Area bounded by curve and horizontal or vertical line

Chapter 8 Class 12 Application of Integrals
Concept wise

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Example 5 Find the area bounded by the ellipse π₯^2/π^2 +π¦^2/π^2 =1 and the ordinates π₯=0 and π₯=ππ, where, π2=π2 (1 β π2) and e < 1 Required Area = Area of shaded region = Area BORQSP = 2 Γ Area OBPS = 2 Γ β«_0^ππβγπ¦.ππ₯γ We know that , π₯^2/π^2 +π¦^2/π^2 =1 (As ellipse is symmetric about its axis ) π¦^2/π^2 =(π^2βγ π₯γ^2)/π^2 π¦^2=π^2/π^2 (π^2βπ₯^2 ) π¦=Β±β(π^2/π^2 (π^2βπ₯^2 ) ) π¦=Β±π/π β((π^2βπ₯^2 ) ) Since OBPS is in 1st quadrant, value of y is positive β΄ π¦=π/π β(π^2βπ₯^2 ) Required Area = 2 Γ β«_0^ππβγπ¦.ππ₯γ = 2β«_0^ππβγπ/π β(π^2βπ₯^2 )γ ππ₯ = 2π/π β«_0^ππββ(π^2βπ₯^2 ) ππ₯ = 2π/π [1/2 π₯β(π^2βπ₯^2 )+π^2/2 sin^(β1)β‘γπ₯/πγ ]_0^ππ =2π/π [(ππ/2 β(π^2β(ππ)^2 )+π^2/2 sin^(β1)β‘γππ/πγ )β(0/2 β(π^2β0)+π^2/2 sin^(β1)β‘(0/π) )] =2π/π [ππ/2 β(π^2βπ^2 π^2 )+π^2/2 sin^(β1)β‘γ(π)β0βπ^2/2 sin^(β1)β‘(0) γ ] =2π/π [ππ/2.πβ(1βπ^2 )+π^2/2 sin^(β1)β‘γπβ0γ ] 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 =2π/π [(π^2 π)/2 β(1βπ^2 )+π^2/2 sin^(β1)β‘π ] =2π/π (π^2/2)[πβ(1βπ^2 )+sin^(β1)β‘π ] =ππ[πβ(1βπ^2 )+sin^(β1)β‘π ] β΄ Required Area =ππ[πβ(πβπ^π )+γπππγ^(βπ)β‘π ] square units