The question
Using integration, find the area of the ellipse x 2 16 + y 2 4 = 1 \dfrac{x^2}{16} + \dfrac{y^2}{4} = 1 16 x 2 + 4 y 2 = 1 x = − 2 x = -2 x = − 2 x = 2 x = 2 x = 2
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The ellipse x 2 16 + y 2 4 = 1 \frac{x^2}{16} + \frac{y^2}{4} = 1 16 x 2 + 4 y 2 = 1 ( ± 4 , 0 ) (\pm4, 0) ( ± 4 , 0 ) ( 0 , ± 2 ) (0, \pm2) ( 0 , ± 2 ) x = − 2 x = -2 x = − 2 x = 2 x = 2 x = 2
Why?
An ellipse x 2 a 2 + y 2 b 2 = 1 \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 a 2 x 2 + b 2 y 2 = 1 ( ± a , 0 ) (\pm a, 0) ( ± a , 0 ) ( 0 , ± b ) (0, \pm b) ( 0 , ± b ) a 2 = 16 a^2 = 16 a 2 = 16 b 2 = 4 b^2 = 4 b 2 = 4 a = 4 a = 4 a = 4 b = 2 b = 2 b = 2
The lines x = ± 2 x = \pm2 x = ± 2 2 < 4 2 < 4 2 < 4
x = −2
x = 2
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The ellipse x 2 16 + y 2 4 = 1 \frac{x^2}{16} + \frac{y^2}{4} = 1 16 x 2 + 4 y 2 = 1 x = − 2 x = -2 x = − 2 x = 2 x = 2 x = 2 From the equation of the ellipse,
y 2 = 4 ( 1 − x 2 16 ) = 16 − x 2 4 \begin{aligned} y^2 &= 4\left(1 - \frac{x^2}{16}\right) \\ &= \frac{16 - x^2}{4} \end{aligned} y 2 = 4 ( 1 − 16 x 2 ) = 4 16 − x 2 For the upper half of the ellipse, y = 1 2 16 − x 2 y = \frac12\sqrt{16 - x^2} y = 2 1 16 − x 2
Why?
To integrate we need y as a function of x. Make y 2 4 \frac{y^2}{4} 4 y 2
y 2 4 = 1 − x 2 16 y 2 = 4 ( 1 − x 2 16 ) = 4 − 4 x 2 16 = 4 − x 2 4 = 16 4 − x 2 4 = 16 − x 2 4 \begin{aligned} \frac{y^2}{4} &= 1 - \frac{x^2}{16} \\ y^2 &= 4\left(1 - \frac{x^2}{16}\right) \\ &= 4 - \frac{4x^2}{16} = 4 - \frac{x^2}{4} \\ &= \frac{16}{4} - \frac{x^2}{4} = \frac{16 - x^2}{4} \end{aligned} 4 y 2 y 2 = 1 − 16 x 2 = 4 ( 1 − 16 x 2 ) = 4 − 16 4 x 2 = 4 − 4 x 2 = 4 16 − 4 x 2 = 4 16 − x 2 The square root of a fraction is the root of the top over the root of the bottom, and 4 = 2 \sqrt4 = 2 4 = 2
y = ± 16 − x 2 2 y = \pm\frac{\sqrt{16 - x^2}}{2} y = ± 2 16 − x 2 The upper half of the curve is the + + +
The region is symmetric about both axes, so
Area = 4 ∫ 0 2 y d x = 4 ∫ 0 2 1 2 4 2 − x 2 d x = 2 ∫ 0 2 4 2 − x 2 d x \begin{aligned} \text{Area} &= 4\int_0^2 y\,dx \\ &= 4\int_0^2\frac12\sqrt{4^2 - x^2}\,dx \\ &= 2\int_0^2\sqrt{4^2 - x^2}\,dx \end{aligned} Area = 4 ∫ 0 2 y d x = 4 ∫ 0 2 2 1 4 2 − x 2 d x = 2 ∫ 0 2 4 2 − x 2 d x Why?
The equation is unchanged when x is replaced by − x -x − x − y -y − y x = ± 2 x = \pm2 x = ± 2
So the shaded region is made of four equal pieces. The piece in the first quadrant lies under the upper curve from x = 0 x = 0 x = 0 x = 2 x = 2 x = 2 ∫ 0 2 y d x \int_0^2 y\,dx ∫ 0 2 y d x
Then 4 ⋅ 1 2 = 2 4\cdot\frac12 = 2 4 ⋅ 2 1 = 2
Using ∫ a 2 − x 2 d x \displaystyle\int\sqrt{a^2 - x^2}\,dx ∫ a 2 − x 2 d x = x 2 a 2 − x 2 + a 2 2 sin − 1 x a + C = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\frac{x}{a} + C = 2 x a 2 − x 2 + 2 a 2 sin − 1 a x + C
Why?
a 2 − x 2 \sqrt{a^2 - x^2} a 2 − x 2 x 2 x^2 x 2 a = 4 a = 4 a = 4 a 2 = 16 a^2 = 16 a 2 = 16 a 2 2 = 8 \frac{a^2}{2} = 8 2 a 2 = 8
For a definite integral the constant C is not needed.
