The area of the plane region bounded by the curves x + 2y² = 0 and x + 3y² = 1 is equal to
(a) 5/3
(

Q: Why did we integrate with respect to $y$ instead of $x$? Isn't $dx$ the standard way?

Great question! While $dx$ is standard, it is not always the most efficient. If you integrate with respect to $x$, you would have to express $y$ in terms of $x$ (involving square roots) and split the integral at the vertex of the parabolas. By choosing $dy$, we treat the parabolas as functions of $y$, which are simple polynomials. Always choose the path that keeps your algebra clean!

Q: How do I know which curve is $x_{right}$ and which is $x_{left}$?

A simple trick is to pick a test point between your intersection limits. For $y=0$, the first curve gives $x=0$ and the second gives $x=1$. Since $1 > 0$, the curve $x = 1 - 3y^2$ is clearly to the right of $x = -2y^2$ in that region. Always verify your boundaries to avoid a negative area result.

Q: Can I just use the formula for the area of a parabolic segment?

You absolutely can! The area of a parabolic segment is $\frac{2}{3} \times \text{base} \times \text{height}$. However, in JEE Advanced, showing the integration process is often safer and helps you avoid errors if the curves are slightly more complex. Use the formula to verify, but master the integral to be bulletproof.

Q: Why did we use the even function property $\int_{-a}^{a} f(y) dy = 2 \int_{0}^{a} f(y) dy$?

It is a time-saver! Whenever you see symmetric limits like $-1$ to $1$ and an even function (where $y^2$ remains $y^2$ when you replace $y$ with $-y$), using this property allows you to plug in $0$ as a limit. It drastically reduces the chance of arithmetic errors during the final subtraction.

Q: What if I get a negative area? Does that mean I failed?

Not at all! A negative area simply means you subtracted the right curve from the left curve. Area is a magnitude, so if you get $-4/3$, you know the absolute value is $4/3$. Just be mindful of your signs next time, but don't panic—the physics and math are still sound.

Question

The area of the plane region bounded by the curves $x + 2y^2 = 0$ and $x + 3y^2 = 1$ is equal to

Options:

(A) 5/3

(B) 1/3

(C) 2/3

(D) 4/3

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a coordinate plane, looking at two paths carved out by the equations $x + 2y^2 = 0$ and $x + 3y^2 = 1$. At first glance, these look like standard algebraic expressions, but to a mathematician, they are trajectories.

They are parabolas, and they are about to enclose a beautiful, symmetric region that we are going to measure.

### Visualizing the Landscape

Let us first rewrite these equations to understand their geometry. The first, $x = -2y^2$, is a parabola with its vertex at the origin $(0,0)$, opening toward the negative $x$-axis.

The second, $x = 1 - 3y^2$, is its sibling, shifted to the right with its vertex at $(1,0)$, also opening toward the negative $x$-axis. Because both open in the same direction, they will eventually 'trap' a region between them.

### Finding the Meeting Point

To find where these two paths collide, we set their $x$-coordinates equal to each other. We are looking for the $y$-values where the curves intersect:

$$-2y^2 = 1 - 3y^2$$

By moving the terms, we find $3y^2 - 2y^2 = 1$, which simplifies to $y^2 = 1$. This tells us the curves meet at $y = 1$ and $y = -1$.

Substituting these back into our equations, we find the intersection points are $(-2, 1)$ and $(-2, -1)$. We have now defined our vertical boundaries for the integration.

### The Elegance of Horizontal Strips

Here is where we make a strategic choice. We could integrate with respect to $x$, but that would force us to deal with square roots and split the region into two separate integrals.

Instead, let us use horizontal strips of thickness $dy$. This allows us to sweep across the region from $y = -1$ to $y = 1$ in one smooth motion. The area $A$ is defined by the integral of the difference between the right-hand curve and the left-hand curve:

$$A = \int_{-1}^{1} (x_{right} - x_{left}) \, dy$$

Substituting our functions, we get:

$$A = \int_{-1}^{1} \left[ (1 - 3y^2) - (-2y^2) ight] \, dy$$

### The Simplification

Look at the integrand: $(1 - 3y^2) + 2y^2 = 1 - y^2$. This is a beautifully simple quadratic.

We are now integrating $1 - y^2$ from $-1$ to $1$. Because $1 - y^2$ is an even function—meaning it is perfectly symmetric across the $y$-axis—we can use the property $\int_{-a}^{a} f(y) \, dy = 2 \int_{0}^{a} f(y) \, dy$.

This turns our integral into:

$$A = 2 \int_{0}^{1} (1 - y^2) \, dy$$

### Final Calculation

Now, we perform the integration. The integral of $1$ is $y$, and the integral of $y^2$ is $\frac{y^3}{3}$. Applying the limits from $0$ to $1$:

$$A = 2 \left[ y - \frac{y^3}{3} ight]_{0}^{1}$$

Substituting the upper limit $y=1$ gives us $1 - \frac{1}{3} = \frac{2}{3}$. The lower limit $y=0$ gives us $0$.

Thus, the final area is:

$$A = 2 	imes \frac{2}{3} = \mathbf{\frac{4}{3}}$$

There it is. Through careful observation and a strategic choice of integration, we have calculated the area to be **$\frac{4}{3}$ square units**.

Animated Solution for Mathematics - Definite Integration: The area of the plane region bounded by the curves $x + 2 y^{2} = 0$ and $x + 3 y^{2} = 1$ is equal to

Select Answer:

Visualized Solution

Analyzing the First Curve

Analyzing the Second Curve

Finding Intersection Points

Solving for $y$

Finding the $x$ -coordinates

Choosing the Axis of Integration

Setting up the Area Integral

Substituting into the Integral

Simplifying the Integrand

Using Even Function Property

Performing the Integration

Applying the Limits

Final Answer

The Sigma Insight: Area Bounded by Curves

Analyzing the Setup

Visualizing the Landscape

Finding the Meeting Point

The Elegance of Horizontal Strips

The Simplification

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 of the plane region bounded by the curves x+2y2=0 and x+3y2=1 is equal to