Ex 8.1

Chapter 8 Class 12 Application of Integrals
Serial order wise

### Transcript

Ex 8.1, 1 Find the area of the region bounded by the ellipse π₯^2/16+π¦^2/9=1Equation Of Given Ellipse is π₯^2/16+π¦^2/9=1 π^π/(π)^π +π^π/(π)^π =π Area of ellipse = Area of ABCD = 2 Γ [Area Of ABC] = 2 Γ β«_(βπ)^πβγπ.γ ππ Finding y We know that π₯^2/16+π¦^2/9=1 π¦^2/9=1βπ₯^2/16 π¦^2/9=(16βπ₯^2)/16 π^π=π/ππ (ππβπ^π ) Taking square root on both sides y = Β± β(9/16 (16βπ₯^2 ) ) y = Β± 3/4 β(16βπ₯^2 ) Since, ABC is above x-axis y will be positive β΄ π=π/π β(ππβπ^π ) Now, Area of ellipse = 2 Γ β«_(β4)^4βγπ¦.γ ππ₯ = 2 Γ β«_(βπ)^πβγ π/π β(ππβπ^π )γ ππ = 2 Γ 3/4 β«_(β4)^4ββ(16βπ₯^2 ) ππ₯ = π/π β«_(βπ)^πββ((π)^πβπ^π ) ππ 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 4 we get = 3/2 [π₯/2 β((4)^2βπ₯^2 )+(4)^2/2 sin^(β1)β‘γ π₯/4γ ]_(β4)^4 = 3/2 [4/2 β((4)^2β(4)^2 )β((β4))/2 β((4)^2β(β4)^2 )+16/2 γ sinγ^(β1)β‘γ(4/4)β16/2γ sin^(β1) ((β4)/4)] = 3/2 [2(0)+2(0)+8 γsin^(β1) (1)γβ‘γβ 8 sin^(β1)β‘(β1) γ ] = 3/2 [0+8 sin^(β1)β‘γ(1)β8 γπππγ^(βπ)β‘(βπ) γ ] = 3/2 [8 sin^(β1)β‘γ(1)β8(βγπππγ^(βπ)β‘(π))γ ] = 3/2 [8 sin^(β1)β‘γ(1)+8 sin^(β1)β‘(1) γ ] = 3/2 Γ 16 γπππγ^(βπ)β‘(π) = 3/2 Γ 16 Γ π/π = 3 Γ 8 Γ π/2 = 12Ο β΄ Area of Ellipse = 12Ο Square units