How do you find the area of the surface generated by rotating the curve about the y-axis $x = t^{2}, y = \frac{1}{3} t^{3}, 0 \leq t \leq 3$ $x=t^2, y=1/3t^3, 0<=t<=3$ ?

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Answer 1 · 2018-07-30T00:18:44

For infinitesimal arc length, $d s$ $ds$ :

$d s = \sqrt{{d x}^{2} + {d y}^{2}}$ $ds = sqrt(dx^2 + dy^2 )$

$:. ds = sqrt(4t^2 + t^4 ) \ dt = t sqrt(4 + t^2 ) \ dt$

Spinning round the y-axis, creating infinitesimal surface area $dS$ , where:

$= 2 pi \ t^3 \ sqrt(4 + t^2 ) \ dt$

$:. S =2 pi int_0^3 \ dt qquad t^3 sqrt(4 + t^2 )$

$= (2 pi)/15 (64 + 247 sqrt13)$

I have skipped the mechanical integration steps. you can start with a sub:

$:. S =2 pi int_4^13 \ (du)/(2t) qquad t^3 \ sqrtu$

$= pi int_4^13 \ du qquad (u - 4) \ sqrtu$

$= pi int_4^13 \ du qquad u^(3/2) - 4u^(1/2)$

Which is trivial but protracted

How do you find the area of the surface generated by rotating the curve about the y-axis x=t^2, y=1/3t^3, 0<=t<=3x=t2,y=13t3,0≤t≤3x=t^2, y=1/3t^3, 0<=t<=3?