Volume of the sphere

The sphere is “sliced” into circular discs of thickness $dz$. At height $z$, the disc has radius $r(z) = \sqrt{R^2-z^2}$ and volume $dV = \pi r^2\,dz = \pi(R^2-z^2)\,dz.$

The sphere sliced into discs: at height $z$ the disc has radius $r(z)=\sqrt{R^2-z^2}$.

Summing the discs from $z=-R$ to $z=R$: $V = \int_{-R}^{R}\pi(R^2-z^2)\,dz = \pi\left[R^2 z - \frac{z^3}{3}\right]_{-R}^{R} = \pi\left[2R^3-\frac{2R^3}{3}\right] = \boxed{\dfrac{4}{3}\pi R^3}.$

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Topics: Integrale
Concepts: Integrale definito · Volume per sezioni
Skills: Calcolare · Integrare