Problem 74 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The area of the plane region bounded by the curves \(x+2 y^{2}=0\) and \(x+3 y^{2}=1\) is equal to (A) \(5 / 3\) (B) \(1 / 3\) (C) \(2 / 3\) (D) \(4 / 3\)

Step-by-Step Solution

Answer

The area of the region is \(\frac{4}{3}\), choice (D).

Step 1: Express the equations in terms of x

Start with both equations: \(x + 2y^2 = 0\) and \(x + 3y^2 = 1\). Solve each for \(x\). For the first equation, we have \(x = -2y^2\). For the second equation, we have \(x = 1 - 3y^2\).

Step 2: Set the equations equal to each other

Since we are looking for the region where the curves intersect, set \(-2y^2 = 1 - 3y^2\).

Step 3: Solve for y

Rearrange the equation \(-2y^2 = 1 - 3y^2\) to find \(y\). Adding \(3y^2\) to both sides gives \(y^2 = 1\). Thus, \(y = \pm 1\).

Step 4: Determine x-values at intersection

Using \(y = 1\) and \(y = -1\), substitute back into \(x = -2y^2\) or \(x = 1 - 3y^2\) to find the corresponding \(x\). For both \(y = 1\) and \(y = -1\), \(x = -2\) for \(x = -2y^2\) and \(x = -2\) for \(x = 1 - 3y^2\).

Step 5: Setup integral for area

The area between the curves is given by the integral \(\int_{-1}^{1} ((1 - 3y^2) - (-2y^2)) \, dy\) which simplifies to \(\int_{-1}^{1} (1 - y^2) \, dy\).

Step 6: Integrate to find the area

Compute the integral: \(\int_{-1}^{1} (1 - y^2) \, dy = \left[y - \frac{y^3}{3}\right]_{-1}^{1}\). Evaluate this as follows: \((1 - \frac{1^3}{3}) - (-1 + \frac{(-1)^3}{3}) = (1 - \frac{1}{3}) - (-1 + \frac{1}{3}) = \frac{2}{3} + \frac{2}{3} = \frac{4}{3}\).

Step 7: Select the correct multiple choice answer

The area is \(\frac{4}{3}\), which corresponds to option (D) from the given choices.

Key Concepts

Integral Calculus

Integral calculus is a branch of mathematics focused on the study of integrals and their properties. It is used to find the total size, length, area, volume, and other measures where an accumulation occurs.
One of the key applications of integral calculus is finding the area under curves. In this case, to find the area between two curves, we use definite integrals.
Here's the basic approach:

• Identify the functions forming the curves.
• Determine the points of intersection, as these are the limits of integration.
• Integrate the difference between the upper and lower function over the given interval.

For the problem at hand, we used the curves defined by the equations \(x + 2y^2=0\) and \(x + 3y^2=1\). By solving them for \(x\), we identified the expressions \(-2y^2\) and \(1 - 3y^2\). The integral \( \int_{-1}^{1} ((1 - 3y^2) - (-2y^2)) \, dy = \int_{-1}^{1} (1 - y^2) \, dy \) represents the area between these curves. This integral simplifies the task of calculating the precise value of this area, emphasizing the utility of integral calculus in determining areas where complex geometrical shapes are involved.

Curve Intersection

The intersection of curves is a significant concept in geometry and calculus. It involves determining the points where two or more curves meet or cross each other. Understanding and finding these intersections helps in understanding the relationship between different curves.
Here's how the intersection process typically works:

• Start by expressing both curves in the same form, such as solving both equations for the same variable.
• Set the expressions equal to each other to find common solutions, which reveals intersecting points.
• Solve this resulting equation to find the values of the variables where intersections occur.

In our problem, the equations \(x = -2y^2\) and \(x = 1 - 3y^2\) were set equal, leading us to solve \(-2y^2 = 1 - 3y^2\). By rearranging, we discovered that \(y^2 = 1\), giving us the roots \(y = \pm 1\). These values were then used to find the \(x\) values at the intersection, which further helped in defining our integral limits.

Solution of Equations

Solving equations is a foundational skill in mathematics, consistently used to discover unknown values or verify solution properties. The process involves manipulating expressions to isolate variables, often used to find specific points such as intersections.
In the context of the given exercise, solving equations was central to several steps:

• Initially, both given equations were solved for \(x\) to facilitate comparison and integration.
• Subsequently, equating these expressions (\(-2y^2\) and \(1 - 3y^2\)) provided a means to solve for \(y\).
• Through algebraic manipulation, specifically isolating terms, we found \(y^2 = 1\), leading to solutions \(y = \pm 1\).

Solving such equations is critical in calculus for determining bounds of integration, which directly influence integral calculations. Mastery of these techniques allows seamless transition from problem setup to the application of calculus methods, particularly in tasks related to geometry or physics where curves and their intersections define much of the analysis.