If the area of the region {(x, y) : |4 - x²| ≤ y ≤ x², y ≤ 4, x ≥ 0} is (80√2/α - β), α, β ∈ N, then

Q: Why did we choose to integrate with respect to $y$ instead of $x$?

Integrating with respect to $x$ would force us to split the region into three distinct parts because the lower boundary changes from $y = 4 - x^2$ to $y = x^2 - 4$ at $x = 2$, and the upper boundary changes from $y = x^2$ to $y = 4$. By integrating with respect to $y$, we simplify the boundaries significantly, making the calculation much more elegant and less prone to algebraic errors.

Q: How do I handle the modulus $|4 - x^2|$ in the integral?

The modulus function $|4 - x^2|$ is defined as $4 - x^2$ when $x^2 \le 4$ (i.e., $x \le 2$) and $x^2 - 4$ when $x^2 > 4$ (i.e., $x > 2$). When you invert these to functions of $y$, you get $x = \sqrt{4 - y}$ for the left part of the region and $x = \sqrt{y + 4}$ for the right part. Always check the domain of your variables to ensure you are using the correct branch of the function.

Q: Why does the integral split at $y = 2$?

The intersection point of the curves $y = x^2$ and $y = 4 - x^2$ occurs at $x = \sqrt{2}$, which corresponds to $y = 2$. Below $y = 2$, the left boundary is defined by the curve $x = \sqrt{4 - y}$. Above $y = 2$, the left boundary shifts to the curve $x = \sqrt{y}$. This transition point is critical for setting up the correct limits of integration.

Question

If the area of the region $\{(x, y) : |4 - x^2| \le y \le x^2, y \le 4, x \ge 0\}$ is $(\frac{80\sqrt{2}}{\alpha} - \beta)$, $\alpha, \beta \in \mathbb{N}$, then $\alpha + \beta$ is equal to _______.

Sigma Percentile · Accepted Answer

### The Geometry of Curves: A Journey into Area

Welcome, student! Today, we are going to unravel a beautiful problem in coordinate geometry. It is not just about finding an area; it is about understanding the dance of curves on a Cartesian plane.

We are tasked with finding the area of the region defined by the set:
$$\{(x, y) : |4 - x^2| \le y \le x^2, y \le 4, x \ge 0\}$$

### Phase 1: Visualizing the Landscape

Before we touch a single integral, let us paint the picture. We are restricted to the first quadrant ($x \ge 0$). We have three primary actors here:

1. The upward-opening parabola $y = x^2$.
2. The reflected parabola $y = |4 - x^2|$.
3. The ceiling, a horizontal line $y = 4$.

Imagine the curve $y = x^2$ rising from the origin. Now, consider $y = |4 - x^2|$. For $x < 2$, this is $4 - x^2$, a downward-opening parabola. For $x > 2$, it becomes $x^2 - 4$, which is an upward-opening parabola.

The region is trapped between these curves. It is a puzzle of shapes, and our job is to measure the space they enclose.

### Phase 2: The Strategic Choice

In many JEE problems, the choice of integration variable is the difference between a five-minute solution and a twenty-minute struggle. If we integrate with respect to $x$, we would need to break the region into multiple integrals because the boundaries change their nature at $x = \sqrt{2}$ and $x = 2$.

Instead, let us look at this horizontally. By integrating with respect to $y$, we can define the region using horizontal strips. The right boundary is consistently defined by the curve $x = \sqrt{y + 4}$.

The left boundary is a bit more dynamic. For $y$ between $0$ and $2$, the left boundary is $x = \sqrt{4 - y}$. For $y$ between $2$ and $4$, the left boundary shifts to $x = \sqrt{y}$.

### Phase 3: The Integral Masterclass

The total area $A$ is the integral of the right boundary minus the left boundary. We can express this as:

$$A = \int_{0}^{4} \sqrt{y + 4} \, dy - \int_{0}^{2} \sqrt{4 - y} \, dy - \int_{2}^{4} \sqrt{y} \, dy$$

Let us tackle these one by one. The first integral, $I_1 = \int_{0}^{4} (y + 4)^{1/2} \, dy$, is a straightforward power rule application.

It evaluates to:
$$\left[ \frac{2}{3}(y + 4)^{3/2} \right]_{0}^{4} = \frac{2}{3}(8^{3/2} - 4^{3/2}) = \frac{2}{3}(16\sqrt{2} - 8) = \frac{32\sqrt{2} - 16}{3}$$

Next, $I_2 = \int_{0}^{2} (4 - y)^{1/2} \, dy$. Remember the chain rule; the derivative of $(4 - y)$ is $-1$, so we get a negative sign.

It evaluates to:
$$\left[ -\frac{2}{3}(4 - y)^{3/2} \right]_{0}^{2} = -\frac{2}{3}(2^{3/2} - 4^{3/2}) = -\frac{2}{3}(2\sqrt{2} - 8) = \frac{16 - 4\sqrt{2}}{3}$$

Finally, $I_3 = \int_{2}^{4} y^{1/2} \, dy$. This is:
$$\left[ \frac{2}{3}y^{3/2} \right]_{2}^{4} = \frac{2}{3}(4^{3/2} - 2^{3/2}) = \frac{2}{3}(8 - 2\sqrt{2}) = \frac{16 - 4\sqrt{2}}{3}$$

### Phase 4: The Grand Finale

Now, we combine them: $A = I_1 - I_2 - I_3$. Substituting our values, we get:

$$A = \frac{32\sqrt{2} - 16}{3} - \frac{16 - 4\sqrt{2}}{3} - \frac{16 - 4\sqrt{2}}{3}$$

Combining the terms over the common denominator of $3$:

$$A = \frac{32\sqrt{2} - 16 - 16 + 4\sqrt{2} - 16 + 4\sqrt{2}}{3} = \frac{40\sqrt{2} - 48}{3} = \frac{40\sqrt{2}}{3} - 16$$

The problem asks us to match this to the form $\frac{80\sqrt{2}}{\alpha} - \beta$. We have $\frac{40\sqrt{2}}{3}$, which is equivalent to $\frac{80\sqrt{2}}{6}$.

Thus, $\alpha = 6$ and $\beta = 16$. The final sum is:
$$\alpha + \beta = 6 + 16 = \mathbf{22}$$

Animated Solution for Mathematics - Definite Integration: If the area of the region ${(x, y) : ∣4 - x^{2} ∣ \leq y \leq x^{2}, y \leq 4, x \geq 0}$ is $(\frac{80 2}{α} - β)$ , $α, β \in N$ , then $α + β$ is equal to _______.

Enter Numerical Value:

Visualized Solution

Identifying the Boundaries

Finding Intersection Points

Visualizing the Shaded Region

Setting up the Integral (Horizontal Strips)

The Integral Expression

Evaluating the First Integral

Evaluating the Second Integral

Evaluating the Third Integral

Calculating Total Area

Comparing with Given Form

Final Answer

The Sigma Insight: Area Bounded by Curves

The Geometry of Curves

Phase 1

Phase 2

Phase 3

Phase 4

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