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.
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.