What is the surface area of the solid created by revolving $f (x) = - 2 x^{3} - 3 x^{2} + 6 x - 12$ $f(x)=-2x^3-3x^2+6x-12$ over $x \in [2, 3]$ $x in [2,3]$ around the x-axis?

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What is the surface area of the solid created by revolving $f (x) = - 2 x^{3} - 3 x^{2} + 6 x - 12$ $f(x)=-2x^3-3x^2+6x-12$ over $x \in [2, 3]$ $x in [2,3]$ around the x-axis?

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Answer 1 · 2016-01-20T18:56:37

Use the surface area of rotation integral:
$S A = \int_{a}^{b} 2 π f (x) \sqrt{(1 + {(\frac{d f (x)}{d x})}^{2})} d x$ $SA = int_a^b 2pif(x) sqrt((1+((df(x))/dx)^2))dx$
Integrate the following:
$S A = 2 π \int_{2}^{3} (- 2 x^{3} - 3 x^{2} + 6 x - 12) \sqrt{1 + (- 6 x^{2} - 6 x + 6)} d x$ $SA =2pi int_2^3(-2x^3 - 3x^2 + 6x -12) sqrt(1+(-6x^2-6x+6))dx$

Explanation:

$S A = \int_{a}^{b} 2 π f (x) \sqrt{(1 + {(\frac{d f (x)}{d x})}^{2})} d x$ $SA = int_a^b 2pif(x) sqrt((1+((df(x))/dx)^2))dx$
may look bad it is not really. Take the derivative of f(x)
$\frac{d f (x)}{d x} = - 6 x^{2} - 6 x + 6$ $(df(x))/dx = -6x^2 -6x+6$
now integrate

$S A = 2 π \int_{2}^{3} (- 2 x^{3} - 3 x^{2} + 6 x - 12) \sqrt{1 + (- 6 x^{2} - 6 x + 6)} d x$ $SA =2pi int_2^3(-2x^3 - 3x^2 + 6x -12) sqrt(1+(-6x^2-6x+6))dx$

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What is the surface area of the solid created by revolving $f (x) = - 2 x^{3} - 3 x^{2} + 6 x - 12$ $f(x)=-2x^3-3x^2+6x-12$ over $x \in [2, 3]$ $x in [2,3]$ around the x-axis?

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Explanation:

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Science

Math

Humanities

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What is the surface area of the solid created by revolving f(x)=−2x3−3x2+6x−12f(x)=-2x^3-3x^2+6x-12 over x∈[2,3]x in [2,3] around the x-axis?

1 Answer

Explanation:

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What is the surface area of the solid created by revolving $f (x) = - 2 x^{3} - 3 x^{2} + 6 x - 12$ $f(x)=-2x^3-3x^2+6x-12$ over $x \in [2, 3]$ $x in [2,3]$ around the x-axis?