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Example 7 - Find area lying above x-axis, included b/w circle - Area between curve and curve

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  1. Chapter 8 Class 12 Application of Integrals
  2. Serial order wise
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Example 7 Find the area lying above x-axis and included between the circle 𝑥2 +𝑦2=8𝑥 and inside of the parabola 𝑦2=4𝑥 Since equation of circle is of form 𝑥−𝑎﷯﷮2﷯+ 𝑦−𝑏﷯﷮2﷯= 𝑟﷮2﷯ , We convert our equation 𝑥﷮2﷯+ 𝑦﷮2﷯=8𝑥 𝑥﷮2﷯−8𝑥+ 𝑦﷮2﷯=0 𝑥﷮2﷯−2 ×4 ×𝑥+ 𝑦﷮2﷯=0 𝑥﷮2﷯−2 ×4 ×𝑥+ 4﷮2﷯− 4﷮2﷯+ 𝑦﷮2﷯=0 𝑥−4﷯﷮2﷯+ 𝑦﷮2﷯= 4﷮2﷯ So, Circle has center 4 , 0﷯ & Radius =4 We need to find Area OPQC First we find point P Point P is point of intersection of circle and parabola Now, equation of circle is 𝑥﷮2﷯+ 𝑦﷮2﷯=8𝑥 Putting 𝑦﷮2﷯=4𝑥 𝑥﷮2﷯+4𝑥=8𝑥 𝑥﷮2﷯=8𝑥−4𝑥 𝑥﷮2﷯=4𝑥 𝑥﷮2﷯−4𝑥=0 𝑥 𝑥−4﷯=0 𝑥=0 & 𝑥=4 Since point P is in 1st quadrant, Coordinates of P = (4, 4) Note that 𝑥-coordinate same as that of center (4, 0) ∴ P lies above point C So, we need to change the figure New figure Area Required Area Required = Area OPC + Area PCQ Area OPC Area OPC = 0﷮4﷮𝑦 𝑑𝑥﷯ For parabola y2 = 4x y = ± ﷮4𝑥﷯ y = ± 2 ﷮𝑥﷯ Since OPC is in 1st quadrant y = 2 ﷮𝑥﷯ ∴ Area OPC = 0﷮4﷮2 ﷮𝑥﷯﷯𝑑𝑥 = 2 0﷮4﷮ 𝑥﷮ 1﷮2﷯﷯﷯ 𝑑𝑥 = 2 𝑥﷮ 1﷮2﷯+1﷯﷮ 1﷮2﷯+1﷯﷯﷮0﷮4﷯ = 2 𝑥﷮ 3﷮2﷯﷯﷮ 3﷮2﷯﷯﷯﷮0﷮4﷯ = 2 × 2﷮3﷯ 4﷯﷮ 3﷮2﷯﷯− 0﷯﷮ 3﷮2﷯﷯﷯ = 4﷮3﷯ 8−0﷯ = 32﷮3﷯ Area PCQ Area PCQ = 4﷮8﷮𝑦 𝑑𝑥﷯ For circle x2 + y2 = 8x y2 = 8x – x2 y = ± ﷮8𝑥− 𝑥﷮2﷯﷯ Since PCQ is in 1st quadrant y = ﷮8𝑥− 𝑥﷮2﷯﷯ ∴ Area PCQ = 4﷮8﷮ ﷮8𝑥− 𝑥﷮2﷯﷯﷯𝑑𝑥 = 4﷮8﷮ ﷮−( 𝑥﷮2﷯−8𝑥)﷯﷯𝑑𝑥 = 4﷮8﷮ ﷮−( 𝑥﷮2﷯−8𝑥+16−16)﷯﷯𝑑𝑥 = 4﷮8﷮ ﷮−( 𝑥﷮2﷯−8𝑥+16)−(−16)﷯﷯𝑑𝑥 = 4﷮8﷮ ﷮16−( 𝑥﷮2﷯−8𝑥+16)﷯﷯𝑑𝑥 = 4﷮8﷮ ﷮16− 𝑥−4﷯﷮2﷯﷯﷯𝑑𝑥 = 4﷮8﷮ ﷮ 4﷮2﷯− 𝑥−4﷯﷮2﷯﷯﷯𝑑𝑥 = (𝑥 − 4)﷮2﷯ ﷮ 4﷮2﷯− (𝑥−4)﷮2﷯﷯+ 4﷮2﷯﷮2﷯ 𝑠𝑖𝑛﷮−1﷯﷮ (𝑥 − 4)﷮4﷯﷯ ﷯﷮4﷮8﷯ = (𝑥 − 4)﷮2﷯ ﷮16−( 𝑥﷮2﷯−8𝑥+ 4﷮2﷯)﷯+ 16﷮2﷯ 𝑠𝑖𝑛﷮−1﷯﷮ (𝑥 − 4)﷮4﷯﷯ ﷯﷮4﷮8﷯ = (𝑥 − 4)﷮2﷯ ﷮−( 𝑥﷮2﷯−8𝑥)﷯+8 𝑠𝑖𝑛﷮−1﷯﷮ (𝑥 − 4)﷮4﷯﷯ ﷯﷮4﷮8﷯ = (8 − 4)﷮2﷯ ﷮−( 8﷮2﷯−8(8))﷯+8 𝑠𝑖𝑛﷮−1﷯﷮ (8 − 4)﷮4﷯﷯﷯ – (4 − 4)﷮2﷯ ﷮−( 4﷮2﷯−8(4))﷯+8 𝑠𝑖𝑛﷮−1﷯﷮ (4 − 4)﷮4﷯﷯﷯ = 4﷮2﷯ ﷮0﷯+8 𝑠𝑖𝑛﷮−1﷯﷮ 1﷯﷯ – 0+8 𝑠𝑖𝑛﷮−1﷯﷮ 0﷯﷯ = 8 𝑠𝑖𝑛﷮−1﷯﷮ 1﷯ – 8 𝑠𝑖𝑛﷮−1﷯﷮ 0﷯ = 8 𝜋﷮2﷯﷯ – 8 × 0 = 4𝜋 Thus, Area Required = Area OPC + Area PCQ = 32﷮3﷯ + 4𝜋 = 4﷮3﷯ (8 + 3𝜋)

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