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Area between curve and line
Ex 8.1, 6 Important You are here
Ex 8.2, 6 (MCQ) Deleted for CBSE Board 2023 Exams
Example 8 Important Deleted for CBSE Board 2023 Exams
Misc 8
Misc 9 Important
Ex 8.2 , 7 (MCQ) Important Deleted for CBSE Board 2023 Exams
Misc 6 Important
Misc 2 Deleted for CBSE Board 2023 Exams
Misc 10
Ex 8.1, 10 Important
Misc 7
Ex 8.1, 9
Misc 12 Important Deleted for CBSE Board 2023 Exams
Ex 8.2, 3 Important Deleted for CBSE Board 2023 Exams
Example 14 Important Deleted for CBSE Board 2023 Exams
Area between curve and line
Last updated at Dec. 12, 2019 by Teachoo
Ex 8.1, 6 Find the area of the region in the first quadrant enclosed by π₯βaxis, line π₯ = β3 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₯=β3 π¦ intersect the circle at point C Therefore, We have to find Area of AOC Finding point C We know that π₯=β3 π¦ Putting value of x in equation of circle π₯^2+π¦^2=4 (β3 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₯=β3 π¦ π₯=β3 Γ 1 π₯=β3 When π=βπ π₯=β3 π¦ π₯=β3 Γ β1 π₯=ββ3 Since point C is in 1st quadrant β΄ C is (β3 , 1) Area OAC Area of OAC = Area OCX + Area XCA Area OCX Area OCX = β«_0^(β3)βγπ¦ ππ₯γ π¦ β Equation of line Now, π₯=β3 π¦ π¦=π₯/β3 Therefore, Area OCX = β«_0^(β3)βγπ¦ ππ₯γ = β«_0^(β3)βγπ₯/β3 ππ₯γ = 1/β3 β«_0^(β3)βγπ₯ ππ₯γ = 1/β3 [π₯^2/2]_0^β3 = 1/(2β3) [π₯^2 ]_0^β3 = (1 )/(2β3) [(β3)^2β(0)^2 ] = (1 )/(2β3) [ 3 ] = β3/2 Area XCA Area XCA = β«_(β3)^2βγπ¦ ππ₯γ π¦ β Equation of circle Now, π₯^2+π¦^2=4 π¦^2=4βπ₯^2 π¦ = Β±β(4βπ₯^2 ) Since XCA is in 1st Quadrant, = (1 )/(2β3) [ 3 ] = β3/2 Area XCA Area XCA = β«_(β3)^2βγπ¦ ππ₯γ π¦ β Equation of circle Now, π₯^2+π¦^2=4 π¦^2=4βπ₯^2 π¦ = Β±β(4βπ₯^2 ) Since XCA is in 1st Quadrant, Value of π¦ will be positive β΄ π¦ = β(4βπ₯^2 ) Area XCA = β«_(β3)^2β β(4βπ₯^2 ) ππ₯ = β«_(β3)^2β β((2)^2βπ₯^2 ) ππ₯ = [1/2 π₯β((2)^2βπ₯^2 )+(2)^2/2 sin^(β1)β‘γπ₯/2γ ]_(β3)^2 = [1/2 π₯β(4βπ₯^2 )+2 sin^(β1)β‘γπ₯/2γ ]_(β3)^2 = 1/2 (2) β(4β2^2 )+2 sin^(β1)β‘γ2/2γβ1/2 (β3) β(4β(β3)^2 )β2 sin^(β1)β‘γβ3/2γ It is of form β(π^2βπ₯^2 ) ππ₯=1/2 π₯β(π^2βπ₯^2 )+π^2/2 γπ ππγ^(β1)β‘γ π₯/π+πγ Replacing a by 2 , we get = 0 + 2 sin^(β1)β‘(1) β β3/2 β(4β3)β2 sin^(β1)β‘γβ3/2γ = 2 sin^(β1)β‘(1) β β3/2 β 2 sin^(β1)β‘γβ3/2γ = (ββ3)/2+2[sin^(β1)β‘γ(1)βπ ππ^(β1) γ β3/2] = (ββ3)/2+2[π/2βπ/3] = (ββ3)/2+2[π/6] = (ββ3)/2+π/3 Therefore, Area of OAC = Area OCX + Area XCA = β3/2ββ3/2+(π )/3 = (π )/3 square units β΄ Required Area = π /π square units