y=4\sqrt{x}$ and lower curve $y=\frac{x}{2}$ from $x=0$ to $x=64$ is

$$A=\int_0^{64}\left(4\sqrt{x}-\frac{x}{2}\right)dx.$$

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5. Evaluate the integral

$$A=\int_0^{64}4x^{1/2}\,dx - \int_0^{64}\frac{x}{2}\,dx.$$

First term:

$$\int_0^{64}4x^{1/2}dx = 4\cdot \frac{2}{3}x^{3/2}\Big|_0^{64} = \frac{8}{3}(64)^{3/2}.$$

Now,

$$64^{3/2}=(\sqrt{64})^3=8^3=512.$$

So

$$\frac{8}{3}\cdot 512=\frac{4096}{3}.$$

Second term:

$$\int_0^{64}\frac{x}{2}dx = \frac{1}{2}\cdot \frac{x^2}{2}\Big|_0^{64} =\frac{64^2}{4}=\frac{4096}{4}=1024.$$

Therefore,

$$A=\frac{4096}{3}-1024 =\frac{4096-3072}{3} =\frac{1024}{3}.$$

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6. Check options

The area is

$$\boxed{\frac{1024}{3}}.$$

So the correct option is D.

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7. Comparison with stored answer

Stored correct answer: D

Our derived answer: D

They agree.