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Area bounded by curve and horizontal or vertical line
Ex 8.1, 7
Ex 8.1, 1
Ex 8.1, 11
Example 11
Ex 8.1, 2 Important
Ex 8.1, 8 Important
Example 3
Misc 3
Ex 8.1, 13 (MCQ) Important
Misc 1 (i)
Ex 8.1, 3
Example 5 Important You are here
Misc 5 Important
Example 13 Important Deleted for CBSE Board 2023 Exams
Example 12
Misc 16 (MCQ)
Misc 17 (MCQ) Important
Area bounded by curve and horizontal or vertical line
Last updated at Dec. 12, 2019 by Teachoo
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₯=ππ, where, π2=π2 (1 β π2) and e < 1 Required Area = Area of shaded region = Area BORQSP = 2 Γ Area OBPS = 2 Γ β«_0^ππβγπ¦.ππ₯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^2βγ π₯γ^2)/π^2 π¦^2=π^2/π^2 (π^2βπ₯^2 ) π¦=Β±β(π^2/π^2 (π^2βπ₯^2 ) ) π¦=Β±π/π β((π^2βπ₯^2 ) ) Since OBPS is in 1st quadrant, value of y is positive β΄ π¦=π/π β(π^2βπ₯^2 ) Required Area = 2 Γ β«_0^ππβγπ¦.ππ₯γ = 2β«_0^ππβγπ/π β(π^2βπ₯^2 )γ ππ₯ = 2π/π β«_0^ππββ(π^2βπ₯^2 ) ππ₯ = 2π/π [1/2 π₯β(π^2βπ₯^2 )+π^2/2 sin^(β1)β‘γπ₯/πγ ]_0^ππ =2π/π [(ππ/2 β(π^2β(ππ)^2 )+π^2/2 sin^(β1)β‘γππ/πγ )β(0/2 β(π^2β0)+π^2/2 sin^(β1)β‘(0/π) )] =2π/π [ππ/2 β(π^2βπ^2 π^2 )+π^2/2 sin^(β1)β‘γ(π)β0βπ^2/2 sin^(β1)β‘(0) γ ] =2π/π [ππ/2.πβ(1βπ^2 )+π^2/2 sin^(β1)β‘γπβ0γ ] 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 =2π/π [(π^2 π)/2 β(1βπ^2 )+π^2/2 sin^(β1)β‘π ] =2π/π (π^2/2)[πβ(1βπ^2 )+sin^(β1)β‘π ] =ππ[πβ(1βπ^2 )+sin^(β1)β‘π ] β΄ Required Area =ππ[πβ(πβπ^π )+γπππγ^(βπ)β‘π ] square units