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Find the volume of the solid generated by revolving around the x-axis the triangle with...

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master451 | Student | Honors

Posted September 24, 2012 at 1:06 PM via web

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Find the volume of the solid generated by revolving around the x-axis the triangle with vertices (0,0), (1,0), (1,1)

 

Use the definition of the definite integral (using Riemann Sum) to find the volume.

Check your answer by evaluating the integral using the FTC.

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lfryerda | High School Teacher | (Level 2) Educator

Posted September 24, 2012 at 8:53 PM (Answer #1)

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To find this volume, we can find vertical cylinders of radius x and height 2y where y=x, and width dx.  Then the volume of each disc is 

`dV=2pi x 2 y dx`

`=4pi x^2dx` 

The volume is then the integral

`V=4pi int_0^1x^2dx`

`={4pi}/3(x^3)_0^1`

`={4pi}/3` 

The volume of the solid is  `{4pi}/3` .

To construct the Riemann sum, we are summing n vertical cylinders, each of width `Delta x={1-0}/n=1/n` , height 2y and radius `x=i Delta x = i/n` . Then the volume of each disc is:

`Delta V=2pi x 2y Delta x`

`=4pi x^2 Delta x`

`={4pi i^2}/n^3`

The the volume of the solid is the sum of all discs in the limit as `n->infty` . This gives

`V_n=sum_{i=0}^n{4pi i^2}/n^3`

`={4pi}/n^3 sum_{i=0}^n i^2`

`={4pi}/n^3{n(n+1)(2n+1)}/6`

` ` `={2pi}/3(2+3/n+1/n^2)`

Then the volume is the limit as `n->infty` .

`V=lim_{n->infty}{2pi}/3(2+3/n+1/n^2)`

`={4pi}/3`

The volume of the solid is `{4pi}/3` .

To check this using the fundamental theorem of calculus, we need to consider the cubic function `g(x)={4pi}/3x^3` . Since `g'(x)=4pix^2` , and it is the integrand of `int_0^1 4pi x^2dx` , we see that the solution is correct.

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