Surface Areas and Volumes | Exercise 12.1

Question 6

A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends (see Fig. 12.10). The length of the entire capsule is $14\text{ mm}$ and the diameter of the capsule is $5\text{ mm}$ . Find its surface area.

Solution

Understand the Question

A medicine capsule consists of a central cylindrical section with a hemisphere attached at each of its two ends.
The total surface area of the capsule is the sum of the curved surface area (CSA) of the cylinder and the curved surface areas of the two hemispheres: $\text{Total Surface Area} = \text{CSA of cylinder} + 2 \times \text{CSA of hemisphere}$
Note that the circular bases of the cylinder and hemispheres are joined internally and are not part of the outer surface.

Step 1 · Find Dimensions of the Cylinder and Hemispheres

Diagram 1

Given the diameter of the capsule is $5\text{ mm}$ , the radius $r$ of both the cylinder and the hemispheres is: $r = \dfrac{5}{2} = 2.5\text{ mm}$

The height (length) of the cylindrical part $h$ is found by subtracting the radius of both hemispherical ends from the total length:

\begin{aligned} h &= 14 - (2.5 + 2.5) \\[0.6em] &= 14 - 5 \\[0.6em] &= 9\text{ mm} \end{aligned}

Step 2 · Calculate the Surface Area of the Capsule

The surface area of the capsule is the sum of the curved surface areas: $\text{Surface area of capsule} = \text{CSA of cylinder} + 2 \times (\text{CSA of hemisphere})$

\begin{aligned} \text{Surface area of capsule} &= 2\pi rh + 2(2\pi r^2) \\[0.6em] &= 2\pi rh + 4\pi r^2 \\[0.6em] &= 2\pi r (h + 2r) \end{aligned}

Substitute $r = 2.5\text{ mm}$ and $h = 9\text{ mm}$ :

\begin{aligned} &= 2 \times \pi \times 2.5 \times (9 + 2 \times 2.5) \\[0.6em] &= 5\pi \times (9 + 5) \\[0.6em] &= 5\pi \times 14 \\[0.6em] &= 70\pi \end{aligned}

Using $\pi = \dfrac{22}{7}$ :

\begin{aligned} &= 70 \times \dfrac{22}{7} \\[0.6em] &= 10 \times 22 \\[0.6em] &= 220\text{ mm}^2 \end{aligned}

Answer

$220\text{ mm}^2$

Common Mistakes

Adding Total Surface Areas: Adding the total surface area (TSA) of the cylinder and hemispheres instead of their curved surface areas (CSA). The circular faces are joined together inside and are not exposed.
Height of the Cylinder: Forgetting to subtract the radius of both hemispheres ( $2 \times 2.5\text{ mm} = 5\text{ mm}$ ) from the total length, erroneously taking $h = 14 - 2.5 = 11.5\text{ mm}$ instead of $9\text{ mm}$ .

Question 6

More questions in Exercise 12.1