What is the surface area of the solid created by revolving $f (x) = \sqrt{4 x}$ $f(x)=sqrt(4x)$ for $x \in [0, 1]$ $x in [0,1]$ around the x-axis?

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Answer 1 · 2016-06-26T11:44:00

$= 2 π$ $= 2pi$

the thin vertical strip width $Delta x$ and between $y = 0$ and $y = sqrt{4x}$ will, when revolved round the x-axis, have volume

$Delta V = pi y^2 Delta x$

So

$V = pi int_0^1 \ y^2 dx$

$= pi int_0^1 \ 4x \ dx$ ........ as $y = sqrt{4x}$

$= 4pi int_0^1 \ x \ dx$

$= 4pi [ x^2/2]_0^1$

$= 2pi$

What is the surface area of the solid created by revolving f(x)=sqrt(4x)f(x)=√4xf(x)=sqrt(4x) for x in [0,1]x∈[0,1]x in [0,1] around the x-axis?