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  • The area of the region <img alt="\mathrm{\mathrm{S}=\left\{(x, y): y^{2} \leq 8 x, y \geq \sqrt{2} x, x \geq 1\right\}}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Cmathrm%7

The area of the region \mathrm{\mathrm{S}=\left\{(x, y): y^{2} \leq 8 x, y \geq \sqrt{2} x, x \geq 1\right\}} is

Option: 1

\frac{13 \sqrt{2}}{6}


Option: 2

\frac{11 \sqrt{2}}{6}


Option: 3

\frac{5 \sqrt{2}}{6}


Option: 4

\frac{19 \sqrt{2}}{6}


Answers (1)

best_answer

Points of intersection of these curves are

\mathrm{\left ( 1,2\sqrt{2} \right ),\left ( 1,-2\sqrt{2} \right ),\left ( 0,0 \right ),\left ( 4,4\sqrt{2} \right ),\left ( 1,\sqrt{2} \right )}.


\mathrm{I= \int_{1}^{4}\left ( 2\, \sqrt{2}\, \sqrt{x}-\sqrt{2}\, x \right )\: dx}
   \mathrm{=\left [ 2\, \sqrt{2}\times \frac{2}{3}x^{3/2} - \sqrt{2}\, \frac{x^{2}}{2}\right ]_{1}^{4}} 
   \mathrm{=\frac{32 \sqrt{2}}{3}-8 \sqrt{2}-\frac{4 \sqrt{2}}{3}+\frac{ \sqrt{2}}{2}= \frac{64 \sqrt{2}-48 \sqrt{2}-8 \sqrt{2}+3 \sqrt{2}}{6}}
   \mathrm{=\frac{11 \sqrt{2}}{6}} 
Option (B)
 

Posted by

rishi.raj

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