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AP EAMCET Engineering 23 Sep 2020 Shift 1 Paper

Section: Mathematics

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Question : 18 of 160

Marks: +1, -0

Calculate the area enclosed by the curves

x^2

2-y

x^2

y

$\frac{2}{3}$
$\frac{4}{3}$
$\frac{8}{3}$
$\frac{11}{3}$

Solution:

Given, curves are

x^{2}=y

....(i)

x^{2}=2-y

....(ii)

From Eqs. (i) and (ii),

y=2-y

\Rightarrow \;\; 2 y=2 \Rightarrow y=1

From Eq. (i),

x^{2}=1 \Rightarrow x=\pm 1

\therefore

Point of intersection are

A(-1,1), B(1,1)

Required area

=2\left[\int\limits_{1}^{2} \sqrt{2-y} d y+\int\limits_{0}^{1} \sqrt{y} d y\right]

=2\left[\left( \;\frac{(2-y)^{3/2}}{-3/2} \right)_{1}^{2}+\left[ \;\frac{y^{3/2}}{3/2} \right]_{0}^{1}\right]

=2\left[ \;\frac{0-1}{-3/2}+ \;\frac{1-0}{3/2} \right] = 2\left( \;\frac{4}{3} \right) = \;\frac{8}{3}

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