The asymptote(s) of the curve parallel to y y y 2 a − x = 0. 2a-x=0. 2 a − x = 0.
Then the vertical asymptote is x = 2 a . x=2a. x = 2 a .
y = x 3 2 a − x y=\sqrt{\dfrac{x^3}{2a-x}} y = 2 a − x x 3 Let x = 2 a sin 2 θ . x=2a\sin^2\theta. x = 2 a sin 2 θ .
d x = 4 a sin θ cos θ d θ dx=4a\sin \theta \cos \theta d\theta d x = 4 a sin θ cos θ d θ
A r e a = A = ∫ 0 2 a x 3 2 a − x d x Area=A=\displaystyle\int_{0}^{2a}\sqrt{\dfrac{x^3}{2a-x}}dx A re a = A = ∫ 0 2 a 2 a − x x 3 d x
= ∫ 0 π / 2 8 a 3 sin 6 θ 2 a − 2 a sin 2 θ 4 a sin θ cos θ d θ =\displaystyle\int_{0}^{\pi/2}\sqrt{\dfrac{8a^3\sin^6\theta}{2a-2a\sin^2\theta}}4a\sin \theta \cos \theta d\theta = ∫ 0 π /2 2 a − 2 a sin 2 θ 8 a 3 sin 6 θ 4 a sin θ cos θ d θ
= 8 a 2 ∫ 0 π / 2 sin 3 θ sin θ cos θ cos θ d θ =8a^2\displaystyle\int_{0}^{\pi/2}\dfrac{\sin^3\theta\sin \theta \cos \theta}{\cos \theta}d\theta = 8 a 2 ∫ 0 π /2 cos θ sin 3 θ sin θ cos θ d θ
= 8 a 2 ∫ 0 π / 2 sin 4 θ d θ = 2 a 2 ∫ 0 π / 2 ( 1 − cos ( 2 θ ) ) 2 d θ =8a^2\displaystyle\int_{0}^{\pi/2}\sin^4\theta d\theta=2a^2\displaystyle\int_{0}^{\pi/2}(1-\cos(2\theta))^2 d\theta = 8 a 2 ∫ 0 π /2 sin 4 θ d θ = 2 a 2 ∫ 0 π /2 ( 1 − cos ( 2 θ ) ) 2 d θ
= 2 a 2 ∫ 0 π / 2 ( 1 − 2 cos ( 2 θ ) + cos 2 ( 2 θ ) ) d θ =2a^2\displaystyle\int_{0}^{\pi/2}(1-2\cos(2\theta)+\cos^2(2\theta)) d\theta = 2 a 2 ∫ 0 π /2 ( 1 − 2 cos ( 2 θ ) + cos 2 ( 2 θ )) d θ
= a 2 ∫ 0 π / 2 ( 2 − 4 cos ( 2 θ ) + 1 + cos ( 4 θ ) ) d θ =a^2\displaystyle\int_{0}^{\pi/2}(2-4\cos(2\theta)+1+\cos(4\theta)) d\theta = a 2 ∫ 0 π /2 ( 2 − 4 cos ( 2 θ ) + 1 + cos ( 4 θ )) d θ
= a 2 [ 3 θ − 2 sin ( 2 θ ) + 1 4 sin 4 θ ( ) ] π / 2 0 =a^2\bigg[3\theta-2\sin(2\theta)+\dfrac{1}{4}\sin4\theta()\bigg]\begin{matrix}
\pi/2 \\
0
\end{matrix} = a 2 [ 3 θ − 2 sin ( 2 θ ) + 4 1 sin 4 θ ( ) ] π /2 0
= 3 π a 2 2 ( u n i t s 2 ) =\dfrac{3\pi a^2}{2} (units^2) = 2 3 π a 2 ( u ni t s 2 ) A r e a = 3 π 2 2 Area=\dfrac{3\pi ^2}{2} A re a = 2 3 π 2
#339914 on Dec 2023