Question

3)
Area bounded by the ellipse
x2
y?
+
= 4 is ....8
4
16
[Part - 2, Chap. - 8]
TT
(A) 6410
[B) 3216
(C) 8T
(D)
64​

Answer 1

Step-by-step explanation:

We have,

Equation of ellipse,

$\frac{ {x}^{2} }{4} + \frac{ {y}^{2} }{16} = 4$

$\implies\frac{ {x}^{2} }{16} + \frac{ {y}^{2} }{64} = 1$

$\implies \frac{ {y}^{2} }{64} = 1 - \frac{ {x}^{2} }{16}$

$\implies {y}^{2} = 64 - 4{x}^{2}$

$\implies y = \sqrt{64 - 4{x}^{2} }$

Now, area under ellipse:

$4 \int \limits^{4}_{0} \sqrt{64 - 4 {x}^{2} } dx$

$= 8 \int \limits^{4}_{0} \sqrt{16-{x}^{2} } dx$

$= 8 [ \frac{x}{2} \sqrt{16 - {x}^{2} } ]^{4} _{0} + 8 \times \frac{16}{2} [ \sin^{ - 1} ( \frac{x}{4} ) ] ^{4}_{0}$

$= 0 + 64( \sin^{ - 1} (1) - \sin ^{ - 1} (0) )$

$= 64 \times \frac{\pi}{2} - 0$

$= 32\pi$