What is the volume of the solid produced by revolving $f (x) = \frac{1}{x} - \frac{1}{x^{2}}, x \in [2, 6]$ $f(x)=1/x-1/x^2, x in [2,6]$ around the x-axis?

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What is the volume of the solid produced by revolving $f (x) = \frac{1}{x} - \frac{1}{x^{2}}, x \in [2, 6]$ $f(x)=1/x-1/x^2, x in [2,6]$ around the x-axis?

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Answer 1 · 2016-12-18T15:25:14

$\frac{49}{324}$ $49/324$

Explanation:

The portion of the graph of f(x) in the interval [2,6] looks like in the figure below:
enter image source here
If this portion is revolved around x-axis, the volume of the solid so generated can be worked out as follows:

Consider an element of the area enclosed by the x-axis and the curve, at a distance x from the origin and length y= f(x). If this is rotated about x-axis a circular disc of area $π y^{2}$ $pi y^2$ If width of this elementary disc is dx, its volume would be $π y^{2} d x$ $pi y^2 dx$ . The volume of the entire solid would be

$\int_{2}^{6} π y^{2} d x = \int_{2}^{6} π {(\frac{1}{x} - \frac{1}{x^{2}})}^{2} d x$ $int_2^6 pi y^2 dx= int_2^6 pi (1/x- 1/x^2)^2 dx$

= $π \int_{2}^{6} (\frac{1}{x^{2}} - \frac{2}{x^{3}} + \frac{1}{x^{4}}) d x$ $pi int_2^6 (1/x^2 -2/x^3 +1/x^4)dx$

= $π {[- \frac{1}{x} + \frac{1}{x^{2}} - \frac{1}{3 x^{3}}]}_{2}^{6}$ $pi[-1/x +1/x^2 -1/(3x^3)]_2^6$

= $π [- \frac{1}{6} + \frac{1}{36} - \frac{1}{648} + \frac{1}{2} - \frac{1}{4} + \frac{1}{24}] = \frac{49}{324}$ $pi[-1/6 +1/36 -1/648 +1/2 -1/4 +1/24]=49/324$

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What is the volume of the solid produced by revolving $f (x) = \frac{1}{x} - \frac{1}{x^{2}}, x \in [2, 6]$ $f(x)=1/x-1/x^2, x in [2,6]$ around the x-axis?

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Explanation:

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What is the volume of the solid produced by revolving f(x)=1x−1x2,x∈[2,6]f(x)=1/x-1/x^2, x in [2,6] around the x-axis?

1 Answer

Explanation:

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What is the volume of the solid produced by revolving $f (x) = \frac{1}{x} - \frac{1}{x^{2}}, x \in [2, 6]$ $f(x)=1/x-1/x^2, x in [2,6]$ around the x-axis?