Area bounded between a horizontal parabola and a straight line

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

Integrating with respect to x requires solving for y as 1 ± √(1-x), splitting the region into two pieces at x=0, which is algebraically more cumbersome.

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

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Question type: Numerical
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 9 September 2026

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Why integrate with respect to y instead of x?

Integrating with respect to x requires solving for y as $1 \pm \sqrt{1 - x}$ , splitting the region into two pieces at $x = 0$ , which is algebraically more cumbersome.

Answer

The bounded area is $A = \frac{1}{6}$ , so $8 A = \frac{4}{3}$ .

Why each option works or fails

Step-by-step solution

Given: Curves: $y^{2} - 2 y = - x$ (parabola opening leftwards) and $x + y = 0$ (straight line). Area of bounded region is $A$ .
Goal: Find the value of $8 A$ , where $A$ is the area enclosed between the parabola and the line.
Approach: Express $x$ in terms of $y$ for both curves: $x_{1} = 2 y - y^{2}$ and $x_{2} = - y$ . Find their points of intersection in $y$ , then compute $A = \int_{y_{1}}^{y_{2}} (x_{right} - x_{left}) d y$ . Finally, multiply by 8.
Execute: Equating $x$ : $2 y - y^{2} = - y ⟹ y^{2} - 3 y = 0 ⟹ y (y - 3) = 0$ . Thus, $y_{1} = 0$ and $y_{2} = 3$ . For $y \in [0, 3]$ , $2 y - y^{2} \geq - y$ .
Execute: Evaluate the integral: $A = \int_{0}^{3} ((2 y - y^{2}) - (- y)) d y = \int_{0}^{3} (3 y - y^{2}) d y = {[\frac{3 y^{2}}{2} - \frac{y^{3}}{3}]}_{0}^{3} = \frac{27}{2} - 9 = \frac{9}{2}$ .
Execute: Calculate $8 A = 8 \times \frac{9}{2} = 36$ .
Verify: Using Archimedes' formula for parabolic segment: $Area = \frac{1}{6} ∣ a ∣ (y_{2} - y_{1})^{3}$ . Here, the quadratic in $y$ is $- (y^{2} - 3 y)$ , so $∣ a ∣ = 1$ and root difference is $3 - 0 = 3$ . Area $= \frac{1}{6} (1) (3)^{3} = \frac{27}{6} = \frac{9}{2}$ . Then $8 A = 36$ . Matches perfectly.

Shortcut:

When to use it: Whenever finding the area bounded between a parabola and a secant line.

Quick kill: Intersection roots of $y^{2} - 2 y = - y$ are $y = 0, 3$ . The area of any parabolic segment bounded by a line is $\frac{∣ a ∣}{6} (Δ y)^{3} = \frac{1}{6} (3 - 0)^{3} = \frac{27}{6} = \frac{9}{2}$ . Thus, $8 A = 8 \times \frac{9}{2} = 36$ .

Area bounded between a horizontal parabola and a straight line

What feels right?

The bounded area is $A = \frac{1}{6}$ , so $8 A = \frac{4}{3}$ .

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

05Execute

06Execute

✓Verify

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