Ex 8.1

Chapter 8 Class 12 Application of Integrals
Serial order wise

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Transcript

Ex 8.1, 6 Find the area of the region in the first quadrant enclosed by π₯βaxis, line π₯ = β3 π¦ and the circle π₯2 + π¦2 = 4. Given Equation of circle π₯^2+π¦^2=4 π₯^2+π¦^2=(2)^2 β΄ Radius π = 2 So, point A is (2, 0) and point B is (0, 2) Let line π₯=β3 π¦ intersect the circle at point C Therefore, We have to find Area of AOC Finding point C We know that π₯=β3 π¦ Putting value of x in equation of circle π₯^2+π¦^2=4 (β3 π¦)^2+π¦^2=4 3π¦^2+π¦^2=4 4π¦^2=4 π¦^2=1 β΄ π¦=Β±1 Now, finding value of x When π=π π₯=β3 π¦ π₯=β3 Γ 1 π₯=β3 When π=βπ π₯=β3 π¦ π₯=β3 Γ β1 π₯=ββ3 Since point C is in 1st quadrant β΄ C is (β3 , 1) Area OAC Area of OAC = Area OCX + Area XCA Area OCX Area OCX = β«_0^(β3)βγπ¦ ππ₯γ π¦ β Equation of line Now, π₯=β3 π¦ π¦=π₯/β3 Therefore, Area OCX = β«_0^(β3)βγπ¦ ππ₯γ = β«_0^(β3)βγπ₯/β3 ππ₯γ = 1/β3 β«_0^(β3)βγπ₯ ππ₯γ = 1/β3 [π₯^2/2]_0^β3 = 1/(2β3) [π₯^2 ]_0^β3 = (1 )/(2β3) [(β3)^2β(0)^2 ] = (1 )/(2β3) [ 3 ] = β3/2 Area XCA Area XCA = β«_(β3)^2βγπ¦ ππ₯γ π¦ β Equation of circle Now, π₯^2+π¦^2=4 π¦^2=4βπ₯^2 π¦ = Β±β(4βπ₯^2 ) Since XCA is in 1st Quadrant, = (1 )/(2β3) [ 3 ] = β3/2 Area XCA Area XCA = β«_(β3)^2βγπ¦ ππ₯γ π¦ β Equation of circle Now, π₯^2+π¦^2=4 π¦^2=4βπ₯^2 π¦ = Β±β(4βπ₯^2 ) Since XCA is in 1st Quadrant, Value of π¦ will be positive β΄ π¦ = β(4βπ₯^2 ) Area XCA = β«_(β3)^2β β(4βπ₯^2 ) ππ₯ = β«_(β3)^2β β((2)^2βπ₯^2 ) ππ₯ = [1/2 π₯β((2)^2βπ₯^2 )+(2)^2/2 sin^(β1)β‘γπ₯/2γ ]_(β3)^2 = [1/2 π₯β(4βπ₯^2 )+2 sin^(β1)β‘γπ₯/2γ ]_(β3)^2 = 1/2 (2) β(4β2^2 )+2 sin^(β1)β‘γ2/2γβ1/2 (β3) β(4β(β3)^2 )β2 sin^(β1)β‘γβ3/2γ It is of form β(π^2βπ₯^2 ) ππ₯=1/2 π₯β(π^2βπ₯^2 )+π^2/2 γπ ππγ^(β1)β‘γ π₯/π+πγ Replacing a by 2 , we get = 0 + 2 sin^(β1)β‘(1) β β3/2 β(4β3)β2 sin^(β1)β‘γβ3/2γ = 2 sin^(β1)β‘(1) β β3/2 β 2 sin^(β1)β‘γβ3/2γ = (ββ3)/2+2[sin^(β1)β‘γ(1)βπ ππ^(β1) γ β3/2] = (ββ3)/2+2[π/2βπ/3] = (ββ3)/2+2[π/6] = (ββ3)/2+π/3 Therefore, Area of OAC = Area OCX + Area XCA = β3/2ββ3/2+(π )/3 = (π )/3 square units β΄ Required Area = π/π square units

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.