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महाराष्ट्र राज्य शिक्षण मंडळएचएससी वाणिज्य (इंग्रजी माध्यम) इयत्ता १२ वी

Find the area of the region bounded by the curve y = 36-x2, the X-axis lying in the first quadrant and the lines x = 0 and x = 6

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प्रश्न

Find the area of the region bounded by the curve y = `sqrt(36 - x^2)`, the X-axis lying in the first quadrant and the lines x = 0 and x = 6

बेरीज
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उत्तर

Let A be the required area.

Given equation of the curve is y = `sqrt(36 - x^2)`

∴ A = `int_0^6 y  "d"x`

= `int_0^6 sqrt(36 - x^2)  "d"x`

= `int_0^6 sqrt((6)^2 + x^2)  "d"x`

= `[x/2 sqrt((6)^2 - x^2) + (6)^2/2 sin^-1 (x/6)]_0^6`

= `[6/2 sqrt((6)^2 - (6)^2) + (6)^2/2 sin^-1 (6/6)] - [0/2 sqrt((6)^2 - 0) + (6)^2/2 sin^-1 (0/6)]`

= `0 + 36/2 sin^-1 (1) - 0`

= `36/2 (pi/2)`

= `(36pi)/4` sq.units

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पाठ 1.7: Application of Definite Integration - Q.2

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