The solid is a cylinder surmounted by a hemispherical bowl. Calculate its (a) total surface area ; (b) volume (Take \(\pi = \frac{22}{7}\))

Answer Details

From the diagram the solid is a closed cylinder capped on top by a hemisphere. The hemisphere has height \(7\text{ cm}\); since a hemisphere's height equals its radius, the radius is \(r = 7\text{ cm}\), which is also the radius of the cylinder. The cylinder height is \(h = 10\text{ cm}\). Take \(\pi = \dfrac{22}{7}\).

(a) Total surface area.

The exposed surfaces are: the curved surface of the cylinder, the flat circular base of the cylinder, and the curved surface of the hemisphere. (The flat top of the cylinder is exactly covered by the flat face of the hemisphere, so it is not counted.)

Curved surface of cylinder:

\[2\pi r h = 2 \times \frac{22}{7} \times 7 \times 10 = 440 \text{ cm}^2\]

Base circle of cylinder:

\[\pi r^2 = \frac{22}{7} \times 7^2 = \frac{22}{7} \times 49 = 154 \text{ cm}^2\]

Curved surface of hemisphere:

\[2\pi r^2 = 2 \times 154 = 308 \text{ cm}^2\]

Total surface area:

\[440 + 154 + 308 = 902 \text{ cm}^2\]

Total surface area \(= 902\text{ cm}^2\).

(b) Volume.

Volume of cylinder:

\[\pi r^2 h = \frac{22}{7} \times 49 \times 10 = 22 \times 7 \times 10 = 1540 \text{ cm}^3\]

Volume of hemisphere:

\[\frac{2}{3}\pi r^3 = \frac{2}{3} \times \frac{22}{7} \times 7^3 = \frac{2}{3} \times \frac{22}{7} \times 343 = \frac{2}{3} \times 22 \times 49 = \frac{2156}{3} \approx 718.67 \text{ cm}^3\]

Total volume:

\[1540 + 718.67 = 2258.67 \text{ cm}^3\]

Total volume \(\approx 2258.67\text{ cm}^3\) (i.e. \(2258\tfrac{2}{3}\text{ cm}^3\)).