# Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified line. y = (1/4)x^2, y = 5-x^2; about the x axis.

`y=1/4x^2`

`y=5-x^2`

First, graph the two functions to determine the bounded region.

*(Green curve graph of y=(1/4)x^2 , Blue curve graph of y=5-x^2, and Red line is the axis of revolution.)*

Notice that there is a gap between the bounded region and the axis of revolution. So, to solve apply the method of rings. And its formula of volume if the axis of revolution is horzontal is:

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

Base on the graph above, the inner radius is the distance from the red line to the green curve. So, the inner radius is:

`r_i=y_g-y_r=1/4x^2-0`

`r_i=1/4x^2`

Also, the outer radius is the distance from the red line to the blue curve. So,

`r_o=y_b-y_r=5-x^2-0`

`r_o=5-x^2`

Moreover, the limits of the integral are the x-coordinates of the intersection of the two curves. So, a=-2 and b=2.

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

`V=int_(-2)^2 pi ((5-x^2)^2-(1/4x^2)^2) dx`

Simplifying the integrand, it becomes:

`V=pi int_(-2)^2 (25-10x^2 +x^4 - 1/16x^4) dx`

`V=pi int_(-2)^2 (25-10x^2 +15/16x^4)dx`

Evaluating the integral, it yields:

`V=pi (25x-10/3x^3+3/16x^5)|_(-2)^2`

`V=pi[(25(2)-10/3(2)^3+3/16(2)^5)-(25(-2)-10/3(-2)^3+3/16(-2)^5)]`

`V=pi[(50-80/3+6)-(-50+80/3-6)]`

`V=pi[88/3-(-88/3)]`

`V=(176pi)/3`

**Therefore, the volume of the solid formed is `(176pi)/3` cubic units.**