1. Interpret the region

We need the area of

$$\{(x,y): y^2\ge 2x,\; x^2+y^2\le 4x,\; x\ge 0,\; y\ge 0\}.$$

So the region lies in the first quadrant, inside the circle

$$x^2+y^2=4x$$

and satisfies

$$y^2\ge 2x.$$

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2. Rewrite the curves

(i) Circle

$$x^2+y^2\le 4x$$ $$(x-2)^2+y^2\le 4$$

This is a circle centered at $(2,0)$ with radius $2$.

(ii) Parabola

$$y^2=2x$$

This is a right-opening parabola. Since $y\ge 0$, we use its upper branch.

Condition $y^2\ge 2x$ means the point is to the left of the parabola.

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3. Find points of intersection

Substitute

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

into the circle:

$$\left(\frac{y^2}{2}\right)^2+y^2=4\cdot \frac{y^2}{2}$$ $$\frac{y^4}{4}+y^2=2y^2$$ $$\frac{y^4}{4}-y^2=0$$ $$y^2\left(\frac{y^2}{4}-1\right)=0$$

So,

$$y=0 \quad \text{or} \quad y=2$$

(since $y\ge 0$).

Corresponding $x$ values:

• for $y=0$, $x=0$
• for $y=2$, $x=2$

Thus the curves intersect at

$$(0,0),\quad (2,2).$$

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4. Set up area using horizontal strips

For a fixed $y\in[0,2]$:

• parabola gives left/right boundary condition $x\le \dfrac{y^2}{2}$,
• circle gives $x=2\pm \sqrt{4-y^2}.$

Since we are in the first quadrant and inside the circle, the relevant $x$-interval inside the circle is

$$2-\sqrt{4-y^2}\le x\le 2+\sqrt{4-y^2}.$$

We also need $x\le \dfrac{y^2}{2}$. In the first quadrant, the required part is between

$$x=2-\sqrt{4-y^2}$$

and

$$x=\frac{y^2}{2}, \qquad 0\le y\le 2.$$

Hence area is

$$A=\int_0^2 \left(\frac{y^2}{2}-\left(2-\sqrt{4-y^2}\right)\right)dy.$$

So,

$$A=\int_0^2 \left(\frac{y^2}{2}-2+\sqrt{4-y^2}\right)dy.$$

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5. Evaluate the integral

Split it:

$$A=\int_0^2 \frac{y^2}{2}\,dy - \int_0^2 2\,dy + \int_0^2 \sqrt{4-y^2}\,dy.$$

First term

$$\int_0^2 \frac{y^2}{2}\,dy=\frac12\cdot \left[\frac{y^3}{3}\right]_0^2=\frac12\cdot \frac{8}{3}=\frac{4}{3}.$$

Second term

$$\int_0^2 2\,dy=4.$$

Third term

$$\int_0^2 \sqrt{4-y^2}\,dy$$

is the area of a quarter-circle of radius $2$, hence

$$=\frac14\pi (2)^2=\pi.$$

Therefore,

$$A=\frac{4}{3}-4+\pi=\pi-\frac{8}{3}.$$

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6. Match with the options

$$A=\pi-\frac{8}{3}$$

which is Option D.

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7. Comparison with stored answer

Stored correct answer: D

Our derived answer: D

So they agree.