Volume of hemisphere – Formula derivation – Examples

Volume of hemisphere

A hemisphere is half part of a complete sphere. Hence its volume is exactly half of the volume of a sphere.

Volume of hemisphere = \frac{2}{3}\pi r^3 (\text{in terms of radius }' r')

                                           = \frac{1}{12} \pi d^{3}(\text{in terms of diameter } ' \mathrm{d}^{\prime})

Derivation

We get two identical hemispheres when we divide a sphere along any plane passing through the centre as shown in the figure given below.

We know that the volume of a sphere having radius ‘r’ is equal to \frac{4}{3} \pi r^{3}.

Volume of the hemisphere + Volume of the hemisphere = Volume of the sphere

\Rightarrow 2 \times \text { Volume of the hemisphere }=\frac{4}{3} \pi r^{3}

⇒ \text{ The volume of the hemisphere }=\frac{4}{3} \pi r^{3} \times \frac{1}{2}=\frac{2}{3} \pi r^{3}

⇒ \text{ The volume of the hemisphere }=\frac{2}{3} \pi \left(\frac{d}{2}\right)^{3}=\frac{2}{3\times 8} \pi d^{3}=\frac{1}{12}\pi d^{3}  (\text{ in terms of diameter } 'd')

Solved Examples

1. A sphere is cut into two halves. Find the volume of each hemisphere if the radius of the sphere is equal to 6 cm?

Solution:

Volume of the hemisphere = \frac{2}{3} \pi r^{3}

= \frac{2}{3} \pi \times 6^{3}

=   \frac{2}{3} \pi \times 216

= 144 \pi

= 452.57 \mathrm{~cm}^{2}   [\text{Where } \pi=\frac{22}{7}]

Hence the volume is equal to =452.57 \mathrm{~cm}^{2}.

2. Find the diameter (in cm ) of a hemisphere having volume (18000 \pi) m m^{2}?

Solution:

The volume of the hemisphere

⇒ 18000 \pi=\frac{2}{3} \pi r^{3}

⇒ 18000 \times \frac{3}{2}=r^{3}

⇒ r^{3}=27000

⇒ r=90 \mathrm{~mm}

Diameter = 2r = 2 × 90 mm = 180 mm = 18 cm  (Since, 10 mm =1 cm)

Hence the diameter is 18 cm.

3. The radius of a hemisphere becomes half. Find the percentage of reduction of volume.

Solution:

Old radius = r

Old volume =\frac{2}{3} \pi r^{3}

New radius =r / 2

New volume =\frac{2}{3} \pi\left(\frac{r}{2}\right)^{3}=\frac{2}{3} \pi \frac{r^{3}}{8}=\pi \frac{r^{3}}{12}

Reduction in volume

=\frac{2}{3} \pi r^{3}-\pi \frac{r^{3}}{12}

=\pi r^{3}\left(\frac{2}{3}\frac{1}{12}\right)=\frac{7}{12} \pi r^{3}

Percentage of reduction in volume

=\frac{\text { Reduction in volume }}{\text { Old volume }} \times 100

=\frac{\frac{7}{12} \pi r^{3}}{\frac{2}{3} \pi r^{3}} \times 100

= 87.5%