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"The base of a certain solid is circular with a radius of 1. Cross-sections of the solid, perpendicular to the base, are square. Find the volume of the solid."

In visualizing this object, i see that it looks like a "blocky" foot ball, where the perimeter is circular, but the corners are sharp because of the volume of the reducing square outward from the center.

I center the circle at zero on the xy plane. Then I note that base of a given slab can be given by [tex]2 \sqrt{1-x^2} [/tex]. Since the area of a sqaure is b^2, [tex]A(x) = 4(1-x^2)[/tex]. The slab has a width [tex]\Delta x[/tex], making the total volume function [tex]4 (1-x^2) \Delta x [/tex]. I then integrate from 0 to 1 and double the resultin volume, as 0 to 1 represents the volume of half this shape. My result is [tex] 2 \int_{0}^{1} 4 (1-x^2) \Delta x = \frac{32}{3}[/tex] cubic units.

Is this logic correct? If not, why?