( x 2 ) 2 3 + ( y 4 ) 2 3 = 1 x 2 12 + y 2 48 = 1 y = ± 48 ( 1 − x 2 12 ) y = ± 2 12 − x 2 \frac{(\frac{x}{2})^2}{3}+\frac{(\frac{y}{4})^2}{3}=1\\
\frac{x^2}{12}+\frac{y^2}{48}=1\\
y=\pm\sqrt{48(1-\frac{x^2}{12})}\\
y=\pm2\sqrt{12-x^2}\\ 3 ( 2 x ) 2 + 3 ( 4 y ) 2 = 1 12 x 2 + 48 y 2 = 1 y = ± 48 ( 1 − 12 x 2 ) y = ± 2 12 − x 2
To find x limit,
put y=0 in the given equation,
x 2 12 + 0 2 48 = 1 x = ± 12 = ± 2 3 \frac{x^2}{12}+\frac{0^2}{48}=1\\
x=\pm\sqrt{12}=\pm2\sqrt{3} 12 x 2 + 48 0 2 = 1 x = ± 12 = ± 2 3
area of the given curve
= 4 × ∫ a b y d x = 4 × ∫ 0 2 3 2 12 − x 2 d x = 8 ∫ 0 2 3 ( 2 3 ) 2 − x 2 d x ( By using integration formula) = 8 [ x 2 12 − x 2 + 12 2 s i n − 1 ( x 2 3 ) + c ] 0 2 3 = 8 [ 0 + 6. π 2 + c − 0 − 0 − c ] = 24 π =4×\int_a^bydx\\
=4×\int_0^{2\sqrt{3}}2\sqrt{12-x^2}dx\\
=8\int_0^{2\sqrt{3}}\sqrt{(2\sqrt{3})^2-x^2}dx\space (\text{By using integration formula)}\\
=8[\frac{x}{2}\sqrt{12-x^2}+\frac{12}{2}sin^{-1}(\frac{x}{2\sqrt{3}})+c]_0^{2\sqrt{3}}\\
=8[0+6.\frac{\pi}{2}+c-0-0-c]\\
=24\pi = 4 × ∫ a b y d x = 4 × ∫ 0 2 3 2 12 − x 2 d x = 8 ∫ 0 2 3 ( 2 3 ) 2 − x 2 d x ( By using integration formula) = 8 [ 2 x 12 − x 2 + 2 12 s i n − 1 ( 2 3 x ) + c ] 0 2 3 = 8 [ 0 + 6. 2 π + c − 0 − 0 − c ] = 24 π π.
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