How do you find the area of the surface generated by rotating the curve about the y-axis $y=1/3(x^2+2)^(3/2), 1<=x<=2$ ?

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Answer 1 · 2017-04-18T20:03:31

The surface area of a curve $y$ rotated around the $x$ -axis on $alt=xlt=b$ is given by:

$S=2piint_a^bysqrt(1+(y')^2)dx$

Where $y=1/3(x^2+2)^(3/2)$ we see that $y'=1/3(3/2(x^2+2)^(1/2))(2x)=xsqrt(x^2+2)$ .

Then the surface area is:

$S=2piint_1^2 1/3(x^2+2)^(3/2)sqrt(1+(xsqrt(x^2+2))^2)dx$

$S=(2pi)/3int_1^2(x^2+2)^(3/2)sqrt(1+x^2(x^2+2))dx$

Note that $sqrt(1+x^2(x^2+2))=sqrt(x^2+2x^2+1)=sqrt((x^2+1)^2)=x^2+1$ :

$S=(2pi)/3int_1^2(x^2+2)^(3/2)(x^2+1)dxapprox69.9055$

How do you find the area of the surface generated by rotating the curve about the y-axis y=1/3(x^2+2)^(3/2), 1<=x<=2y=1/3(x^2+2)^(3/2), 1<=x<=2?