Name: Area Enclosed by a Curve and a Line: Illustration
Uploaded: 2019-07-19
Duration: 3 min 35 s
Description: How would you find the area enclosed between a parabola and a straight line? Watch the video to see the method as an application of definite integrals.

Question

Find the area enclosed by the ellipse

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1

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Accepted Answer

We have $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ , which represents an ellipse with length of major axis $= a$ & length of minor axis $= b$ ; Ends of major axis $= (a, 0) & (- a, 0)$ .

Simplifying the equation

$\Rightarrow \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 \Rightarrow \frac{y^{2}}{b^{2}} = 1 - \frac{x^{2}}{a^{2}} \Rightarrow y = \pm \frac{b}{a} \sqrt{a^{2} - x^{2}}$

Area of ellipse $= 2 \int_{- a}^{a} y d x$

We have to find $\int_{- a}^{a} y d x$

As we know, $\int \sqrt{a^{2} - x^{2}} d x = \frac{x}{2} \sqrt{a^{2} - x^{2}} + \frac{a^{2}}{2} \sin^{- 1} (\frac{x}{a}) + c$

Area of ellipse $=$ $2 \int_{- a}^{a} \frac{b}{a} \sqrt{a^{2} - x^{2}} d x = \frac{2 b}{a} \int_{- a}^{a} \sqrt{a^{2} - x^{2}} d x = \frac{2 b}{a} {[\frac{x}{2} \sqrt{a^{2} - x^{2}} + \frac{a^{2}}{2} \sin^{- 1} (\frac{x}{a})]}_{- a}^{a} = \frac{2 b}{a} [\frac{a}{2} (0) + \frac{a^{2}}{2} \sin^{- 1} (1) - [\frac{- a}{2} (0) + \frac{a^{2}}{2} \sin^{- 1} (- 1)]] = \frac{2 b}{a} [\frac{a^{2}}{2} \frac{π}{2} - \frac{a^{2}}{2} (\frac{- π}{2})] = \frac{2 b}{a} [\frac{a^{2} π}{2}] = πab$

Find the area enclosed by the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ .

Important Questions on Application of Integrals

Learn and Prepare for any exam you want !

Find the area enclosed by the ellipse x2a2+y2b2=1.

Important Questions on Application of Integrals

Learn and Prepare for any exam you want !

Find the area enclosed by the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ .