Question

The area enclosed by the curve $x = \sqrt{3} \cos \theta, y = \sqrt{3} \sin \theta$ is

• A. $\sqrt{3}\pi$ sq. units
• B. $9\pi$ sq. units
• C. $6\pi$ sq. units
• D. $3\pi$ sq. units

Hint:

Parametric equations of the form $x = a \cos \theta, y = a \sin \theta$ always describe a circle of radius '$a$'. If the coefficients are different, like $x = a \cos \theta, y = b \sin \theta$, it describes an ellipse with area $\pi ab$. Recognizing these standard parametric forms saves the effort of formal integration.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The given equations are parametric: \[ x = \sqrt{3}\cos\theta, y = \sqrt{3}\sin\theta \] We convert them into Cartesian form to identify the curve.
Step 2: Eliminate the Parameter:
Square both equations: \[ x^2 = 3\cos^2\theta, y^2 = 3\sin^2\theta \] Add them: \[ x^2 + y^2 = 3(\cos^2\theta + \sin^2\theta) \] Using identity $\cos^2\theta + \sin^2\theta = 1$: \[ x^2 + y^2 = 3 \]
Step 3: Identify the Curve:
The equation $x^2 + y^2 = 3$ represents a circle centered at origin with radius: \[ r = \sqrt{3} \]
Step 4: Area of the Curve:
Area of a circle is: \[ A = \pi r^2 \] \[ A = \pi (\sqrt{3})^2 = 3\pi \]
Step 5: Final Answer:
\[ \boxed{3\pi \text{ sq. units}} \]