# 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 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.

lfryerda | High School Teacher | (Level 2) Educator

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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.