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प्रश्न
Find the area of the ellipse `x^2/36 + y^2/64` = 1, using integration
उत्तर
By the symmetry of the ellipse, required area of the ellipse is 4 times the area of the region OPQO.
For the region OPQO, the limits of integration are x = 0 and x = 6.
Given equation of the ellipse is `x^2/36 + y^2/64` = 1
∴ `y^2/64 = 1 - x^2/36`
∴ y2 = `64(1 - x^2/36)`
∴ y2 = `64/36(36 - x^2)`
∴ y = `+- 8/6 sqrt(36 - x^2)`
∴ y = `4/3 sqrt(36 - x^2)` ......[∵ In first quadrant, y > 0]
∴ Required area = 4(area of the region OPQO)
= `4int_0^6 y "d"x`
= `4 int_0^6 4/3 sqrt(36 - x^2) "d"x`
= `16/3[x/2 sqrt(36 - x^2) + 36/2 sin^-1 (x/6)]_0^6`
= `16/3[6/2 sqrt(36 - 36) + 36/2 sin^-1 (1) - {0 +36/2 sin^-1 (0)}]`
= `16/3(0 + 36/2* pi/2 - 0)`
= 48π sq.units
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