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

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Answer 1 · 2016-06-23T09:04:03

$\frac{7 π}{12}$ $(7 pi) /12$

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Small element of width $Delta x$ has volume $Delta V = pi f^2(x) Delta x$ so

$V = pi \int_0^1 \ (x(sqrt(x+1))^2 \ dx$

$= pi \int_0^1 \ x^2(x+1) \ dx = pi \int_0^1 \ x^3+ x^2 \ dx$

$= pi [ (x^4)/4+ (x^3)/3]_0^1$

$= (7 pi) /12$

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