Question

Using integration, find the area of the region enclosed between the circle \( x^2 + y^2 = 16 \) and the lines \( x = -2 \) and \( x = 2 \).

Hint:

For definite integrals of the form $\int \sqrt{r^2-x^2}\,dx$ over symmetric limits, first convert the boundary value of $x$ into an angle using $\sin\theta = \frac{x}{r}$; this angle directly gives the sweep angle for the sector formula $A=\frac{1}{2}r^2\theta$, saving you from expanding $\cos^2\theta$ separately.

Answer 1

The given circle is: \[ x^2 + y^2 = 16 \quad \Rightarrow \quad y = \pm \sqrt{16 - x^2}. \] The lines \( x = -2 \) and \( x = 2 \) are vertical lines. We need to find the area of the region between these vertical lines and the circle. 1. Area under the curve \( y = \sqrt{16 - x^2} \): The area between \( x = -2 \) and \( x = 2 \) above the \( x \)-axis is: \[ A_1 = \int_{-2}^{2} \sqrt{16 - x^2} \, dx. \] 2. Use symmetry: The total area enclosed is twice the area above the \( x \)-axis: \[ {Total Area} = 2A_1 = 2 \int_{-2}^{2} \sqrt{16 - x^2} \, dx. \] 3. Solve the integral: Substitute \( x = 4 \sin \theta \), so \( dx = 4 \cos \theta \, d\theta \) and \( \sqrt{16 - x^2} = 4 \cos \theta \): \[ \int_{-2}^{2} \sqrt{16 - x^2} \, dx = \int_{-\pi/6}^{\pi/6} 4 \cos \theta \cdot 4 \cos \theta \, d\theta = 16 \int_{-\pi/6}^{\pi/6} \cos^2 \theta \, d\theta. \] 4. Simplify \( \cos^2 \theta \): Using the identity \( \cos^2 \theta = \frac{1 + \cos 2\theta}{2} \): \[ 16 \int_{-\pi/6}^{\pi/6} \cos^2 \theta \, d\theta = 16 \int_{-\pi/6}^{\pi/6} \frac{1 + \cos 2\theta}{2} \, d\theta. \] Separate the terms: \[ 16 \int_{-\pi/6}^{\pi/6} \frac{1}{2} \, d\theta + 16 \int_{-\pi/6}^{\pi/6} \frac{\cos 2\theta}{2} \, d\theta. \] 5. Integrate: - First term: \[ 16 \int_{-\pi/6}^{\pi/6} \frac{1}{2} \, d\theta = 16 \cdot \frac{1}{2} \cdot \left(\frac{\pi}{6} - \left(-\frac{\pi}{6}\right)\right) = 16 \cdot \frac{\pi}{6}. \] - Second term: \[ 16 \int_{-\pi/6}^{\pi/6} \frac{\cos 2\theta}{2} \, d\theta = 16 \cdot \frac{1}{2} \cdot \left[\frac{\sin 2\theta}{2}\right]_{-\pi/6}^{\pi/6} = 0. \] Total area above the \( x \)-axis: \[ A_1 = \frac{16\pi}{6} = \frac{8\pi}{3}. \] Total enclosed area: \[ {Total Area} = 2A_1 = 2 \cdot \frac{8\pi}{3} = \frac{16\pi}{3}. \] Final Answer: The area is \( \boxed{\frac{16\pi}{3}} \).

Answer 2

Concept:

• The area swept by the circle boundary between two symmetric vertical lines can be found directly using the sector-sweep formula $A = \frac{1}{2}r^2\theta$, without expanding $\cos^2\theta$ term by term.

Step 1: Find the sweep angle.
The line $x = 2$ meets the circle $x^2+y^2=16$ where $4+y^2=16$, so $y = \pm 2\sqrt{3}$. Using the standard circle substitution $x = r\sin\theta$ with $r = 4$: $\sin\theta = \frac{2}{4} = \frac{1}{2}$, so $\theta = \frac{\pi}{6}$.

Step 2: Apply the sector-sweep formula for the upper half.
As $x$ runs from $-2$ to $2$ along the upper boundary of the circle, the radius vector effectively sweeps through an angle from $-\frac{\pi}{6}$ to $\frac{\pi}{6}$, a total angle of $\frac{\pi}{3}$. Using $A = \frac{1}{2}r^2\theta$ directly:
$A_1 = \frac{1}{2}(4)^2 \left(\frac{\pi}{3}\right) = \frac{1}{2}(16)\left(\frac{\pi}{3}\right) = \frac{8\pi}{3}$

Step 3: Use symmetry for the lower half.
The circle is symmetric about the $x$-axis, so the lower half between $x=-2$ and $x=2$ sweeps out an equal area. Total enclosed area:
$\text{Total Area} = 2 \times A_1 = 2 \times \frac{8\pi}{3} = \frac{16\pi}{3}$

Final Answer: The area is $\frac{16\pi}{3}$.