Ex 8.2, 1 - Find area of circle 4x2 + 4y2 =9 interior to parabola - Ex 8.2

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  1. Chapter 8 Class 12 Application of Integrals
  2. Serial order wise
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Ex 8.2 , 1 Find the area of circle 4 𝑥﷮2﷯+4 𝑦﷮2﷯=9 which is interior to the parabola x2 = 4𝑦 Given equations 4 𝑥﷮2﷯+4 𝑦﷮2﷯=9 𝑥﷮2﷯=4𝑦 We can write (1) as 4 𝑥﷮2﷯+4 𝑦﷮2﷯=9 4 𝑥﷮2﷯+4 𝑦﷮2﷯﷮4﷯= 9﷮4﷯ 𝑥﷮2﷯+ 𝑦﷮2﷯= 9﷮4﷯ 𝑥﷮2﷯+ 𝑦﷮2﷯= 3﷮2﷯﷯﷮2﷯ Comparing with (𝑥−ℎ)﷮2﷯+ (𝑦−𝑘)﷮2﷯= 𝑟﷮2﷯ , It is a circle with radius (r) = a & center = (h, k) = (0, 0) Also, 𝑥﷮2﷯=4𝑦 This is equation of a parabola with a vertical axis Step 1: Make the figure Required Area = Area ABCO Finding point of intersection A and C Solving (1) and (2) 4 𝑥﷮2﷯+4 𝑦﷮2﷯=9 …(1) 𝑥﷮2﷯=4𝑦 …(2) Putting value of 𝑥﷮2﷯ from (2) in (1) 4 𝑥﷮2﷯+4 𝑦﷮2﷯=9 4 4𝑦﷯+4 𝑦﷮2﷯=9 16𝑦+4 𝑦﷮2﷯−9=0 4𝑦﷮2﷯+16𝑦−9=0 4𝑦﷮2﷯+18𝑦−2𝑦−9=0 2y 2y+9﷯−1 2𝑦+9﷯=0 (2y −1)(2𝑦+9)=0 Hence, 𝑦= 1﷮2﷯ & 𝑦= −9﷮2﷯ Putting values of y in (2) Hence the points are A= − ﷮2﷯ , 1﷮2﷯﷯ & C= ﷮2﷯ , 1﷮2﷯﷯ Step 3: Finding Area Area required = Area ABCO Since ABCO is symmetric in y – axis, Area ABCO = 2 × Area BOC Area BOC = Area BCDO − Area OCD Area BCDO Area BCDO = 0﷮ ﷮2﷯﷮𝑦 𝑑𝑥﷯ y → equation of circle 4x2 + 4y2 = 9 4y2 = 9 − 4x2 y2 = 9﷮4﷯ − x2 y = ± ﷮ 9﷮4﷯ − x2﷯ Since BCDO is above x − axis, we take y positive ∴ y = ﷮ 9﷮4﷯ − x2﷯ So Area BCDO = 0﷮ ﷮2﷯﷮ ﷮ 9﷮4﷯ − x2﷯﷯ dx = 0﷮ ﷮2﷯﷮ ﷮ 3﷮2﷯﷯﷮2﷯ − x﷮2﷯﷯﷯ dx = 𝑥﷮2﷯ ﷮ 3﷮2﷯﷯﷮2﷯ − x﷮2﷯﷯+ 3﷮2﷯﷯﷮2﷯﷮2﷯ sin﷮−1﷯﷮ 𝑥﷮ 3﷮2﷯﷯﷯﷯ ﷯﷮0﷮ ﷮2﷯﷯ = 𝑥﷮2﷯ ﷮ 3﷮2﷯﷯﷮2﷯ − x﷮2﷯﷯+ 9﷮8﷯ sin﷮−1﷯﷮ 2𝑥﷮3﷯﷯ ﷯﷮0﷮ ﷮2﷯﷯ = ﷮2﷯﷮2﷯ ﷮ 9﷮4﷯ − (2)﷮2﷯﷯ + 9﷮8﷯ sin﷮−1﷯﷮ 2 ﷮2﷯﷮3﷯﷯﷯ ﷯ − 0﷮2﷯ ﷮ 9﷮4﷯ − 0﷮2﷯﷯+ 9﷮8﷯ sin﷮−1﷯﷮ 2×0﷮3﷯﷯﷯ ﷯ = ﷮2﷯﷮2﷯ ﷮ 9﷮4﷯ − 4﷯+ 9﷮8﷯ sin﷮−1﷯﷮ 2 ﷮2﷯﷮3﷯﷯− 9﷮8﷯ sin﷮−1﷯﷮(0)﷯﷯ = ﷮2﷯﷮2﷯× ﷮ 1﷮4﷯﷯+ 9﷮8﷯ sin﷮−1﷯﷮ 2 ﷮2﷯﷮3﷯﷯− 9﷮8﷯×0﷯ = ﷮2﷯﷮4﷯+ 9﷮8﷯ sin−1 2 ﷮2﷯﷮3﷯﷯ Area OCD Area OCD = 0﷮ ﷮2﷯﷮𝑦﷯ dx y → equation of parabola x2 = 4y y = 𝑥﷮2﷯﷮4﷯ So, Area OCD = 0﷮ ﷮2﷯﷮𝑦﷯ dx = 0﷮ ﷮2﷯﷮ 𝑥﷮2﷯﷮4﷯﷯ dx = 1﷮4﷯ 0﷮ ﷮2﷯﷮ 𝑥﷮2﷯﷯ dx = 1﷮4﷯ 𝑥﷮3﷯﷮3﷯﷯﷮0﷮ ﷮2﷯﷯ = 1﷮4﷯ ﷮2﷯﷯﷮3﷯﷮3﷯− 0﷮3﷯﷮3﷯﷯ = 1﷮4﷯ ﷮2﷯× ﷮2﷯× ﷮2﷯﷮3﷯−0﷯ = 1﷮4﷯ 2 ﷮2﷯﷮3﷯﷯ = ﷮2﷯﷮6﷯ So Area BOC = Area BCDO − Area OCD = ﷮2﷯﷮4﷯ + 9﷮8﷯ sin﷮−1﷯﷮ 2 ﷮2﷯﷮3﷯﷯﷯ – ﷮2﷯﷮6﷯ = ﷮2﷯﷮4﷯ − ﷮2﷯﷮6﷯+ 9﷮8﷯ sin﷮−1﷯﷮ 2 ﷮2﷯﷮3﷯﷯﷯ = ﷮2﷯﷮12﷯ + 9﷮8﷯ sin﷮−1﷯﷮ 2 ﷮2﷯﷮3﷯﷯﷯ Area required = Area ABCO = 2 × Area BOC = 2 × ﷮2﷯﷮12﷯+ 9﷮8﷯ sin﷮−1﷯﷮ 2 ﷮2﷯﷮3﷯﷯﷯﷯ = ﷮2﷯﷮6﷯+ 9﷮4﷯ sin﷮−1﷯﷮ 2 ﷮2﷯﷮3﷯﷯﷯

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.