SATHEE: Area Question 48

Area Question 48

Question 48

Find the area bounded by the curves $x^{2} + y^{2} = 25, 4 y = | 4 - x^{2} |$ and $x = 0$ above the $X$ -axis.

$(1987, 6 M)$

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Solution:

Given curves, $x^{2} + y^{2} = 25, 4 y = | 4 - x^{2} |$ could be sketched as below, whose points of intersection are

$x^{2} + \frac{{(4 - x^{2})}^{2}}{16} = 25$

$\Rightarrow (x^{2} + 24) (x^{2} - 16) = 0$

$\Rightarrow x = \pm 4$

$∴$ Required area $= 2 \int_{0}^{4} \sqrt{25 - x^{2}} d x - \int_{0}^{2} \frac{4 - x^{2}}{4} d x$

$$ \begin{gathered} -\int_{2}^{4} \frac{x^{2}-4}{4} d x \ =2 \frac{x}{2} \sqrt{25-x^{2}}+\frac{25}{2} \sin ^{-1} \frac{x}{5}{ }{0}^{4} \ -\frac{1}{4} 4 x-\frac{x^{3}}{3}{ }{0}^{2}-\frac{1}{4} \frac{x^{3}}{3}-4 x_{2}^{4} \end{gathered} $$

$= 2 6 + \frac{25}{2} \sin^{- 1} \frac{4}{5} - \frac{1}{4} 8 - \frac{8}{3}$

$- \frac{1}{4} \frac{64}{3} - 16 - \frac{8}{3} - 8$

$= 2 6 + \frac{25}{2} \sin^{- 1} \frac{4}{5} - \frac{4}{3} - \frac{4}{3} - \frac{4}{3}$

$= 4 + 25 \sin^{- 1} \frac{4}{5}$ sq units

Area Question 48

Question 48

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Area Question 48

Question 48

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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