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