KCET · Maths · Area Under Curves
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
Answer & Solution
Correct Answer
(D) \(3 \pi\) sq. units
Step-by-step Solution
Detailed explanation
Given parametric equations are \(x = \sqrt{3}\cos\theta\) and \(y = \sqrt{3}\sin\theta\).
Squaring and adding both equations, we get:
\(x^2 + y^2 = (\sqrt{3}\cos\theta)^2 + (\sqrt{3}\sin\theta)^2\)
\(x^2 + y^2 = 3(\cos^2\theta + \sin^2\theta)\)
\(x^2 + y^2 = 3\)
This represents a circle with center at the origin \((0,0)\) and radius \(r = \sqrt{3}\).
The area of the circle is given by \(\pi r^2\).
Area \(= \pi (\sqrt{3})^2 = 3\pi\) sq. units.
Answer: \(3 \pi\) sq. units
Squaring and adding both equations, we get:
\(x^2 + y^2 = (\sqrt{3}\cos\theta)^2 + (\sqrt{3}\sin\theta)^2\)
\(x^2 + y^2 = 3(\cos^2\theta + \sin^2\theta)\)
\(x^2 + y^2 = 3\)
This represents a circle with center at the origin \((0,0)\) and radius \(r = \sqrt{3}\).
The area of the circle is given by \(\pi r^2\).
Area \(= \pi (\sqrt{3})^2 = 3\pi\) sq. units.
Answer: \(3 \pi\) sq. units
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