Ellipse

Calculus Level 2

{ x = a cos ⁡ t y = b sin ⁡ t \large \begin{cases} x = a \cos t \\ y = b \sin t \end{cases}

The equation above shown are equation of an ellipse . Find the area enclosed by ellipse in terms of a , b a , b .

Details :

  • 0 ≤ t ≤ 2 π 0 \leq t \leq 2\pi .
  • a , b a , b are positive constants.
π 4 a b \dfrac{\pi}{4}ab π a b \pi ab π 2 a b \dfrac{\pi}{2}ab 4 π a b 4\pi ab a b ab 2 π a b 2\pi ab

Akhil Bansal by Akhil Bansal , 22, India

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1 solution

The area of a function given by parametric equations x = f ( t ) , x=f(t), y = g ( t ) , y = g(t), is given by A = ∫ a b g ( t ) f ′ ( t ) d t . A=\int_{a}^{b} g(t)f'(t)dt. In this problem, g ( t ) = a cos ⁡ ( t ) , g(t) = a \cos (t), f ( t ) = a sin ⁡ ( t ) , f(t) = a \sin (t), f ′ ( t ) = a cos ⁡ ( t ) , f'(t)= a \cos (t), so the area is given by: A = ∫ 0 2 π a b cos ⁡ 2 ( t ) d t = a b ∫ 0 2 π cos ⁡ 2 ( t ) d t = a b [ 2 π 2 + 1 4 sin ⁡ ( 4 π ) − 0 2 − 1 4 sin ⁡ ( 0 ) ] = π a b A=\int_{0}^{2 \pi} a b \cos^2 (t) dt = ab \int_{0}^{2 \pi} \cos^2 (t) dt = ab\left [ \frac{2\pi}{2 } + \frac{1}{4} \sin(4\pi)-\frac{0}{2}-\frac{1}{4}\sin(0) \right ]=\pi ab

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