What is the surface area of the solid created by revolving $f (x) = \sqrt{x^{3}}$ $f(x)=sqrt(x^3)$ for $x \in [1, 2]$ $x in [1,2]$ around the x-axis?

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What is the surface area of the solid created by revolving $f (x) = \sqrt{x^{3}}$ $f(x)=sqrt(x^3)$ for $x \in [1, 2]$ $x in [1,2]$ around the x-axis?

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Answer 1 · 2016-03-17T18:02:57

24.93

Explanation:

Surface area of solid of revolution about x axis is given by the formula

$S = \int_{a}^{b} 2 π y d s = \int_{a}^{b} 2 π y \sqrt{1 + {(\frac{d y}{d x})}^{2}} d x$ $S= int_a^b 2pi y ds = int_a^b 2piy sqrt (1+(dy/dx)^2) dx$ . In the present case ,
$\frac{d y}{d x} = \frac{3}{2} x^{\frac{1}{2}}$ $dy/dx = 3/2 x^(1/2)$ , hence $\sqrt{1 + {(\frac{d y}{d x})}^{2}} = \sqrt{1 + 9 \frac{x}{4}}$ $sqrt (1+(dy/dx)^2)=sqrt ( 1+9x/4)$

$S = 2 π \int_{1}^{2} x^{\frac{3}{2}} \sqrt{1 + 9 \frac{x}{4}} d x$ $S= 2pi int_1^2 x^(3/2)sqrt (1+9x/4) dx$

= $π \int_{1}^{2} x^{\frac{3}{2}} \sqrt{9 x + 4} d x$ $pi int_1^2 x^(3/2) sqrt (9x+4) dx$

This integration is too long to write it here. hence
use integral calculator to solve the integral

= 24.93

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What is the surface area of the solid created by revolving $f (x) = \sqrt{x^{3}}$ $f(x)=sqrt(x^3)$ for $x \in [1, 2]$ $x in [1,2]$ around the x-axis?

1 Answer

Explanation:

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What is the surface area of the solid created by revolving $f (x) = \sqrt{x^{3}}$ $f(x)=sqrt(x^3)$ for $x \in [1, 2]$ $x in [1,2]$ around the x-axis?