The area of the region A = {(x, y) : 4x² + y² ≤ 8 and y² ≤ 4x} is:

Q: Wait, why do we reject the negative root $x = -2$ when solving the intersection?

That is a fantastic question! Always look back at the original equations. We have the parabola $y^2 = 4x$. Since $y^2$ is always non-negative (for real $y$), $4x$ must also be non-negative. This forces $x \ge 0$. The value $x = -2$ would imply $y^2 = -8$, which is impossible in the real plane. Always check your domain constraints before proceeding!

Q: Why do we have to split the integral at $x = 1$? Can't we just integrate one function?

Think of the boundary as a 'ceiling'. If you walk along the $x$-axis from $0$ to $\sqrt{2}$, look up. From $0$ to $1$, your ceiling is the parabola $y = 2\sqrt{x}$. But once you cross $x = 1$, the parabola is no longer the upper boundary; the ellipse $y = \sqrt{8 - 4x^2}$ takes over. If you don't split the integral, you are essentially calculating the area under the wrong curve for part of the journey.

Q: Is it possible to integrate with respect to $y$ instead of $x$?

Yes, absolutely! You could express $x$ in terms of $y$ for both curves ($x = y^2/4$ and $x = \sqrt{2 - y^2/4}$) and integrate with respect to $y$. However, you would still need to split the integral because the 'right-hand' boundary changes. Often, one variable is easier to integrate than the other. In this case, $x$ is quite straightforward, but feel free to try $y$ as a practice exercise to see which one you find more intuitive!

Question

The area of the region $A = \{(x, y) : 4x^2 + y^2 \le 8 	ext{ and } y^2 \le 4x\}$ is:

Options:

(A) $\pi + 4$

(B) $\pi + \frac{2}{3}$

(C) $\frac{\pi}{2} + 2$

(D) $\frac{\pi}{2} + \frac{1}{3}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are tasked with finding the area of the region $A$ defined by the inequalities $4x^2 + y^2 \le 8$ and $y^2 \le 4x$.

The first inequality represents the interior of an ellipse. Rewriting it as:

$$ \frac{x^2}{2} + \frac{y^2}{8} = 1 $$

We identify an ellipse centered at the origin with semi-axes $a = \sqrt{2}$ and $b = \sqrt{8} = 2\sqrt{2}$.

The second inequality, $y^2 \le 4x$, represents the interior of a parabola opening to the right. The region $A$ is the intersection of these two interiors.

### The Intersection

To find the points of intersection, we substitute $y^2 = 4x$ into the ellipse equation $4x^2 + y^2 = 8$:

$$ 4x^2 + 4x - 8 = 0 \Rightarrow x^2 + x - 2 = 0 $$

Factoring the quadratic equation gives $(x + 2)(x - 1) = 0$. This yields potential solutions at $x = 1$ and $x = -2$.

Since the parabola $y^2 = 4x$ requires $x \ge 0$, we reject $x = -2$. At $x = 1$, we find $y^2 = 4$, which gives $y = \pm 2$. Thus, the curves intersect at $(1, 2)$ and $(1, -2)$.

### Symmetry and Strategy

Both curves are symmetric about the $x$-axis. We can calculate the area in the first quadrant and multiply the result by $2$.

In the first quadrant, the upper boundary changes at $x = 1$. From $x = 0$ to $x = 1$, the boundary is the parabola $y = 2\sqrt{x}$. From $x = 1$ to $x = \sqrt{2}$, the boundary is the ellipse $y = \sqrt{8 - 4x^2}$.

### The Execution

The total area $A$ is given by $2 	imes (	ext{Area}_1 + 	ext{Area}_2)$.

For the first part, the area under the parabola is:

$$ 	ext{Area}_1 = \int_{0}^{1} 2\sqrt{x} \, dx = 2 \left[ \frac{x^{3/2}}{3/2} ight]_0^1 = \frac{4}{3} $$

For the second part, the area under the ellipse is:

$$ 	ext{Area}_2 = \int_{1}^{\sqrt{2}} \sqrt{8 - 4x^2} \, dx = 2 \int_{1}^{\sqrt{2}} \sqrt{2 - x^2} \, dx $$

Using the standard integral formula $\int \sqrt{a^2 - x^2} \, dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}(\frac{x}{a})$ with $a = \sqrt{2}$:

$$ 2 \left[ \frac{x}{2}\sqrt{2 - x^2} + \frac{2}{2}\sin^{-1}\left(\frac{x}{\sqrt{2}}ight) ight]_1^{\sqrt{2}} = 2 \left[ \left(0 + \sin^{-1}(1)ight) - \left(\frac{1}{2} + \sin^{-1}\left(\frac{1}{\sqrt{2}}ight)ight) ight] $$

Evaluating this expression:

$$ 2 \left[ \frac{\pi}{2} - \frac{1}{2} - \frac{\pi}{4} ight] = 2 \left[ \frac{\pi}{4} - \frac{1}{2} ight] = \frac{\pi}{2} - 1 $$

### Final Calculation

Summing the components and multiplying by the symmetry factor of $2$:

$$ A = 2 	imes \left( \frac{4}{3} + \frac{\pi}{2} - 1 ight) = 2 	imes \left( \frac{\pi}{2} + \frac{1}{3} ight) $$

The final area of the region is:

$$ \mathbf{A = \pi + \frac{2}{3}} $$

Animated Solution for Mathematics - Definite Integration: The area of the region $A = {(x, y) : 4 x^{2} + y^{2} \leq 8 and y^{2} \leq 4 x}$ is:

Select Answer:

Visualized Solution

Visualizing the Region $A$

Finding Intersection Points

Solving for Coordinates

Symmetry and Splitting

Setting up the Integrals

Integrating the Parabola Part

Integrating the Ellipse Part

Applying the Standard Formula

Evaluating the Limits

Final Summation

The Sigma Insight: Area Bounded by Curves

Analyzing the Setup

The Intersection

Symmetry and Strategy

The Execution

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 region A={(x,y):4x2+y2≤8 and y2≤4x} is: