find the bolume of the solid generated by revolving the region bounded by the graphs of y=sqrt x , y=0, and x=4 about the x axis
To find the volume of the solid generated by revolving the region bounded by the graphs of y = √x, y = 0, and x = 4 about the x-axis, we can use the method of cylindrical shells.
First, let's visualize the region and the solid we are trying to find the volume of:
```
^
y /|\
/ | \
/ | \
| | |
| | _|_ x
```
To use cylindrical shells, we need to consider an infinitesimally thin vertical strip within the region and rotate it about the x-axis to form a cylindrical shell. The height of this strip will be given by the difference in y-values of the two curves: √x (upper curve) and 0 (lower curve).
The formula for the volume of a cylindrical shell is given by:
V = 2π * integral of (radius * height * thickness) with respect to x
In this case, the radius of the cylindrical shell will be x (distance from x-axis), the height will be √x - 0 = √x, and the thickness will be dx (infinitesimal thickness along x-axis).
Now, let's set up and evaluate the integral:
V = 2π * integral from x = 0 to x = 4 of (x * √x * dx)
To solve this integral, we can rewrite it as:
V = 2π * integral from x = 0 to x = 4 of x^(3/2)dx
To integrate x^(3/2), we add 1 to the exponent and divide by the new exponent:
V = 2π * [ (2/5) * x^(5/2) ] evaluated from x = 0 to x = 4
Evaluating this expression, we get:
V = 2π * [(2/5) * 4^(5/2) - (2/5) * 0^(5/2)]
V = 2π * [(2/5) * 32 - (2/5) * 0]
V = 2π * (64/5)
V = (128π/5) or approximately 80.53 units cubed
Therefore, the volume of the solid generated by revolving the region bounded by the graphs of y = √x, y = 0, and x = 4 about the x-axis is approximately 80.53 cubic units.