If the area of the region bounded by the curves y² - 2y = -x, x + y = 0 is A, then 8A is equal to

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

Great question! If you try to integrate with respect to $x$, you would have to solve $y^2 - 2y + x = 0$ for $y$ using the quadratic formula, which introduces messy square roots. By rewriting the equations as $x = 2y - y^2$ and $x = -y$, we can integrate with respect to $y$ directly, which is much cleaner and avoids those radicals entirely.

Q: How do I know which curve is on the right and which is on the left?

A simple trick is to pick a test point between your intersection limits. Our limits are $y=0$ and $y=3$. Let's pick $y=1$. At $y=1$, the parabola gives $x = 2(1) - (1)^2 = 1$. The line gives $x = -1$. Since $1 > -1$, the parabola is on the right and the line is on the left. Thus, we subtract the line from the parabola.

Q: Wait, is the final answer $4.5$ or $36$?

The area $A$ of the region is indeed $4.5$ (or $\frac{9}{2}$). However, the question specifically asks for the value of $8A$. Always double-check the final line of the question to ensure you are providing the exact value requested! In this case, $8 \times 4.5 = 36$.

Question

If the area of the region bounded by the curves $y^2 - 2y = -x, x + y = 0$ is $A$, then $8A$ is equal to

Sigma Percentile · Accepted Answer

### The Geometry of Perspective

Welcome, fellow traveler on the path to JEE mastery. Today, we are going to unravel a beautiful problem involving the area bounded by curves.

Often, when we see equations like $y^2 - 2y = -x$ and $x + y = 0$, our instinct is to immediately try to graph them as functions of $x$. But here lies the first lesson of the day: **Perspective is everything.**

### Phase 1: The Shift

Look at the equation $y^2 - 2y = -x$. If we try to solve for $y$, we are forced into the quadratic formula, which brings in square roots and potential for error.

But what if we look at it from the perspective of the $y$-axis? If we rewrite it as $x = 2y - y^2$, suddenly, it becomes a simple parabola opening towards the negative $x$-axis.

Similarly, the line $x + y = 0$ becomes $x = -y$. By shifting our perspective, we have transformed a potentially messy problem into a clean, elegant exercise in integration.

### Phase 2: The Dance of Intersection

To find the area enclosed, we must first find where these two dancers meet. We set the two expressions for $x$ equal to each other:

$$2y - y^2 = -y$$

Rearranging this, we get:

$$y^2 - 3y = 0$$

Factoring this is straightforward: $y(y - 3) = 0$. This tells us that our curves intersect at $y = 0$ and $y = 3$. These are our limits of integration.

Imagine the region bounded between these two values; this is the stage where our area calculation will take place.

### Phase 3: The Integral

Now, we set up our integral. Since we are integrating with respect to $y$, we use horizontal strips of thickness $dy$. The area $A$ is defined as the integral of the right-hand curve minus the left-hand curve:

$$A = \int_{0}^{3} (x_{	ext{right}} - x_{	ext{left}}) dy$$

Substituting our expressions, we get:

$$A = \int_{0}^{3} ((2y - y^2) - (-y)) dy$$

Simplifying the integrand, we have:

$$A = \int_{0}^{3} (3y - y^2) dy$$

This is a standard polynomial integral. Let's solve it step-by-step:

$$A = \left[ \frac{3y^2}{2} - \frac{y^3}{3} ight]_0^3$$

Evaluating at the upper limit $y=3$:

$$A = \left( \frac{3(3)^2}{2} - \frac{3^3}{3} ight) = \left( \frac{27}{2} - 9 ight) = 13.5 - 9 = 4.5$$

So, the area $A$ is $\frac{9}{2}$.

### Phase 4: The Final Reveal

We have found the area $A = 4.5$. But remember, in the heat of a JEE exam, the final step is where the traps are often laid.

The question asks for $8A$. We must not stop at $A$. We take our result and multiply:

$$8A = 8 	imes \frac{9}{2} = 36$$

And there it is. The elegance of the solution lies not just in the calculation, but in the choice of perspective.

By choosing to integrate with respect to $y$, we turned a complex problem into a simple, satisfying victory. Keep this mindset—always look for the path of least resistance, and the math will reward you.

Animated Solution for Mathematics - Definite Integration: If the area of the region bounded by the curves $y^{2} - 2 y = - x, x + y = 0$ is $A$ , then $8 A$ is equal to

Enter Numerical Value:

Visualized Solution

Analyze the Curves

Graphing the Parabola

Graphing the Line

Intersection Points Setup

Solving for $y$

Intersection Coordinates

The Bounded Region

Area Integral Setup

Substituting the Curves

Simplifying the Integrand

Integrating with respect to $y$

Evaluating the Limits

Final Area Value

Calculating $8 A$

The Sigma Insight: Area Bounded by Curves

The Geometry of Perspective

Phase 1

Phase 2

Phase 3

Phase 4

Similar Questions

The area of the region bounded by $y - x = 2$ and $x^{2} = y$ is equal to :-

If the area of the region ${(x, y) : ∣ x^{2} - 2∣ \leq y \leq x}$ is $A$ , then $6 A + 162$ is equal to ______________

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

If the area bounded by the curve $2 y^{2} = 3 x$ , lines $x + y = 3, y = 0$ and outside the circle $(x - 3)^{2} + y^{2} = 2$ is $A$ , then $4 (π + 4 A)$ is equal to _________.

If the area of the region ${(x, y) : 0 \leq y \leq min {2 x, 6 x - x^{2}}}$ is $A$ , then $12 A$ is equal to.

The area (in sq. units) of the region enclosed between the parabola $y^{2} = 2 x$ and the line $x + y = 4$ is ____.

Let $A_{1}$ be the bounded area enclosed by the curves $y = x^{2} + 2, x + y = 8$ and $y$ -axis that lies in the first quadrant. Let $A_{2}$ be the bounded area enclosed by the curves $y = x^{2} + 2, y^{2} = x, x = 2$ , and $y$ -axis that lies in the first quadrant. Then $A_{1} - A_{2}$ is equal to

The area of the region bounded by the curves $x (1 + y^{2}) = 1$ and $y^{2} = 2 x$ is:

The area of the region enclosed by the parabola $(y - 2)^{2} = x - 1$ , the line $x - 2 y + 4 = 0$ and the positive coordinate axes is_________.

If the area of the region bounded by the curves $y = 4 - \frac{x ^{2}}{4}$ and $y = \frac{x - 4}{2}$ is equal to $α$ , then $6 α$ equals

Animated Solution for Mathematics - Definite Integration: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the area of the region bounded by the curves y2−2y=−x,x+y=0 is A, then 8A is equal to