volume of a sphere
$$V_{\text{sphere}} = \frac{4}{3}\pi r^3$$
for proof see Cavalieri’s principle and proof with a triangular prism
surface area
$$A_{\text{sphere}} = 4\pi r^2$$
proof
The sphere is a curved surface that cannot be developed onto a plane (like the nets of cones and cylinders). Here, we need another way to derive the surface area.
- first divide the surface in “curved” triangles and connect the vertices with the centre of the sphere M. This way the sphere is made up of many small solids
$K_1,K_2,K_3, ...,K_n$that approximate to pyramids with height r.
$$ V_{\text{sphere}}= V_{K_1} + V_{K_2} + ... + V_{K_n}$$Using the formula for the volume of a pyramid, we get$$ \begin{align}V & \approx \frac{1}{3} A_{G_1}\cdot r + \frac{1}{3} A_{G_2}\cdot r + ... + \frac{1}{3}A_{G_n}\cdot r \\ & \approx \frac{1}{3} r \cdot \left[ A_{G_1} + A_{G_2} + ... + A_{G_n}\right]\end{align}$$the expression in the brackets is the surface area of the sphere$A_{\text{sphere}}$.- if we now divide the surface into “infinitely” small triangles their base area approaches a “flat” surface and the volume of the solids becomes more and more like that of a true pyramid. So,
$$ V_{\text{sphere}}= \frac{1}{3} r \cdot A_{\text{sphere}}$$ - we also know, that
$$V_{\text{sphere}} = \frac{4}{3}\pi r^3$$Hence,$$ \frac{4}{3}\pi r^3 = \frac{1}{3}r \cdot A_{\text{sphere}}$$ - So,
$$A_{\text{sphere}} = \frac{3}{r} \cdot\frac{4}{3}\pi r^3 = 4 \pi r^2 $$
exercises
There are plenty of exercises on Aufgabenfuchs on the sphere.