How do you find the volume of $y = \frac{3}{x + 1}$ $y=3/(x+1)$ , $y = 0$ $y=0$ ; $x = 0$ $x=0$ ; $x = 8$ $x= 8$ rotated around the x-axis?

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How do you find the volume of $y = \frac{3}{x + 1}$ $y=3/(x+1)$ , $y = 0$ $y=0$ ; $x = 0$ $x=0$ ; $x = 8$ $x= 8$ rotated around the x-axis?

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Answer 1 · 2016-03-30T15:57:17

$8 π$ $8 pi$

Explanation:

The region to be rotated around x axis is shown, shaded blue, in the picture shown below

enter image source here . In this region consider an element of width dx, at a distance x from y axis. Its length is y. If it is rotated about x axis, the volume of this elementary disc would be $π y^{2} d x$ $pi y^2 dx$

The volume of the solid so generated by the rotation of the whole shaded region would be

$\int_{0}^{8} π y^{2} d x$ $int_0^8 pi y^2 dx$ = $9 π \int_{0}^{8} \frac{1}{{(x + 1)}^{2}} d x$ $9pi int_0^8 1/(x+1)^2 dx$

= $9 π {[- \frac{1}{x + 1}]}_{0}^{8}$ $9 pi [-1/(x+1)]_0^8$

= $9 π [1 - \frac{1}{9}]$ $9pi [1-1/9]$ = $8 π$ $8 pi$

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How do you find the volume of $y = \frac{3}{x + 1}$ $y=3/(x+1)$ , $y = 0$ $y=0$ ; $x = 0$ $x=0$ ; $x = 8$ $x= 8$ rotated around the x-axis?

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How do you find the volume of y=3/(x+1)y=3x+1y=3/(x+1), y=0y=0y=0; x=0x=0x=0; x= 8x=8x= 8 rotated around the x-axis?

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Explanation:

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How do you find the volume of $y = \frac{3}{x + 1}$ $y=3/(x+1)$ , $y = 0$ $y=0$ ; $x = 0$ $x=0$ ; $x = 8$ $x= 8$ rotated around the x-axis?