Volume of a solid with square cross sections over a circular base

Key concept: This kind of volume problem is all about reading the geometry carefully. The base is a disk, and each slice is a square whose side length comes from the circle equation.

Step by step

Let the base be the region

$x^2+y^2=a^2.$

Cross sections perpendicular to the $x$-axis are squares.

1) Find the side length of a square slice

For a fixed $x$, the circle gives

$y=\pm\sqrt{a^2-x^2}.$

So the vertical width of the base is

$s(x)=2\sqrt{a^2-x^2}.$

That is the side length of the square.

2) Area of each cross section

$A(x)=s(x)^2=\left(2\sqrt{a^2-x^2}\right)^2=4(a^2-x^2).$

3) Integrate over the full interval

The disk runs from $-a$ to $a$, so

$V=\int_{-a}^{a}4(a^2-x^2)\,dx.$

Compute:

$V=4\left[ a^2x-\frac{x^3}{3}\right]_{-a}^{a}.$

At $x=a$:

$a^3-\frac{a^3}{3}=\frac{2a^3}{3}.$

At $x=-a$:

$-a^3+\frac{a^3}{3}=-\frac{2a^3}{3}.$

So

$V=4\left(\frac{2a^3}{3}-\left(-\frac{2a^3}{3}\right)\right)=4\cdot\frac{4a^3}{3}=\frac{16a^3}{3}.$

Answer

$\boxed{\frac{16a^3}{3}}$

So the correct choice is B.

Pitfall alert

The usual slip is forgetting that the square side length is the full width of the circle, not just the top half. If you use only $\sqrt{a^2-x^2}$, the volume comes out too small by a factor of 4.

Try different conditions

If the cross sections were perpendicular to the $y$-axis instead, the setup would look the same by symmetry and the volume would still be $\frac{16a^3}{3}$. If the slices were semicircles instead of squares, the area formula would change completely, so the integral would not match this answer.

Further reading

cross section, disk region, volume by integration