Question

Consider the following curve in polar coordinates.
\[ r = 2 - 2\sin\theta \]
Which one of the following is the area enclosed by the curve for \(0 \le \theta \le 2\pi\)?

• A. \(3\pi\)
• B. \(4\pi\)
• C. \(5\pi\)
• D. \(6\pi\)

Hint:

Use the polar area formula A = (1/2) integral of r-squared d-theta over 0 to 2pi, expand (2-2sin(theta))-squared, and integrate term by term.

Answer 1

The Correct Option is D

Step 1: Area formula.
\[ A = \frac{1}{2}\int r^2\, d\theta \]
Step 2: Square the radius.
\[ r^2 = 4-8\sin\theta+4\sin^2\theta \]
Step 3: Integrate.
\[ \int_0^{2\pi}r^2 d\theta = 8\pi-0+4\pi=12\pi \]
Step 4: Apply area formula.
\[ A = 6\pi \]
Step 5: Sanity check with cardioid formula.
Standard cardioid $r=a(1\pm\sin\theta)$ encloses $\frac{3}{2}\pi a^2$; with a=2, $A=6\pi$.
\[ \boxed{A = 6\pi} \]