SATHEE: Area Question 49

Area Question 49

Question 49

Find the area bounded by the curves $x^{2} + y^{2} = 4$ , $x^{2} = - \sqrt{2} y$ and $x = y$ .

$(1986, 5$ M)

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Solution:

Given curves are $x^{2} + y^{2} = 4, x^{2} = - \sqrt{2} y$ and $x = y$ .

Thus, the required area

$\begin{aligned} = | \int_{- \sqrt{2}}^{\sqrt{2}} \sqrt{4 - x^{2}} d x | - | \int_{- \sqrt{2}}^{0} x d x | - | \int_{0}^{\sqrt{2}} \frac{- x^{2}}{\sqrt{2}} d x | & = 2 \int_{0}^{\sqrt{2}} \sqrt{4 - x^{2}} d x - \frac{x^{2}}{2} - \frac{x^{3}}{3 \sqrt{2}} 0 & = 2 \frac{x}{2} \sqrt{4 - x^{2}} - \frac{4}{2} \sin^{- 1} {\frac{x}{2}}_{0}^{\sqrt{2}} - 1 - \frac{2}{3} & = (2 - π) - \frac{5}{3} & = \frac{1}{3} - π sq units \end{aligned}$

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Area Question 49

Question 49

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