Question

The area of region represented by $\{(x, y): x^2 + y^2 \geq c^2, y^2 + 4x^2 \leq 4c^2, |y| < c, c > 0\}$

• A. $(2\pi + 3^{\frac{3}{2}})c^2$
• B. $(2\pi - 3^{\frac{3}{2}})c^2$
• C. $\frac{c^2(3^{\frac{3}{2}} - \pi)}{3}$
• D. $\frac{c^2(3^{\frac{3}{2}} - \pi)}{6}$

Answer 1

Answer: (C)

$\frac{c^2(3^{\frac{3}{2}} - \pi)}{3}$

Answer 2

The region is defined by the following inequalities:

1. $x^2 + y^2 \geq c^2$ (Region outside or on the circle $C: x^2 + y^2 = c^2$)
2. $y^2 + 4x^2 \leq 4c^2$ (Region inside or on the ellipse $E: \frac{x^2}{c^2} + \frac{y^2}{(2c)^2} = 1$)
3. $|y| < c$ (Region between the lines $y = -c$ and $y = c$, exclusive of the lines)
4. $c > 0$

Let $A_E$ be the area of the ellipse $E$ bounded by the lines $y=-c$ and $y=c$. Let $A_C$ be the area of the circle $C$.

The problem asks for the area of the region that is inside the ellipse, outside the circle, and between $y=-c$ and $y=c$.

Any point $(x,y)$ such that $x^2+y^2 < c^2$ (i.e., inside the circle) automatically satisfies $y^2 < c^2$, which implies $|y| < c$. Also, if $x^2+y^2 < c^2$, then $4x^2+y^2 < 4c^2$ (since $4x^2 \ge x^2$), which means the circle is entirely contained within the ellipse.

Therefore, the desired area is the area of the part of the ellipse between $y=-c$ and $y=c$, minus the area of the circle.

Step 1: Calculate the area of the ellipse $y^2 + 4x^2 \leq 4c^2$ bounded by $|y| < c$.

From the ellipse equation, $4x^2 = 4c^2 - y^2$, so $x^2 = c^2 - \frac{y^2}{4}$, and $x = \pm \frac{1}{2}\sqrt{4c^2 - y^2}$.

The area is given by the integral:

$A_{E, |y|<c} = \int_{-c}^{c} 2x \, dy = \int_{-c}^{c} \sqrt{4c^2 - y^2} \, dy$.

To evaluate this integral, let $y = 2c \sin\theta$. Then $dy = 2c \cos\theta \, d\theta$.

When $y = -c$, $\sin\theta = -1/2 \implies \theta = -\pi/6$. When $y = c$, $\sin\theta = 1/2 \implies \theta = \pi/6$.

Substitute these into the integral:

$A_{E, |y|<c} = \int_{-\pi/6}^{\pi/6} \sqrt{4c^2 - (2c \sin\theta)^2} (2c \cos\theta) \, d\theta$

$= \int_{-\pi/6}^{\pi/6} \sqrt{4c^2(1 - \sin^2\theta)} (2c \cos\theta) \, d\theta$

$= \int_{-\pi/6}^{\pi/6} \sqrt{4c^2 \cos^2\theta} (2c \cos\theta) \, d\theta$

Since $\theta \in [-\pi/6, \pi/6]$, $\cos\theta \geq 0$, so $\sqrt{\cos^2\theta} = \cos\theta$.

$= \int_{-\pi/6}^{\pi/6} (2c \cos\theta) (2c \cos\theta) \, d\theta$

$= 4c^2 \int_{-\pi/6}^{\pi/6} \cos^2\theta \, d\theta$

Using the identity $\cos^2\theta = \frac{1 + \cos(2\theta)}{2}$:

$= 4c^2 \int_{-\pi/6}^{\pi/6} \frac{1 + \cos(2\theta)}{2} \, d\theta$

$= 2c^2 \left[ \theta + \frac{\sin(2\theta)}{2} \right]_{-\pi/6}^{\pi/6}$

$= 2c^2 \left[ \left( \frac{\pi}{6} + \frac{\sin(\pi/3)}{2} \right) - \left( -\frac{\pi}{6} + \frac{\sin(-\pi/3)}{2} \right) \right]$

$= 2c^2 \left[ \frac{\pi}{6} + \frac{\sqrt{3}/2}{2} + \frac{\pi}{6} + \frac{\sqrt{3}/2}{2} \right]$

$= 2c^2 \left[ \frac{2\pi}{6} + \frac{\sqrt{3}}{2} \right]$

$= \frac{2\pi c^2}{3} + \sqrt{3}c^2$.

Step 2: Calculate the area of the circle $x^2 + y^2 < c^2$.

The area of a circle with radius $c$ is $\pi c^2$.

Step 3: Subtract the area of the circle from the area of the ellipse part.

Desired Area $= A_{E, |y|<c} - A_C$

Desired Area $= \left( \frac{2\pi c^2}{3} + \sqrt{3}c^2 \right) - \pi c^2$

Desired Area $= \sqrt{3}c^2 + \left( \frac{2\pi}{3} - \pi \right)c^2$

Desired Area $= \sqrt{3}c^2 - \frac{\pi}{3}c^2$

Desired Area $= c^2 \left( \sqrt{3} - \frac{\pi}{3} \right)$

This can be written as $\frac{c^2(3\sqrt{3} - \pi)}{3}$.

Since $3\sqrt{3} = 3 \cdot 3^{1/2} = 3^{3/2}$, the area is $\frac{c^2(3^{3/2} - \pi)}{3}$.