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

To find the area enclosed by the curve given in the parametric form \(x = \sqrt{3} \cos \theta\) and \(y = \sqrt{3} \sin \theta\), we first need to recognize the type of curve. This curve represents an ellipse. Let's derive the general form of this ellipse and calculate its area.

Start by considering the parametric equations:

• \(x = \sqrt{3} \cos \theta\)
• \(y = \sqrt{3} \sin \theta\)

To find the Cartesian equation of the ellipse, we solve for \(\cos \theta\) and \(\sin \theta\) and substitute:

• \(\cos \theta = \frac{x}{\sqrt{3}}\)
• \(\sin \theta = \frac{y}{\sqrt{3}}\)

Using the Pythagorean identity \((\cos \theta)^2 + (\sin \theta)^2 = 1\), we have:

\(\left(\frac{x}{\sqrt{3}}\right)^2 + \left(\frac{y}{\sqrt{3}}\right)^2 = 1\)

Simplifying, we get:

\(\frac{x^2}{3} + \frac{y^2}{3} = 1\)

Which can be rewritten as:

\(\frac{x^2}{3} + \frac{y^2}{3} = 1\)

This is the equation of a circle with radius \(\sqrt{3}\).

The general formula for the area of a circle is:

\(\pi r^2\)

Substituting the value of \(r = \sqrt{3}\), we find the area:

\(\pi (\sqrt{3})^2 = 3\pi\) square units.

Therefore, the correct answer is:

\(3\pi\) square units.