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  • If the area of the region <img alt="\mathrm{\left\{(x, y): x^{\frac{2}{3}}+y^{\frac{2}{3}} \leq 1, x+y \geq 0, y \geq 0\right\}}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5

If the area of the region \mathrm{\left\{(x, y): x^{\frac{2}{3}}+y^{\frac{2}{3}} \leq 1, x+y \geq 0, y \geq 0\right\}} is \mathrm{A}, then \mathrm{\frac{256A}{\pi}} is equal to _______.

Option: 1

36


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

Desired area in second quadrant =\frac{1}{2} Area in first quadrant

So total Area   \mathrm{\frac{3}{2} \int_{0}^{1}\left(1-x^{2 / 3}\right)^{3 / 2} d x}

\mathrm{Let\: x^{1 / 3}=\sin \theta \Rightarrow 1-x^{2 / 3}=\cos ^{2} \theta, }\\

\mathrm{\Rightarrow x \cdot \sin ^{3} \theta \Rightarrow d x=3 \sin ^{2} \theta \cos \theta d \theta}.

\mathrm{Area= \frac{3}{2} \int_{0}^{\pi / 2}\left(\cos ^{2} \theta\right)^{3 / 2} \times 3 \sin ^{2} \theta \cos \theta d \theta}.

\mathrm{=\frac{9}{2} \int_{0}^{\pi / 2} \sin ^{2} \theta 4s^{4} \theta d \theta=\frac{9}{2} \times \frac{1 \times 3 \times 1}{6 \times 4 \times 2} \times \frac{\pi}{2}=\frac{9 \pi}{64} }\\

\mathrm{\Rightarrow \frac{256 \mathrm{~A}}{\pi}=36 }

Hence the answer is \mathrm{36 }

 

Posted by

Riya

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