The area of the region in the first quadrant inside the circle x² + y² = 8 and outside the parabola

Q: Wait, why do we split the integral at $x = 2$?

Great question! The boundary of the region under the curves changes at the intersection point. From $x = 0$ to $x = 2$, the 'floor' of the area is defined by the parabola $y = \sqrt{2x}$. However, from $x = 2$ to $x = 2\sqrt{2}$, the 'floor' is defined by the circle $y = \sqrt{8-x^2}$. Because the function defining the boundary changes, we must split the integral to accurately calculate the area.

Q: Can we just integrate with respect to $y$ instead of $x$?

Absolutely! You could integrate with respect to $y$. If you do, the limits would be from $y = 0$ to $y = 2$. The area would be $\int_{0}^{2} (x_{circle} - x_{parabola}) dy$, where $x_{circle} = \sqrt{8-y^2}$ and $x_{parabola} = \frac{y^2}{2}$. This is actually a very elegant approach and might even be faster! Always look for the path of least resistance in your integration.

Q: Why is the area of the quarter-circle $2\pi$?

Remember that the area of a full circle is $\pi R^2$. Here, the equation is $x^2 + y^2 = 8$, so $R^2 = 8$. The area of the full circle is $8\pi$. Since we are only interested in the first quadrant, we take one-fourth of that area: $\frac{1}{4} \times 8\pi = 2\pi$. It is a simple but crucial step—never rush the basics!

Question

The area of the region in the first quadrant inside the circle $x^2 + y^2 = 8$ and outside the parabola $y^2 = 2x$ is equal to :

Options:

(A) $\frac{\pi}{2} - \frac{1}{3}$

(B) $\pi - \frac{1}{3}$

(C) $\frac{\pi}{2} - \frac{2}{3}$

(D) $\pi - \frac{2}{3}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are working with two primary curves in the first quadrant: the circle defined by $x^2 + y^2 = 8$ (with radius $R = 2\sqrt{2}$) and the parabola defined by $y^2 = 2x$.

Our objective is to calculate the area that lies inside the circle but outside the parabola within the first quadrant.

### The Meeting Point: Finding the Intersection

To define our limits of integration, we must find where these two curves intersect. By substituting the parabola's equation, $y^2 = 2x$, into the circle's equation, $x^2 + y^2 = 8$, we obtain:

$$x^2 + 2x - 8 = 0$$

Factoring this quadratic equation yields $(x+4)(x-2) = 0$. Since we are restricted to the first quadrant, we discard the negative root and accept $x = 2$.

Substituting $x = 2$ back into $y^2 = 2x$, we find $y = 2$. Thus, the curves intersect at the point $(2, 2)$.

### The Strategy: The Art of Subtraction

The total area of the quarter-circle in the first quadrant is given by:

$$	ext{Area}_{	ext{quarter}} = \frac{1}{4} \pi R^2 = \frac{1}{4} \pi (8) = 2\pi$$

To find the target area, we subtract the area under the curves from this total. The region under the curves is split into two distinct parts along the $x$-axis: the area under the parabola from $x = 0$ to $x = 2$, and the area under the circle from $x = 2$ to $x = 2\sqrt{2}$.

### The Calculus: Heavy Lifting with Elegance

First, we calculate the area under the parabola from $0$ to $2$:

$$\int_{0}^{2} \sqrt{2x} \, dx = \sqrt{2} \int_{0}^{2} x^{1/2} \, dx = \sqrt{2} \left[ \frac{2}{3} x^{3/2} ight]_0^2 = \frac{8}{3}$$

Next, we calculate the area under the circle from $2$ to $2\sqrt{2}$ using the standard integral $\int \sqrt{a^2-x^2} \, dx = \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}(\frac{x}{a})$:

$$\int_{2}^{2\sqrt{2}} \sqrt{8-x^2} \, dx = \left[ \frac{x}{2}\sqrt{8-x^2} + 4\sin^{-1}\left(\frac{x}{2\sqrt{2}}ight) ight]_2^{2\sqrt{2}}$$

Evaluating at the limits, the upper limit yields $4(\pi/2) = 2\pi$, and the lower limit yields $2 + 4(\pi/4) = 2 + \pi$. The resulting area is $\pi - 2$.

### Final Calculation

Summing the two areas under the curves, we get:

$$	ext{Area}_{	ext{under}} = \frac{8}{3} + (\pi - 2) = \pi + \frac{2}{3}$$

Finally, we subtract this from the total area of the quarter-circle:

$$	ext{Area}_{	ext{target}} = 2\pi - \left( \pi + \frac{2}{3} ight) = \pi - \frac{2}{3}$$

The final area of the region is **$\pi - \frac{2}{3}$** square units.

Animated Solution for Mathematics - Definite Integration: The area of the region in the first quadrant inside the circle $x^{2} + y^{2} = 8$ and outside the parabola $y^{2} = 2 x$ is equal to :

Select Answer:

Visualized Solution

Visualizing the Curves

Finding the Intersection Point

Solving for $x$

Solving for $y$

Strategy for the Required Area

Total Area of the Quarter Circle

Setting up the 'Inside' Area Integral

Integrating the Parabola Part

The Circle Integral Formula

Evaluating the Circle Integral

Total Area Inside the Parabola

The Final Calculation

The Sigma Insight: Area Bounded by Curves

Analyzing the Setup

The Meeting Point

The Strategy

The Calculus

Final Calculation

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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 of the region in the first quadrant inside the circle x2+y2=8 and outside the parabola y2=2x is equal to :