Find the volume of the solid generated by rotating the region 0<y<5-x^2 about the x-axis

the volume would be

∫[-√5,√5] πr^2 dx
where r=y=(5-x^2)
Because of symmetry, that is
2∫[0,√5] π(5-x^2)^2 dx = 80π√5/3

using shells, that is
2∫[0,5] 2πrh dy
where r=y and h=x=√(5-y)
2∫[0,5] 2πy√(5-y) dy = 80π√5/3

To find the volume of the solid generated by rotating a region about the x-axis, we can use the method of cylindrical shells.

The region given in the problem is the area between the curve y = 5 - x^2 and the x-axis, bounded by the values of x.

First, let's find the interval on the x-axis where the region exists. To do this, set the equation 5 - x^2 = 0 and solve for x:
x^2 = 5
x = ±√5

So, the region exists for -√5 ≤ x ≤ √5.

Next, for each x value in this interval, we'll consider a vertical strip or shell with thickness Δx. The radius of each shell is equal to the y-value of the curve at that x-value (5 - x^2), and the height of each shell is equal to the length of the x-interval (Δx).

The volume of each shell can be approximated as the product of the circumference of the shell (2π(radius)) and the height (Δx).

Therefore, the volume of each shell is given by:
V_shell = 2π(5 - x^2) * Δx

To find the total volume, we need to sum up the volumes of all the shells using integration. We'll integrate the expression for the volume of each shell over the interval -√5 ≤ x ≤ √5:

V_total = ∫[√5, -√5] (2π(5 - x^2)) dx

Evaluating this integral will give us the final result for the volume.

Please note that if you want an exact solution, you can evaluate the integral mathematically. However, if you want an approximate value, you can use numerical methods or software.