3.7 Q-13

Question Statement

Find the area between the $x$-axis and the curve $y = \sqrt{2 a x - x^2}$ when $a > 0$.

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Background and Explanation

To solve this problem, we need to calculate the area between the curve and the $x$-axis for the given equation. The key concepts involved are:

1. Curve Intersection: Identifying the points where the curve intersects the $x$-axis.
2. Definite Integration: Using integration to find the area under the curve over the specified interval.
3. Trigonometric Substitution: Simplifying the integral using trigonometric identities for easier calculation.

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Solution

Step 1: Finding the points of intersection

We first find the $x$-values where the curve intersects the $x$-axis, i.e., when $y = 0$.

Given the equation:

$$y = \sqrt{2 a x - x^2}$$

Set $y = 0$:

$$\sqrt{2 a x - x^2} = 0$$

Squaring both sides:

$$2 a x - x^2 = 0$$

Factorizing:

$$x(2a - x) = 0$$

Thus, the curve intersects the $x$-axis at $x = 0$ and $x = 2a$.

Step 2: Setting up the integral

The curve is above the $x$-axis between $x = 0$ and $x = 2a$. Therefore, we need to integrate the function $y = \sqrt{2 a x - x^2}$ from $x = 0$ to $x = 2a$:

$$\text{Area} = \int_0^{2a} \sqrt{2 a x - x^2} , dx$$

Step 3: Simplifying the integrand

We rewrite the integrand in a more convenient form:

$$\sqrt{2 a x - x^2} = \sqrt{a^2 - (a - x)^2}$$

Thus, the integral becomes:

$$\int_0^{2a} \sqrt{a^2 - (a - x)^2} , dx$$

Step 4: Trigonometric substitution

Let $a - x = a \sin \theta$, which implies:

$$dx = -a \cos \theta , d\theta$$

When $x = 0$, $\theta = \frac{\pi}{2}$, and when $x = 2a$, $\theta = -\frac{\pi}{2}$.

Substituting into the integral:

$$\int_{\frac{\pi}{2}}^{-\frac{\pi}{2}} \sqrt{a^2 - a^2 \sin^2 \theta} \cdot (-a \cos \theta) , d\theta$$

Simplifying the square root:

$$\sqrt{a^2 \cos^2 \theta} = a \cos \theta$$

Thus, the integral becomes:

$$\int_{\frac{\pi}{2}}^{-\frac{\pi}{2}} a \cos \theta \cdot a \cos \theta , d\theta = a^2 \int_{\frac{\pi}{2}}^{-\frac{\pi}{2}} \cos^2 \theta , d\theta$$

Step 5: Solving the integral

Use the identity $\cos^2 \theta = \frac{1 + \cos 2\theta}{2}$:

$$a^2 \int_{\frac{\pi}{2}}^{-\frac{\pi}{2}} \frac{1 + \cos 2\theta}{2} , d\theta$$

This separates into two integrals:

$$\frac{a^2}{2} \left( \int_{\frac{\pi}{2}}^{-\frac{\pi}{2}} 1 , d\theta + \int_{\frac{\pi}{2}}^{-\frac{\pi}{2}} \cos 2\theta , d\theta \right)$$

The first integral evaluates to:

$$\int_{\frac{\pi}{2}}^{-\frac{\pi}{2}} 1 , d\theta = \pi$$

The second integral is:

$$\int_{\frac{\pi}{2}}^{-\frac{\pi}{2}} \cos 2\theta , d\theta = 0$$

since $\sin 2\theta$ is zero at both limits.

Thus, the area becomes:

$$\frac{a^2}{2} \cdot \pi = \frac{\pi}{2} a^2$$

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Key Formulas or Methods Used

• Definite Integral: Used to calculate the area under the curve: $\text{Area} = \int_{a}^{b} f(x) , dx$

• Trigonometric Substitution: To simplify the integral, we used the substitution $a - x = a \sin \theta$.

• Trigonometric Identity: $\cos^2 \theta = \frac{1 + \cos 2\theta}{2}$

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Summary of Steps

1. Find intersection points: Set $y = 0$ to find $x = 0$ and $x = 2a$.
2. Set up the integral: Integrate the function $\sqrt{2 a x - x^2}$ from $x = 0$ to $x = 2a$.
3. Simplify the integrand: Express $\sqrt{2 a x - x^2}$ as $\sqrt{a^2 - (a - x)^2}$.
4. Apply trigonometric substitution: Let $a - x = a \sin \theta$ to simplify the integral.
5. Solve the integral: Use the identity $\cos^2 \theta = \frac{1 + \cos 2\theta}{2}$ and compute the result.
6. Final result: The area is $\frac{\pi}{2} a^2$ square units.