The graph of the equation 2x=y^2 from A (0,0) to B (2,2) is revolved about the x-axis. The surface area of the resulting solid( in square units and to three decimal places) is ?

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The graph of the equation 2x=y^2 from A (0,0) to B (2,2) is revolved about the x-axis. The surface area of the resulting solid( in square units and to three decimal places) is ?

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Answer 1 · 2018-06-06T00:29:15

$54.454$ $54.454$ units^2

Explanation:

Surface area of revolution is given by, $A= 2piint_a^bf[x]sqrt[1+[f'[x]]^2$ dx.

From the question, $2x=y^2$ , differentiating implicitly, $2=2ydy/dx$ , i.e, $dy/dx= 1/y$ , but $y=sqrt2x,$ so $dy/dx=1/sqrt[2x]$ and so $[dy/dx]^2 = 1/[2x]$ .

Therefore $A= 2piint_a^b sqrt2xsqrt[1+1/[2x]$ dx= $2piint_a^bsqrt2xsqrt[[2x+1]/[2x]$ dx.

$A= 2piint_a^b[sqrt2x]/[sqrt2x]sqrt[2x+1]dx$ = $2piint_a^bsqrt[2x+1]$ dx......... $[1]$ .

Let $u= 2x+1$ , so $[du]/[dx]=2$ , ie, $dx=[du]/[2]$ .

substituting for $u$ and $du$ in ...... $[1]$ we have ,
$A= 2piintsqrtu[du]/[2]$ = $piintsqrtudu$ , and after integrating by the power rule,

In terms of $u$ area $A=[2pi]/[3]sqrt[u^3$ , and substituting back for $u = 2x+1$

$A = [2pi]/[3]sqrt[[2x+1]^3]$ , evaluated for $x= 2, x=0$ , will give the answer above.

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The graph of the equation 2x=y^2 from A (0,0) to B (2,2) is revolved about the x-axis. The surface area of the resulting solid( in square units and to three decimal places) is ?

1 Answer

Explanation:

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