The area (in sq. units) of the region A = {(x, y) : y²/2 ≤ x ≤ y+4} is :- (A) 53/3 (B) 18 (C) 30 (D)

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

Great question! If you integrate with respect to $x$, you would have to split the region into two parts because the lower boundary of the region changes at $x=2$. By integrating with respect to $y$, we treat the line as the right boundary and the parabola as the left boundary throughout the entire interval, making the calculation much simpler and less prone to errors.

Q: How do I know which function is the 'right' one?

A simple trick is to pick a test point inside the region. For example, at $y=0$, the parabola gives $x=0$ and the line gives $x=4$. Since $0 < 4$, the line is to the right of the parabola. Thus, $x_{\text{right}} = y+4$ and $x_{\text{left}} = \frac{y^2}{2}$.

Q: What if I made a mistake with the signs in the quadratic equation?

It happens to the best of us! Always double-check your rearrangement. Moving terms from one side to the other changes their signs. $y^2/2 = y+4$ becomes $y^2 = 2y+8$, which leads to $y^2 - 2y - 8 = 0$. Factoring this correctly gives $(y-4)(y+2)=0$. Take a deep breath, re-verify your steps, and you will catch it!

Question

The area (in sq. units) of the region $A = \{(x, y) : y^2/2 \le x \le y+4\}$ is :-

Options:

(A) $53/3$

(B) $18$

(C) $30$

(D) $16$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

To find the area of the region trapped between the parabola $x = \frac{y^2}{2}$ and the line $x = y+4$, we must first determine the points of intersection. At these points, the $x$-coordinates are identical.

Setting the equations equal to each other, we have:

$$\frac{y^2}{2} = y + 4$$

Multiplying by $2$ to clear the fraction yields $y^2 = 2y + 8$. Rearranging this into the standard quadratic form, we obtain:

$$y^2 - 2y - 8 = 0$$

Factoring the quadratic gives $(y-4)(y+2) = 0$. This reveals our vertical limits of integration: $y = 4$ and $y = -2$.

### The Elegance of the Integral

We choose to integrate with respect to $y$ because the right boundary is consistently the line $x = y+4$ and the left boundary is consistently the parabola $x = \frac{y^2}{2}$. This allows us to calculate the area using a single integral.

The area $A$ is defined by the integral of the difference between the right and left boundaries:

$$A = \int_{-2}^{4} \left( y + 4 - \frac{y^2}{2} \right) dy$$

### The Final Calculation

We perform the integration term by term:

$$\int \left( y + 4 - \frac{y^2}{2} \right) dy = \left[ \frac{y^2}{2} + 4y - \frac{y^3}{6} \right]_{-2}^{4}$$

Evaluating at the upper limit $y = 4$:

$$\frac{16}{2} + 16 - \frac{64}{6} = 8 + 16 - \frac{32}{3} = 24 - \frac{32}{3} = \frac{40}{3}$$

Evaluating at the lower limit $y = -2$:

$$\frac{4}{2} - 8 - \frac{-8}{6} = 2 - 8 + \frac{4}{3} = -6 + \frac{4}{3} = -\frac{14}{3}$$

Subtracting the lower limit from the upper limit, we find the total area:

$$A = \frac{40}{3} - \left( -\frac{14}{3} \right) = \frac{54}{3} = 18$$

The area of the region is exactly **$18$ square units**.

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

Select Answer:

Visualized Solution

Identifying the Boundaries

Finding Intersection Points

Forming the Quadratic Equation

Solving for $y$

Setting up the Integral

The Definite Integral

Integration Step

Upper Limit Evaluation

Lower Limit Evaluation

Final Calculation

The Sigma Insight: Area Bounded by Curves

Analyzing the Setup

The Elegance of the Integral

The Final Calculation

Similar Questions

The area (in sq. units) of the region $A = {(x, y) : x^{2} \leq y \leq x + 2}$ is

The area of the region ${(x, y) : x^{2} + 4 x + 2 \leq y \leq ∣ x + 2∣}$ is equal to

The area (in sq. units) of the region $A = {(x, y) \in R \times R ∣0 < x < 3, 0 < y < 4, y < x^{2} + 3 x}$ is :

The area (in sq. units) of the region ${(x, y) : x \geq 0, x + y \leq 3, x^{2} \leq 4 y and y \leq 1 + x}$ is:

The area of the region ${(x, y) : ∣ x - y ∣ \leq y \leq 4 x}$ is

Area of the region ${(x, y) : x^{2} + (y - 2)^{2} \leq 4, x^{2} \geq 2 y}$ is

The area of the region given by $A = {(x, y) : x^{2} \leq y \leq min {x + 2, 4 - 3 x}}$ is:

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

The area of the region $x, y : x^{2} \leq y \leq ∣ x^{2} - 4∣, y \geq 1$ is

The area of the region $S = {(x, y) : 3 x^{2} \leq 4 y \leq 6 x + 24}$ is .

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 (in sq. units) of the region A={(x,y):y2/2≤x≤y+4} is :-