Application of Integrals — Part 2: Areas of circles and ellipses: visual lesson, CBSE Class 12 Mathematics

Lesson

Application of Integrals, Part 2

Circles and ellipses: the area formulas you know, now proved.

In this lesson you will

Write the upper half of a circle or an ellipse as a function
Use the standard integral of the square root of a² − x²
Prove the areas πa² and πab by integration
Find areas of pieces cut off by lines parallel to an axis
Choose the slicing and symmetry factor for a region

The shaded quarter has area $\displaystyle\int_0^4 \sqrt{16 - x^2}\,dx = 4\pi$.

Concept

What you will be able to do

Write the upper half of a circle or ellipse as $y = \ldots$
Use $\displaystyle\int \sqrt{a^2 - x^2}\,dx$
Prove $\pi a^2$ and $\pi ab$
Find pieces cut off by lines $x = h$
Choose the symmetry factor: $2$ or $4$

The one integral this part needs.
Integral	Antiderivative
$\displaystyle\int \sqrt{a^2 - x^2}\,dx$	$\dfrac x2\sqrt{a^2 - x^2} + \dfrac{a^2}{2}\sin^{-1}\dfrac xa + C$
$\displaystyle\int_0^a \sqrt{a^2 - x^2}\,dx$	$\dfrac{\pi a^2}{4}$

Concept

A circle as two functions

$x^2 + y^2 = a^2$ is not a function, but its upper half is: $y = \sqrt{a^2 - x^2}$ for $-a \le x \le a$ . The lower half is $y = -\sqrt{a^2 - x^2}$ .

$y = \sqrt{36 - x^2}$
$y = -\sqrt{36 - x^2}$

Four equal quarters: area $= 4 \times 9\pi = 36\pi$.

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