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Areas of Bounded Regions — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsAreas of Bounded Regions5 Marks
Question
Find the area bounded by the ellipse $\frac{\text{x}^{2}}{\text{a}^{2}}+\frac{\text{y}^{2}}{\text{b}^{2}}=1$ and the ordinated x = ae and $x = 0$, where$ b^2= a^2(1 - e^2)$ and $e < 1.$
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Answer

$\frac{\text{x}^{2}}{\text{a}^{2}}+\frac{\text{y}^{2}}{\text{b}^{2}}=1$ represents a parabola, symmetrical about both the acis.
It cuts a x-axis at A(a, 0) and A'(-a, 0)
It cuts a y-axis at B(0, b) and B'(0, -b)
x = ae is a line parallel to y-axis
Consider a vertical strip of length = |y| and width = dx, in the first
Area of approximating rectangle in the first = |y|dx
Area of thr shaded region = 2 area in the first quadrant
$\Rightarrow \text{A}=2\int\limits_{0}^{\text{ae}}|\text{y}|\text{dx} $
$\Rightarrow\text{A}=2\int\limits_{0}^{\text{ae}}\text{y}\text{dx} $
$\Rightarrow \text{A}=2\int\limits_{0}^{\text{ae}}\frac{\text{b}}{\text{a}}\sqrt{\text{a}^{2}-\text{x}^{2}}\text{dx}$
$\Rightarrow \text{A}=\frac{2\text{b}}{\text{a}}\int\limits_{0}^{\text{ae}}\sqrt{\text{a}^{2}-\text{x}^{2}}\text{dx}$
$\Rightarrow \text{A}=\frac{2\text{b}}{\text{a}}\int\limits_{0}^{\text{ae}}\bigg[\frac{1}{2}\text{x}\sqrt{\text{a}^{2}-\text{x}^{2}}+\frac{1}{2}\text{a}^{2}\sin^{-1}\frac{\text{ae}}{\text{a}}-0\bigg]$
$\Rightarrow \text{A}=\frac{\text{b}}{\text{a}}\text{a}^{2}\Big[\text{e}\sqrt{1-\text{e}^{2}}+\frac{1}{2}\sin^{-1}\text{e}\Big]$
$\Rightarrow \text{A}=\text{ab}\Big[\text{e}\sqrt{1-\text{e}^{2}}+\frac{1}{2}\sin^{-1}\text{e}\Big]\ \text{sq.}\ \text{units}$

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