Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the y-axis. `y = 9 - x^2` `y = 0, x = 2, x = 3`
First, graph the given equations to determine the bounded region formed by their graphs.
(Note: The blue curve is the graph of y=9-x^2. The red line is the graph of y=0. The green line is the graph of x=2. The purple line is the graph of x=3. And the yellow dashed line is the axis of revolution.)
Notice that there is a gap between the bounded region and the axis of revolution. So to get the volume of solid formed when the bounded region is revolved about the y-axis, apply method of rings.Its formula when the axis of revolution is vertical is:
where r_o is the outer radius and the r_i is the inner radius.
Base on the graph above, r_i is the distance of the green line from the yellows line.
Also, r_o is the distance of the blue curve from the yellow line.
Since the integral should be expressed in terms of y variable, then r_o should be expressed in y variable. To do so, refer to the equation of the blue curve.
Isolate the x in this equation.
Since the portion of the blue curve that we consider is at the positive x-axis, consider only `x=sqrt(9-y)` . Hence, the outer radius when expressed in terms of y is:
Then, substitute the inner and outer radius to the formula of volume.
The limits of the integral are the y-coordinates of the intersections or corners of the bounded region. So a = 0 and b=5.
`V=pi[5*5 - 5^2/2]`
Hence, the volume of the solid generated by revolving the region bounded by the graphs of the equations about the y-axis is `(25pi)/2` cubic units.