Chapter 8 Class 12 Application of Integrals

Class 12
Important Questions for exams Class 12

### Transcript

Question 4 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