The area (in sq. units) of the region {(x, y) : y² ≥ 2x and x² + y² ≤ 4x, x ≥ 0, y ≥ 0} is:
(a) π -

Q: Wait, why do we reject the negative root when solving for y?

Great observation! The problem explicitly states $y \ge 0$. This constraint restricts our region to the first quadrant, so we only care about the positive branch of the parabola and the upper semicircle.

Q: Why can't we just use the area of a full circle?

You absolutely can use the circle's properties, but remember that the integral $\int_{0}^{2} \sqrt{4x - x^2} dx$ only represents the area under the upper arc of the circle from $x=0$ to $x=2$. Geometrically, this is exactly one-quarter of the circle's total area, which is $\frac{1}{4} \pi r^2$. Since $r=2$, the area is $\pi$.

Q: Is there a trick to integrating $\sqrt{4x - x^2}$?

You could use trigonometric substitution ($x = 2 + 2\sin\theta$), but that is the long way! Recognizing the geometric shape as a quarter-circle is the 'JEE-smart' way to save time and avoid complex integration.

Question

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

Options:

(A) $\pi - \frac{4\sqrt{2}}{3}$

(B) $\frac{\pi}{2} - \frac{2\sqrt{2}}{3}$

(C) $\pi - \frac{4}{3}$

(D) $\pi - \frac{8}{3}$

Sigma Percentile · Accepted Answer

### Analyzing the Geometry of the Curves

We begin by examining the two given curves on the coordinate plane. The first is the parabola defined by $y^2 = 2x$, which opens to the right with its vertex at the origin.

The second curve is the circle $x^2 + y^2 = 4x$. By completing the square, we rewrite the equation as:

$$(x-2)^2 + y^2 = 2^2$$

This reveals a circle centered at $(2, 0)$ with a radius of $r = 2$.

### Finding the Intersection Points

To determine the region bounded by these curves, we identify their points of intersection. We substitute $y^2 = 2x$ into the circle's equation:

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

This simplifies to the quadratic equation $x^2 - 2x = 0$, which yields solutions at $x = 0$ and $x = 2$.

At $x = 2$, the parabola gives $y^2 = 4$, implying $y = 2$ in the first quadrant. Thus, the curves intersect at $(0, 0)$ and $(2, 2)$.

### The Integration Strategy

The area of the region trapped between the curves is found by integrating the difference between the upper boundary (the circle) and the lower boundary (the parabola). The area $A$ is given by:

$$A = \int_{0}^{2} (\sqrt{4x - x^2} - \sqrt{2x}) \, dx$$

We split this into two distinct integrals to simplify the calculation.

### Evaluating the Components

The first integral, $\int_{0}^{2} \sqrt{4x - x^2} \, dx$, represents the area under the upper arc of the circle. Recognizing this as a quarter-circle with radius $r = 2$, we calculate:

$$Area_{circle} = \frac{1}{4} \pi (2)^2 = \pi$$

The second integral, $\int_{0}^{2} \sqrt{2x} \, dx$, is solved using the power rule:

$$\int_{0}^{2} \sqrt{2} x^{1/2} \, dx = \sqrt{2} \left[ \frac{2}{3} x^{3/2} \right]_{0}^{2}$$

Evaluating this at the limits, we obtain:

$$\sqrt{2} \cdot \frac{2}{3} \cdot (2)^{3/2} = \sqrt{2} \cdot \frac{2}{3} \cdot 2\sqrt{2} = \frac{8}{3}$$

### Final Calculation

Subtracting the area under the parabola from the area under the circle, we arrive at the final result:

**Area** $= \pi - \frac{8}{3}$

This result represents the precise area of the region bounded by the two curves. By visualizing the geometry first, we successfully navigated the intersection of algebra and calculus.

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

Select Answer:

Visualized Solution

Visualizing the Region

Analyzing the Circle

Finding Intersection Points

Solving for Intersection

Defining the Required Area

Area under the Circle

Computing Circle Area

Area under the Parabola

Computing Parabola Area

Final Result

The Sigma Insight: Area Bounded by Curves

Analyzing the Geometry of the Curves

Finding the Intersection Points

The Integration Strategy

Evaluating the Components

Final Calculation

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Animated Solution for Mathematics - Definite Integration: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The area (in sq. units) of the region {(x,y):y2≥2x and x2+y2≤4x,x≥0,y≥0} is: