The area (in sq. units) of the region $$\{(x, y) : y^2 \geq 2x$$ and $$x^2 + y^2 \leq 4x$$, $$x \geq 0$$, $$y \geq 0\}$$ is

We are asked to find the area of the set of all points $$\,(x,y)\,$$ in the first quadrant that satisfy simultaneously

$$y^2 \ge 2x,\qquad x^2 + y^2 \le 4x,\qquad x \ge 0,\qquad y \ge 0.$$

First, let us interpret each inequality geometrically in the $$xy$$-plane.

1. The parabola. From $$y^2 = 2x$$ we obtain $$x = \dfrac{y^2}{2}.$$ Because the inequality is $$y^2 \ge 2x,$$ all points lie to the left of (or on) this parabola that opens to the right from the origin.

2. The circle. Starting with $$x^2 + y^2 \le 4x,$$ we complete the square in $$x$$:

$$x^2 - 4x + y^2 \le 0$$

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

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

This is the interior (and the boundary) of the circle with centre $$(2,0)$$ and radius $$2.$$ Because of $$x \ge 0,\;y \ge 0,$$ we restrict ourselves to the first quadrant portion of this circle.

3. The required region. The region is therefore the part of the first-quadrant circle that lies to the left of the parabola. Both the circle and the parabola pass through the origin $$O(0,0)$$ and through the point $$A(2,2),$$ since

$$y = 2 \;\;\Longrightarrow\;\; x = \dfrac{2^2}{2} = 2\quad\text{(parabola)},$$

$$(x-2)^2 + y^2 = 0 + 4 = 4 = 2^2\quad\text{(circle)}.$$

Thus for every $$y$$ between $$0$$ and $$2$$ the vertical slice of the desired region extends from the left border of the circle to the parabola. For a fixed $$y$$ (with $$0 \le y \le 2$$)

• Equation of the circle gives

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

The leftmost $$x$$ on the circle is

$$x_{\text{left}} = 2 - \sqrt{\,4 - y^2\,}.$$

• The parabola supplies the right boundary

$$x_{\text{right}} = \dfrac{y^2}{2}.$$

(One can check algebraically that $$x_{\text{right}} \ge x_{\text{left}}$$ for $$0 \le y \le 2,$$ so the slice is indeed non-empty.)

Hence, the horizontal thickness of an elemental strip at height $$y$$ is

$$\bigl[x_{\text{right}} - x_{\text{left}}\bigr] = \dfrac{y^2}{2} \;-\;\Bigl(2 - \sqrt{\,4 - y^2\,}\Bigr) = \dfrac{y^2}{2} - 2 + \sqrt{\,4 - y^2\,}.$$

4. Setting up the definite integral. The area $$A$$ is obtained by integrating this width from $$y = 0$$ to $$y = 2$$:

$$A = \int_{0}^{2}\!\left(\dfrac{y^2}{2} - 2 + \sqrt{\,4 - y^2\,}\right)dy.$$

5. Evaluating each term separately.

(i) Using $$\displaystyle\int y^2\,dy = \dfrac{y^3}{3},$$ we have

$$\int_{0}^{2}\dfrac{y^2}{2}\,dy = \dfrac{1}{2}\left[\dfrac{y^3}{3}\right]_{0}^{2} = \dfrac{1}{2}\left(\dfrac{8}{3} - 0\right) = \dfrac{4}{3}.$$

(ii) For the constant term $$-2,$$

$$\int_{0}^{2}(-2)\,dy = -2[y]_{0}^{2} = -2(2-0) = -4.$$

(iii) The integral $$\displaystyle\int_{0}^{2}\sqrt{\,4 - y^2\,}\,dy$$ represents the area of a quarter-circle of radius $$2,$$ because in polar form $$x^2 + y^2 = r^2$$ and the limits $$y = 0$$ to $$y = 2$$ with $$x \ge 0$$ sweep the first-quadrant sector. The known formula for the area of a sector of angle $$\theta$$ in a circle of radius $$r$$ is $$\dfrac{\theta}{2\pi}\cdot\pi r^2.$$ Here $$\theta = \dfrac{\pi}{2}$$ (a right angle), so

$$\int_{0}^{2}\sqrt{\,4 - y^2\,}\,dy = \dfrac{\pi}{2\pi}\,\pi(2)^2 = \dfrac{1}{4}\,\pi(4) = \pi.$$

(iv) Adding the three results:

$$A = \left(\dfrac{4}{3}\right) + (-4) + \pi = \pi - 4 + \dfrac{4}{3} = \pi - \dfrac{12}{3} + \dfrac{4}{3} = \pi - \dfrac{8}{3}.$$

Thus the required area equals $$\pi - \dfrac{8}{3}$$ square units.

Hence, the correct answer is Option D.