The parabolas y² = 4x and x² = 4y divide the square region bounded by the lines x=4, y=4 and the coo

Q: Wait, why do we reject the negative root for $y^2 = 4x$?

Great observation! We are working within the first quadrant of the Cartesian plane, where both $x$ and $y$ are non-negative. Since the region is bounded by the coordinate axes ($x=0, y=0$) and the lines $x=4, y=4$, we only care about the positive branch $y = 2\sqrt{x}$.

Q: Why can't we just use symmetry to say all areas are equal?

Symmetry is a powerful intuition, but in JEE Advanced, you must justify it. The parabolas $y^2 = 4x$ and $x^2 = 4y$ are reflections of each other across the line $y=x$. Because the bounding square is also symmetric about $y=x$, the regions $S_1$ and $S_3$ must be equal. Calculating them confirms this, but the calculation is the proof.

Q: Is the order of integration important for $S_2$?

Absolutely. When finding the area between two curves, you must subtract the 'lower' function from the 'upper' function to ensure the area is positive. In the interval $[0,4]$, $2\sqrt{x}$ is greater than $\frac{x^2}{4}$, so the integrand must be $2\sqrt{x} - \frac{x^2}{4}$.

Question

The parabolas $y^2 = 4x$ and $x^2 = 4y$ divide the square region bounded by the lines $x=4, y=4$ and the coordinate axes. If $S_1, S_2, S_3$ are respectively the areas of these parts numbered from top to bottom; then $S_1 : S_2 : S_3$ is

Options:

(A) 1:2:1

(B) 1:2:3

(C) 2:1:2

(D) 1:1:1

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine a coordinate plane containing a square defined by the vertices $(0,0), (4,0), (4,4),$ and $(0,4)$. The total area of this square is $4 	imes 4 = 16$ square units.

Within this square, two parabolas, $y^2 = 4x$ and $x^2 = 4y$, intersect at the origin $(0,0)$ and the point $(4,4)$. These curves divide the square into three distinct regions: $S_1$ (top), $S_2$ (middle), and $S_3$ (bottom).

### The Bottom Region: $S_3$

The region $S_3$ is bounded below by the $x$-axis ($y=0$) and above by the parabola $x^2 = 4y$, which is equivalent to $y = \frac{x^2}{4}$.

To find the area of $S_3$, we integrate this function from $x=0$ to $x=4$:

$$S_3 = \int_{0}^{4} \frac{x^2}{4} dx$$

Performing the integration:

$$S_3 = \frac{1}{4} \left[ \frac{x^3}{3} ight]_{0}^{4} = \frac{1}{4} \left( \frac{64}{3} - 0 ight) = \frac{16}{3}$$

Thus, the area of $S_3$ is $\frac{16}{3}$ square units.

### The Middle Region: $S_2$

The middle region, $S_2$, is trapped between the upper boundary $y = 2\sqrt{x}$ (derived from $y^2 = 4x$) and the lower boundary $y = \frac{x^2}{4}$. We calculate this by integrating the difference of these functions:

$$S_2 = \int_{0}^{4} \left( 2\sqrt{x} - \frac{x^2}{4} ight) dx$$

Integrating term by term, the integral of $2\sqrt{x}$ is $\frac{4}{3}x^{3/2}$, and the integral of $\frac{x^2}{4}$ is $\frac{x^3}{12}$. Evaluating from $0$ to $4$:

$$S_2 = \left[ \frac{4}{3} x^{3/2} - \frac{x^3}{12} ight]_{0}^{4} = \left( \frac{4}{3} 	imes 8 - \frac{64}{12} ight) = \frac{32}{3} - \frac{16}{3} = \frac{16}{3}$$

Remarkably, the area of $S_2$ is also $\frac{16}{3}$ square units.

### The Top Region: $S_1$

Since the total area of the square is $16$ and the sum of the three regions must equal this total, we can determine $S_1$ using subtraction:

$$S_1 = 16 - (S_2 + S_3)$$

Substituting the known values:

$$S_1 = 16 - \left( \frac{16}{3} + \frac{16}{3} ight) = 16 - \frac{32}{3} = \frac{48 - 32}{3} = \frac{16}{3}$$

### The Revelation

The calculations reveal that $S_1 = \frac{16}{3}$, $S_2 = \frac{16}{3}$, and $S_3 = \frac{16}{3}$.

All three regions possess the exact same area. Consequently, the ratio $S_1 : S_2 : S_3$ is **$1 : 1 : 1$**.

Select Answer:

Visualized Solution

The Square Bounding Box

First Parabola: $x^{2} = 4 y$

Second Parabola: $y^{2} = 4 x$

Regions $S_{1}, S_{2}, S_{3}$

Setup for Area $S_{3}$

Evaluating $S_{3}$

Setup for Area $S_{2}$

Evaluating $S_{2}$

Logic for Area $S_{1}$

Evaluating $S_{1}$

Final Ratio $S_{1} : S_{2} : S_{3}$

The Sigma Insight: Area Bounded by Curves

Analyzing the Setup

The Bottom Region: $S_{3}$

The Middle Region: $S_{2}$

The Top Region: $S_{1}$

The Revelation

Similar Questions

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Let $A_{1}$ be the area of the region bounded by the curves $y = sin x$ , $y = cos x$ and $y$ -axis in the first quadrant. Also, let $A_{2}$ be the area of the region bounded by the curves $y = sin x$ , $y = cos x$ , $x$ -axis and $x = \frac{π}{2}$ in the first quadrant. Then,

Select Answer:

Visualized Solution

The Square Bounding Box

First Parabola: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x2=4y

Second Parabola: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y2=4x

Regions .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S1​,S2​,S3​

Setup for Area .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S3​

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S3​

Setup for Area .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S2​

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S2​

Logic for Area .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S1​

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S1​

Final Ratio .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S1​:S2​:S3​

The Sigma Insight: Area Bounded by Curves

Analyzing the Setup

The Bottom Region: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S3​

The Middle Region: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S2​

The Top Region: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S1​