# 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`

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`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:

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

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.

`r_i=2`

Also, r_o is the distance of the blue curve from the yellow line.

`r_o= x`

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.

`y=9-x^2`

Isolate the x in this equation.

`y-9=-x^2`

`-y+9=x^2`

`9-y=x^2`

`+-sqrt(9-y)=x`

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:

`r_o=x=sqrt(9-y)`

Then, substitute the inner and outer radius to the formula of volume.

`V=int_a^b pi((sqrt(9-y))^2-2^2)dy`

`V=int_a^bpi(9-y-4)dy`

`V=int_a^bpi(5-y)dy`

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=int_0^5 pi(5-y)dy`

`V=pi int_0^5(5-y)dy`

`V=pi(5y-y^2/2)|_0^5`

`V=pi[5*5 - 5^2/2]`

`V=pi(25/2)`

`V=(25pi)/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.**