Area bounded by curve and horizontal or vertical line

Chapter 8 Class 12 Application of Integrals
Concept wise

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### Transcript

Ex 8.1, 3 Find the area of the region bounded by π₯2= 4π¦ , π¦ = 2 , π¦=4 and the π¦-axis in the first quadrant. Here, The curve is π₯^2=4π¦ We have to find area between y = 2 and y = 4 β΄ We have to find area of BCFE Area of BCFE = β«_2^4βπ₯ ππ¦ We know that π₯^2=4π¦ Taking square root on both sides π₯="Β±" β4π¦β π₯="Β±" 2βπ¦ Since, BCEF is in 1st Quadrant We take positive value of x β΄ π₯=2βπ¦ Area of BCFE = β«_2^4βπ₯ ππ¦ = β«_2^4βγ2βπ¦γ ππ¦ = 2β«_2^4βγ(π¦)^(1/2) ππ¦γ = 2 [π¦^(3/2)/(3/2)]_2^4 = 2 Γ 2/3 [π¦^(3/2) ]_2^4 = 4/3 [(4)^(3/2 )β(2)^(3/2) ] = 4/3 [((4)^(1/2) )^3β((2)^(1/2) )^3 ] = 4/3 [(2)^3β(β2)^3 ] = 4/3 [8 β2β2] = (32 β 8β2)/3 Thus, Area = (ππ β πβπ)/π square units