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GUJCET 2020 Math Solved Paper

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Question : 34 of 40

Marks: +1, -0

The area enclosed by the curve

x=4\cos\theta, y=3\sin\theta

is ____________.

$4\pi$
$6\pi$
$8\pi$
$12\pi$

Solution:

Here,

x=4\cos\theta, \;\; y=3\sin\theta

\cos\theta=\;\frac{x}{4}, \;\; \sin\theta=\;\frac{y}{3}

We know that,

\sin^{2}\theta+\cos^{2}\theta=1

\left(\;\frac{y}{3}\right)^{2}+\left(\;\frac{x}{4}\right)^{2}=1

\;\frac{x^{2}}{16}+\;\frac{y^{2}}{9}=1

\Rightarrow \;\; \frac{y}{9^{2}}=1-\;\frac{x^{2}}{16}

\Rightarrow \;\; y^{2}=\;\frac{9}{16}(16-x^{2})

Here,Area

=4 \times \text{A of I}

\;\; =4 \int\limits_{0}^{4} y \, dx

\;\; =4 \times \int\limits_{0}^{4} \;\frac{3}{4} \sqrt{16-x^{2}} \, dx

\;\; =3 \int\limits_{0}^{4} \sqrt{16-x^{2}} \, dx

\;\; =3 \int\limits_{0}^{4} \sqrt{(4)^{2}-x^{2}} \, dx

\left[\int \sqrt{a^{2}-x^{2}} \, dx = \frac{x}{2} \sqrt{a^{2}-x^{2}} + \frac{a^{2}}{2} \sin^{-1}\left(\frac{x}{a}\right) + c \right]

\;\; =3\left[ \frac{x}{2} \sqrt{16-x^{2}} + \frac{16}{2} \sin^{-1}\left(\frac{x}{4}\right) \right]_{0}^{4}

\;\; =3\left[ \frac{4}{2} \sqrt{16-16} + 8 \sin^{-1}\left(\frac{4}{4}\right) - 0 - 8 \sin^{-1}(0) \right]

\;\; =3\left[ 0 + 8 \sin^{-1}(1) - 0 - 8 \sin^{-1}(0) \right]

\;\; =3\left[ \frac{4}{8}\left[ \frac{\pi}{z} - 0 \right] \right]

\;\; =12\pi

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