I'm not sure how you arrived at 2 - x for the radius.
If you integrate along the y-axis, then your radius needs to be in terms of y.
The vertical cross-section yields a right-triangle with height 5 and base 2. The line passing through the hypotenuse has the following equation.
y = -(5/2)*x + 5
Taking the inverse gives us the distance x in terms of the height along the y-axis (i.e., the radius of the cylindrical disks).
r = -(2/5)*y + 2
The circular area of each disk is Pi*(-2/5*y + 2)^2
The value of V below is 20*Pi/3.
The line passing through the hypotenuse can be shifted downward so that it passes through the origin. The relationship between x and y will not change. In other words, we could have started with the following equation and then taken its inverse to find an expression for r in terms of y.
y = -(5/2)x
This is why the integrand in your text contains only the y-squared term.