What is the surface area of the solid created by revolving $f (x) = \frac{e^{x^{2} + x - 1}}{x + 1}$ $f(x)=e^(x^2+x-1)/(x+1)$ over $x \in [0, 1]$ $x in [0,1]$ around the x-axis?

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Answer 1 · 2017-01-16T01:33:02

$\approx 1.45099$ $~~1.45099$

The Volume of Revolution about $O x$ $Ox$ is given by:

$V= int_(x=a)^(x=b) \ pi y^2 \ dx$

So for for this problem:

$V= int_0^1 \ pi (e^(x^2+x-1)/(x+1))^2 \ dx$

There is no elementary anti-derivative. The solution can be found numerically as $~~1.45099$

What is the surface area of the solid created by revolving f(x)=e^(x^2+x-1)/(x+1)f(x)=ex2+x−1x+1f(x)=e^(x^2+x-1)/(x+1) over x in [0,1]x∈[0,1]x in [0,1] around the x-axis?