Find the area of the region bounded by the ellipse 16x2+9y2=1.

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Find the area of the region bounded by the ellipse {x^2}{16}

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Find the area of the region bounded by the ellipse $\frac{x ^{2}}{16} + \frac{y ^{2}}{9} = 1$ .

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Find the area of the region bounded by the ellipse

\frac{x ^{2}}{16} + \frac{y ^{2}}{9} = 1

13 mins ago

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13 mins ago

The given equation of the ellipse,

\frac{x ^{2}}{16} + \frac{y ^{2}}{9} = 1

the standard form of ellipse is given by

\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1

by comparing we get,

a = 4 an d b = 3

It can be observed that the ellipse is symmetrical about

x

-axis and

y

-axis.

∴

Area bounded by ellipse

= 4 \times

Area of

O A B

Area of

O A B

= \int_{0}^{a} y d x

= \int_{0}^{4} y d x

= \int_{0}^{4} 3 1 - \frac{x ^{2}}{16} d x

= \frac{3}{4} \int_{0}^{4} 16 - x^{2} d x

= \frac{3}{4} [\frac{x}{2} 16 - x^{2} + \frac{16}{2} sin^{- 1} \frac{x}{4}]_{0}^{4}

= \frac{3}{4} [2 16 - 16 + 8 sin^{- 1} (1) - 0 - 8 sin^{- 1} (0)]

= \frac{3}{4} [\frac{8 π}{2}]

= \frac{3}{4} [4 π] = 3 π

Therefore, are bounded by the ellipse

= 4 \times 3 π = 12 π u ni t s

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Question Text	Find the area of the region bounded by the ellipse $\frac{x ^{2}}{16} + \frac{y ^{2}}{9} = 1$ .
Answer Type	Text solution:1
Upvotes	150