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

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

1 Answer

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Answer 1 · 2018-01-22T12:52:41

I got as far as the definite integral but then I got stuck at the solution of the integral....probably there is a mistake or there is an easier way...

Explanation:

I considered the function and the rotation:

enter image source here

then I used the standard technique:

enter image source here

giving the expression of the surface area $S$ $S$ as:

$S = \int_{0}^{1} (2 π) (- \frac{\sqrt{x}}{e^{x}}) \cdot \sqrt{1 + \frac{4 x^{2} - 4 x + 1}{4 e^{2 x} \cdot x}} d x$ $S=int_0^1(2pi)(-sqrt(x)/e^x)*sqrt(1+(4x^2-4x+1)/(4e^(2x)*x))dx$

which I am not able to solve!!!

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

1 Answer

Explanation:

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

1 Answer

Explanation:

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