# Find the volume of the solid formed by rotating the region enclosed by y=e^(5x) , \ y=0, \ x=0, \ x=0.8 about the x-axis. Use integrals to find the volume.

*print*Print*list*Cite

Plot the given equations to determine the region enclosed by the curves.

So the bounded region is:

(Red-Graph of `y=e^(5x)` ; Green-Graph of `x=0.8` ; Blue-Graph of `x=0` and; Purple-Graph of `y=0` )

To solve for the volume of the bounded region when it is revolved about the x-axis, we may use the disk method. The formula is:

`V=pi int_a^b (r_o^2 - r_i^2)dx`

Base on the graph, the outer radius is `y=e^(5x)` .

`r_o = e^(5x)`

`r_o^2 = (e^(5x))^2`

And the inner radius is y=0.

`r_i=0`

`r_i^2 =0`

Then, let's determine the limits of the integral.

Note that `y=0` and `y=e^(5x)` does not intersect base on the properties of exponential function, wherein the "a" in `a=b^m` should always be greater than zero.

So our limits of the integral are the x-coordinates of intersection between `x=0` & `y=e^(5x)` and between ` x=0.8` & `y=e^(5x)` .

Hence, a=0 and b=0.8.

The volume then is:

`V=pi int_0^0.8 (e^(10x)-0)dx = pi int_0^0.8 e^(10x) dx`

`V= pi e^(10x)/10` `|_0^0.8` `= pi(e^(10*0.8) - e^(10*0)) =pi(e^8-1)`

`V = 297.996`

**Hence the volume of the solid formed by rotating the enclosed region of the given equations about x-axis is 297.996 cubic units.**