1. The problem asks why the formula for the volume of a sphere is $\frac{4}{3}\pi r^3$ even though a sphere can be thought of as made by adding infinitely many circles. 2. First, let's clarify the formulas: - The **surface area** of a sphere is $4\pi r^2$. - The **volume** of a sphere is $\frac{4}{3}\pi r^3$. 3. The volume formula comes from calculus, specifically integration, where we sum up infinitely many infinitesimally thin disks (circles with thickness) stacked along the radius. 4. The volume of each thin disk is approximately the area of the circle times its thickness: $\pi y^2 \Delta x$, where $y$ is the radius of the disk at position $x$. 5. Using the equation of a circle $y=\sqrt{r^2 - x^2}$, the volume is the integral: $$V = \int_{-r}^r \pi (r^2 - x^2) dx$$ 6. Evaluating this integral: $$V = \pi \left[ r^2 x - \frac{x^3}{3} \right]_{-r}^r = \pi \left( r^3 - \frac{r^3}{3} - (-r^3 + \frac{r^3}{3}) \right) = \pi \left( \frac{4}{3} r^3 \right) = \frac{4}{3} \pi r^3$$ 7. So, the volume formula $\frac{4}{3}\pi r^3$ is the exact sum of infinitely many infinitesimally thin circular disks making up the sphere. 8. The key is that these are not just circles, but disks with thickness, and integration sums their volumes, not just areas. Final answer: The volume of a sphere is $\frac{4}{3}\pi r^3$ because it is the integral sum of infinitely many thin disks (circles with thickness) stacked to form the sphere.