Area = 2 [ x 2 16 − x 2 + 8 sin − 1 x 4 ] 0 2 = 2 [ 12 + 8 ⋅ π 6 ] − 2 ( 0 + 0 ) = 2 12 + 8 π 3 \begin{aligned} \text{Area} &= 2\bigg[\frac{x}{2}\sqrt{16 - x^2} \\ &\qquad + 8\sin^{-1}\frac{x}{4}\bigg]_0^2 \\ &= 2\left[\sqrt{12} + 8\cdot\frac{\pi}{6}\right] - 2(0 + 0) \\ &= 2\sqrt{12} + \frac{8\pi}{3} \end{aligned} Area = 2 [ 2 x 16 − x 2 + 8 sin − 1 4 x ] 0 2 = 2 [ 12 + 8 ⋅ 6 π ] − 2 ( 0 + 0 ) = 2 12 + 3 8 π Why?
At x = 2 x = 2 x = 2 2 2 16 − 4 = 12 \frac22\sqrt{16 - 4} = \sqrt{12} 2 2 16 − 4 = 12 sin − 1 2 4 = sin − 1 1 2 = π 6 \sin^{-1}\frac24 = \sin^{-1}\frac12 = \frac{\pi}{6} sin − 1 4 2 = sin − 1 2 1 = 6 π
2 2 16 − ( 2 ) 2 + 8 sin − 1 2 4 = 1 ⋅ 16 − 4 + 8 sin − 1 1 2 = 12 + 8 ⋅ π 6 \begin{aligned} &\frac{2}{2}\sqrt{16 - (2)^2} + 8\sin^{-1}\frac{2}{4} \\ &= 1 \cdot \sqrt{16 - 4} + 8\sin^{-1}\frac12 \\ &= \sqrt{12} + 8\cdot\frac{\pi}{6} \end{aligned} 2 2 16 − ( 2 ) 2 + 8 sin − 1 4 2 = 1 ⋅ 16 − 4 + 8 sin − 1 2 1 = 12 + 8 ⋅ 6 π (sin − 1 1 2 = π 6 \sin^{-1}\frac12 = \frac{\pi}{6} sin − 1 2 1 = 6 π sin π 6 = 1 2 \sin\frac{\pi}{6} = \frac12 sin 6 π = 2 1
At x = 0 x = 0 x = 0 sin − 1 0 = 0 \sin^{-1}0 = 0 sin − 1 0 = 0
0 2 16 − 0 + 8 sin − 1 0 = 0 + 0 \frac{0}{2}\sqrt{16 - 0} + 8\sin^{-1}0 = 0 + 0 2 0 16 − 0 + 8 sin − 1 0 = 0 + 0 Finally 2 12 = 2 ⋅ 2 3 = 4 3 2\sqrt{12} = 2\cdot 2\sqrt3 = 4\sqrt3 2 12 = 2 ⋅ 2 3 = 4 3 2 ⋅ 8 π 6 = 8 π 3 2\cdot\frac{8\pi}{6} = \frac{8\pi}{3} 2 ⋅ 6 8 π = 3 8 π
12 = 4 × 3 = 4 3 = 2 3 2 12 = 2 × 2 3 = 4 3 2 ⋅ 8 ⋅ π 6 = 16 π 6 = 16 π ÷ 2 6 ÷ 2 = 8 π 3 \begin{aligned} \sqrt{12} &= \sqrt{4 \times 3} = \sqrt4\,\sqrt3 = 2\sqrt3 \\ 2\sqrt{12} &= 2 \times 2\sqrt3 = 4\sqrt3 \\[4pt] 2 \cdot 8 \cdot \frac{\pi}{6} &= \frac{16\pi}{6} \\ &= \frac{16\pi \div 2}{6 \div 2} = \frac{8\pi}{3} \end{aligned} 12 2 12 2 ⋅ 8 ⋅ 6 π = 4 × 3 = 4 3 = 2 3 = 2 × 2 3 = 4 3 = 6 16 π = 6 ÷ 2 16 π ÷ 2 = 3 8 π ∴ \therefore ∴ = ( 4 3 + 8 π 3 ) = \left(4\sqrt3 + \dfrac{8\pi}{3}\right) = ( 4 3 + 3 8 π )
Where this comes from Chapter 8, Application of Integrals : Area of part of an ellipse between two lines.
Question type: Ellipses
How to spot it: click to see Type 4. Ellipses: read off a a a b b b
Spot it: x 2 a 2 + y 2 b 2 = 1 \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 a 2 x 2 + b 2 y 2 = 1 x 2 x^2 x 2 y 2 y^2 y 2
Here: a = 4 a = 4 a = 4 b = 2 b = 2 b = 2 y = 1 2 16 − x 2 y = \frac12\sqrt{16 - x^2} y = 2 1 16 − x 2 x = ± 2 x = \pm2 x = ± 2
Quick lesson · Worked NCERT solutions · Interactive lesson
Also uses Chapter 7, Integrals .