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The surface area of the solid of revolution of the region bounded by $y=2x,x=0$ and $x=2$ about $x$-axis is

$\begin{array}{1 1}(1)8\sqrt{5}\pi&(2)2\sqrt{5}\pi\\(3)\sqrt{5}\pi&(4)4\sqrt{5}\pi\end{array}$

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Solvig $x=2$ & $y=2x$
We have $y=4$
$OA= 2\qquad AB= 4$
$OB =\sqrt {4+ 16} =2 \sqrt {5}$
Required surface area = surface area of the cone of radius 4 height 2.
$\qquad= \pi rl=\pi \times 4 \times 2 \sqrt 5$
$\qquad= 8 \sqrt 5\; \pi$
Hence 1 is the correct answer.
answered May 21, 2014 by

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