Question #326261

find the surface area of the object obtained by rotating y=4+3x^2, 1<=x<=2 about the y axis

1
Expert's answer
2022-04-11T17:57:53-0400

If the continuous curve is described by the function y = f(x), axb, then the area is equal to integral

A=2πabx1+(dydx)2dxA=2\pi\int_{a}^{b}{x\sqrt{1+\left(\frac{dy}{dx}\right)^2}dx}

For given function

A=2π12x1+(6x)2dx=π36122(6x)1+(6x)2d(6x)A=2\pi\int_{1}^{2}{x\sqrt{1+\left(6x\right)^2}dx}=\frac{\pi}{36}\int_{1}^{2}{2\left(6x\right)\sqrt{1+\left(6x\right)^2}d\left(6x\right)}

Substitution t=6x:

A=π366122t1+t2dtA=\frac{\pi}{36}\int_{6}^{12}{2t\sqrt{1+t^2}dt}

Substitution  u=t2:

A=π36361441+udu=π3623(1+u)314436=π54(1451453737)=88.486A=\frac{\pi}{36}\int_{36}^{144}{\sqrt{1+u}du}=\frac{\pi}{36}\frac{2}{3}\left(\sqrt{1+u}\right)^3\left|\begin{matrix}144\\36\\\end{matrix}=\frac{\pi}{54}\left(145\sqrt{145}-37\sqrt{37}\right)\right.=88.486

Answer

A = 88.486


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