# Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the x-axis. `y = e^(x/2) , y = 0 , x = 0 , x = 4`

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### 1 Answer

Sketch the region bounded by the graphs of the given equations.

`y=e^(x/2) ` (Blue curve)

`y=0` (Yellow line)

`x=0` (Red line)

`x=4` (Green line)

To solve for the volume of the solid formed by revolving the bounded region around the x-axis, apply the disk method. The formula when the axis of rotation is horizontal is:

`V=int_a^b pi r^2 dx`

In our graph above, the radius is the distance between the axis of rotation and the blue curve.

`r=y_U-y_L= e^(x/2)-0`

`r=e^(x/2)`

And the limits of the integral are the end x values of the bounded region.

`a=0`

`b=4`

Plugging them to the formula of volume, we would have:

`V=int_0^4 pi (e^(x/2))^2 dx`

`V=pi int_0^4 e^(x/2*2) dx`

`V=pi int_0^4 e^x dx`

Evaluating the integral yields:

`V= pi (e^x)|_0^4`

`V=pi (e^4-e^0)`

`V=pi(e^4-1)`

`V=168.4`

**Therefore, the volume of the solid formed by revolving the bounded region around the x-axis is 168.4 cubic units.**