The area of the region, enclosed by the circle x² + y² = 2 which is not common to the region bounded

Q: Wait, why do we subtract the common region from the circle? The question is confusing.

Think of the circle as a full pizza. The region bounded by the parabola and the line is a specific slice of that pizza. The question asks for the area of the circle that is *not* that slice. So, you calculate the entire circle's area and remove the area of that specific slice.

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

You *could* integrate with respect to $x$, but it would be more complex because the upper and lower boundaries of the common region change at $x=1$. Integrating with respect to $y$ allows you to use a single, simple integral from $y=0$ to $y=1$, which is much cleaner and less prone to errors!

Q: How do I know the radius of the circle is $\sqrt{2}$?

Great catch! The standard equation of a circle centered at the origin is $x^2 + y^2 = r^2$. Comparing this to $x^2 + y^2 = 2$, we see that $r^2 = 2$, which means the radius $r$ is $\sqrt{2}$. The area is $\pi r^2$, so $\pi (2) = 2\pi$.

Question

The area of the region, enclosed by the circle $x^2 + y^2 = 2$ which is not common to the region bounded by the parabola $y^2 = x$ and the straight line $y = x$, is

Options:

(A) $\frac{1}{3}(12\pi - 1)$

(B) $\frac{1}{6}(12\pi - 1)$

(C) $\frac{1}{3}(6\pi - 1)$

(D) $\frac{1}{6}(24\pi - 1)$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are tasked with finding the area of the circle $x^2 + y^2 = 2$ that lies outside the region bounded by the parabola $y^2 = x$ and the line $y = x$.

The circle is centered at the origin $(0,0)$ with a radius of $r = \sqrt{2}$. The total area of this circle is given by:

$$Area_{circle} = \pi r^2 = 2\pi$$

### The Intersection Dance

To define the region trapped between the parabola and the line, we must first identify their points of intersection. By substituting $x = y$ into the parabola equation $y^2 = x$, we obtain:

$$y^2 = y \implies y^2 - y = 0$$

Factoring this quadratic equation yields $y(y - 1) = 0$. Thus, the curves intersect at $y = 0$ and $y = 1$.

Correspondingly, since $x = y$, the intersection points are $(0,0)$ and $(1,1)$. These points serve as the limits of integration for our trapped region.

### The Heart of the Integral

We calculate the area of the common region, denoted as $A_c$, by integrating with respect to $y$. In this interval, the line $x = y$ acts as the right boundary, while the parabola $x = y^2$ acts as the left boundary.

The area is determined by the following integral:

$$A_c = \int_{0}^{1} (y - y^2) dy$$

Evaluating this integral, we find:

$$A_c = \left[ \frac{y^2}{2} - \frac{y^3}{3} \right]_{0}^{1} = \left( \frac{1}{2} - \frac{1}{3} \right) - 0 = \frac{1}{6}$$

### Final Calculation

To find the area of the circle not common to the region bounded by the parabola and the line, we subtract the trapped area $A_c$ from the total area of the circle.

$$Area = 2\pi - \frac{1}{6}$$

By finding a common denominator, we arrive at the final result:

**$$Area = \frac{12\pi - 1}{6}$$**

Animated Solution for Mathematics - Definite Integration: The area of the region, enclosed by the circle $x^{2} + y^{2} = 2$ which is not common to the region bounded by the parabola $y^{2} = x$ and the straight line $y = x$ , is

Select Answer:

Visualized Solution

Visualize the Curves

Identify the Target Region

Find Intersection Points

Solve for Intersection

Set up the Integral

Compute the Integration

Evaluate Limits

Calculate Circle Area

Final Subtraction

The Sigma Insight: Area Bounded by Curves

Analyzing the Setup

The Intersection Dance

The Heart of the Integral

Final Calculation

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