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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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Question

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

Sum
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Solution

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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Chapter 1.7: Application of Definite Integration - Q.2

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SCERT Maharashtra Mathematics and Statistics (Commerce) [English] 12 Standard HSC
Chapter 1.7 Application of Definite Integration
Q.2 | Q 11

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