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TS EAMCET · Maths · Area Under Curves

If the area lying in the first quadrant and bounded by the circle \(x^2+y^2-4 x=0\), the parabola \(y^2=x\) and the \(X\)-axis is \(A\), then \(6 A-9 \sqrt{3}=\)

  1. A \(\pi\)
  2. B \(2 \pi\)
  3. C \(3 \pi\)
  4. D \(4 \pi\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(4 \pi\)

Step-by-step Solution

Detailed explanation

Given curve \(x^2+y^2-4 x=0 \Rightarrow(x-2)^2+y^2=4 \text { and } y^2=x\) intersecting point of circle and parabola is \((0,0)\) and \((3, \sqrt{3})\) Required area \(A=\int_0^3 \sqrt{x} d x+\int_3^4 \sqrt{4-(x-2)^2} d x\)…