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The area enclosed by $\mathrm{y^{2}=8 x \: and \: y=\sqrt{2} x}$ that lies outside the triangle formed by $\mathrm{y=\sqrt{2} x, x=1, y=2 \sqrt{2},}$ is equal to:
Option: 1
$\frac{16 \sqrt{2}}{6}$

Option: 2
$\frac{11 \sqrt{2}}{6}$

Option: 3
$\frac{13 \sqrt{2}}{6}$

Option: 4
$\frac{5 \sqrt{2}}{6}$

Answers (1)

best_answer

Point of intersection of $\mathrm{y=\sqrt{2}x}$ and $\mathrm{y^{2}=8x}$

$\mathrm{\Rightarrow(\sqrt{2} x)^{2}=8 x \Rightarrow 2 x^{2}=8 x \Rightarrow x=0,4} \\$

$\mathrm{\Rightarrow y=\sqrt{2} \cdot 4=4 \sqrt{2}} \\$

$\mathrm{\text { At } x=1 \Rightarrow y^{2}=8 x \Rightarrow y^{2}=8 \Rightarrow y=2 \sqrt{2}}$

We need shaded area

$\mathrm{\left(\text { As } y^{2}=8 x \Rightarrow y=2 \sqrt{2 x}\right)}\\$

$\mathrm{=\int_{0}^{4}(2 \sqrt{2 x}-\sqrt{2} x) d x-\frac{1}{2} \cdot \sqrt{2} \cdot 1}\\$

$\mathrm{=\left.2 \sqrt{2} \cdot \frac{x^{3 / 2}}{3 / 2}\right|_{0} ^{4}-\left.\sqrt{2} \cdot \frac{x^{2}}{2}\right|_{0} ^{4}-\frac{1}{\sqrt{2}}}\\$

$\mathrm{=\frac{8 \sqrt{2}}{3}-\frac{1}{\sqrt{2}}=\frac{16 \sqrt{2}-3 \sqrt{2}}{6}=\frac{13 \sqrt{2}}{6} }$

Hence the answer is option 3.

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