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 \(6 A-9 \sqrt{3}=\)
- A \(\pi\)
- B \(2 \pi\)
- C \(3 \pi\)
- D \(4 \pi\)
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 \(A=\int_0^3 \sqrt{x} d x+\int_3^4 \sqrt{4-(x-2)^2} d x\)…
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