Question

The area enclosed by the curve $x=3\cos\theta$, $y=2\sin\theta$, $0\le\theta\le\pi$, is (in square units)

• A. $9\pi$
• B. $6\pi$
• C. $4\pi$
• D. $3\pi$
• E. $2\pi$

Hint:

Geometry Tip: You can also evaluate the area using the parametric integral formula $A = \int_{0}^{\pi} y(\theta) |x^{\prime}(\theta)| d\theta$. Here it evaluates to $\int_0^\pi (2\sin\theta)(3\sin\theta) d\theta = 6 \int_0^\pi \sin^2\theta d\theta = 3\pi$.

Answer 1

The Correct Option is D

Concept:
The given parametric equations $x = a\cos\theta$ and $y = b\sin\theta$ represent an ellipse. The total area of a full ellipse is $\pi a b$. The parameter $\theta$ running from $0$ to $\pi$ traces exactly one half of this ellipse.

Step 1: Identify the shape from the parametric equations.
Given $x = 3\cos\theta$ and $y = 2\sin\theta$, rewrite them as $\frac{x}{3} = \cos\theta$ and $\frac{y}{2} = \sin\theta$. Squaring and adding them yields the Cartesian equation: $$\frac{x^2}{9} + \frac{y^2}{4} = 1$$ This is an ellipse with semi-major axis $a=3$ and semi-minor axis $b=2$.

Step 2: Calculate the total area of the full ellipse.
The formula for the total area of an ellipse is $A_{total} = \pi a b$: $$A_{total} = \pi (3)(2) = 6\pi$$

Step 3: Determine the fraction of the ellipse traced.
The parameter $\theta$ goes from $0$ to $\pi$ radians ($180^\circ$). Since a full ellipse requires $\theta$ to range from $0$ to $2\pi$ ($360^\circ$), the given domain traces exactly half of the ellipse (the top half).

Step 4: Calculate the enclosed area.
Since it is exactly half of the ellipse, we divide the total area by 2: $$Area = \frac{1}{2} A_{total}$$

Step 5: Compute the final value.
$$Area = \frac{1}{2} (6\pi) = 3\pi$$ Hence the correct answer is (D) $3\pi$.