The parabola y² = 4x divides the area of the circle x² + y² = 5 in two parts. The area of the smalle

Q: Why do we split the integral at $x=1$?

Because the upper boundary of the region changes. From $x=0$ to $x=1$, the boundary is the parabola. From $x=1$ to $x=\sqrt{5}$, the boundary is the circle. Calculus requires a single function definition for each interval to compute the area correctly.

Q: Why do we multiply by 2?

Symmetry! The region is symmetric about the x-axis. Calculating the area of the top half and doubling it is a classic JEE trick to save time and reduce the risk of calculation errors.

Q: How do we handle the inverse trig identity?

Remember that $\frac{\pi}{2} - \sin^{-1} x = \cos^{-1} x$. This is a powerful tool to simplify your final answer to match the given options. Always look for ways to condense your expression using these fundamental identities.

Question

The parabola $y^2 = 4x$ divides the area of the circle $x^2 + y^2 = 5$ in two parts. The area of the smaller part is equal to:

Options:

(A) $\frac{1}{3} + 5 \sin^{-1}(\frac{2}{\sqrt{5}})$

(B) $\frac{1}{3} + \sqrt{5} \sin^{-1}(\frac{2}{\sqrt{5}})$

(C) $\frac{2}{3} + 5 \sin^{-1}(\frac{2}{\sqrt{5}})$

(D) $\frac{2}{3} + \sqrt{5} \sin^{-1}(\frac{2}{\sqrt{5}})$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler on the path to JEE excellence! Today, we are not just solving a problem; we are choreographing a dance between two fundamental shapes: the parabola and the circle.

We begin with a circle defined by $x^2 + y^2 = 5$, which has a radius of $\sqrt{5}$. We also have a parabola defined by $y^2 = 4x$, which is a curve opening to the right.

To find the area of the region enclosed by these curves, we must first determine their points of intersection. We substitute $y^2 = 4x$ into the circle's equation:

$$x^2 + 4x - 5 = 0$$

This quadratic factors into $(x + 5)(x - 1) = 0$. Since the parabola only exists for $x \ge 0$, we discard the negative root and identify the intersection at $x = 1$.

### The Calculus of Slicing

To calculate the area, we must split the integral because the "ceiling" of the region changes at the intersection point. From $x = 0$ to $x = 1$, the boundary is the parabola $y = \sqrt{4x}$.

From $x = 1$ to $x = \sqrt{5}$, the boundary is the circle $y = \sqrt{5 - x^2}$. Because the region is symmetric about the $x$-axis, we calculate the area of the top half and multiply by two.

The total area $A$ is given by the following integral setup:

$$A = 2 \left[ \int_{0}^{1} \sqrt{4x} \, dx + \int_{1}^{\sqrt{5}} \sqrt{5-x^2} \, dx ight]$$

### The Integration

First, we evaluate the area under the parabola. Applying the power rule:

$$2 \int_{0}^{1} 2\sqrt{x} \, dx = 4 \left[ \frac{x^{3/2}}{3/2} ight]_{0}^{1} = \frac{8}{3}$$

Next, we integrate the circular component using the standard form $\int \sqrt{a^2 - x^2} \, dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}(\frac{x}{a})$. With $a^2 = 5$, we evaluate from $1$ to $\sqrt{5}$:

$$\left[ \frac{x}{2}\sqrt{5-x^2} + \frac{5}{2}\sin^{-1}\left(\frac{x}{\sqrt{5}}ight) ight]_{1}^{\sqrt{5}}$$

Evaluating at the limits, the upper limit $\sqrt{5}$ yields $\frac{5}{2} \sin^{-1}(1) = \frac{5\pi}{4}$. The lower limit $1$ yields $\frac{1}{2}(2) + \frac{5}{2}\sin^{-1}(\frac{1}{\sqrt{5}}) = 1 + \frac{5}{2}\sin^{-1}(\frac{1}{\sqrt{5}})$.

### Final Calculation

Combining these results and multiplying by the factor of $2$ outside the brackets:

$$A = 2 \left[ \frac{4}{3} + \left( \frac{5\pi}{4} - 1 - \frac{5}{2}\sin^{-1}\left(\frac{1}{\sqrt{5}}ight) ight) ight]$$

Simplifying the constants, we obtain $\frac{8}{3} - 2 + \frac{5\pi}{2} - 5\sin^{-1}(\frac{1}{\sqrt{5}})$, which reduces to $\frac{2}{3} + 5(\frac{\pi}{2} - \sin^{-1}(\frac{1}{\sqrt{5}}))$.

Using the identity $\frac{\pi}{2} - \sin^{-1}(	heta) = \cos^{-1}(	heta)$, we get $\frac{2}{3} + 5\cos^{-1}(\frac{1}{\sqrt{5}})$. Converting $\cos^{-1}(\frac{1}{\sqrt{5}})$ to $\sin^{-1}(\frac{2}{\sqrt{5}})$, we reach the final result:

**Area** $= \frac{2}{3} + 5\sin^{-1}\left(\frac{2}{\sqrt{5}}ight)$

Animated Solution for Mathematics - Definite Integration: The parabola $y^{2} = 4 x$ divides the area of the circle $x^{2} + y^{2} = 5$ in two parts. The area of the smaller part is equal to:

Select Answer:

Visualized Solution

Visualizing the Curves

Identifying the Target Region

Finding Intersection Points - Setup

Solving for x

Splitting the Area

Setting up the Integrals

Integrating the Parabola Part

Integrating the Circle Part - Formula

Applying the Formula

Evaluating the Limits

Circle Area Result

Combining the Areas

Trigonometric Simplification

Final Conversion

The Final Answer

The Sigma Insight: Area Bounded by Curves

Analyzing the Setup

The Calculus of Slicing

The Integration

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 parabola y2=4x divides the area of the circle x2+y2=5 in two parts. The area of the smaller part is equal to: