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Find the area of the region bounded by the ellipse x216+y29=1.

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प्रश्न

Find the area of the region bounded by the ellipse  `x^2/16 + y^2/9 = 1.`

योग
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उत्तर

Given equation of ellipse `x^2/16 + y^2/9 = 1`

The given ellipse is symmetric about both axes and has identical x and y axes.

`= y^2/9 = 1 - x^2/16`

`= y = pm 3/4 (sqrt(16 - x^2))`

Area enclosed by the ellipse = 4(Area of ​​sector) = 4(Area OAC)

Ellipse in the first quadrant `= 4 int_0^4 y dx = int_0^4 3/4 sqrt(16 - x^2)  dx`

Let `x = 4 sin theta ; dx = 4 cos theta  d theta`

Hence, when x = 0, `theta = 0 ;` when x = 4, `theta = pi/2`

Required Area `= (4 xx 3)/4 int_0^(pi//2) sqrt(16 - 16 sin^2 theta). 4 cos theta  d theta.`

`= 3 int_0^(pi/2) 4sqrt(1 - sin^2 theta). 4 cos theta  d theta`

`= 48 int_0^(pi/2) cos^2 theta  d theta`

`= 24 int_0^(pi/2)  (1 + cos 2 theta)d theta`

`= 24 [theta + (sin 2 theta)/2]_0^(pi/2)`

`= 12π  square unit

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अध्याय 8: Application of Integrals - Exercise 8.1 [पृष्ठ ३६६]

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एनसीईआरटी Mathematics Part 1 and 2 [English] Class 12
अध्याय 8 Application of Integrals
Exercise 8.1 | Q 4 | पृष्ठ ३६६

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