Applications and Combinations of Shapes

Most real-world objects are not pure shapes — they combine cylinders, cones, hemispheres, and cuboids. A tent is a cylinder topped by a cone. A vitamin capsule is a cylinder with hemispheres on each end. A grain silo is a cylinder with a hemispherical or conical roof. This lesson teaches how to handle such composite objects: identify the parts, compute each separately, and add (or subtract) appropriately.

The general approach

1. Identify the basic shapes that make up the object.
2. Compute the relevant quantity (surface area or volume) for each part separately.
3. Add or subtract as appropriate — typically add for total volume, add for surface areas while omitting shared interfaces (faces hidden inside the object).

For surface areas, the most common mistake is including a face that is inside the composite object and should not be counted.

Five typical configurations

Configuration 1: Cone on top of a cylinder (a tent). The bottom of the cone is the same as the top of the cylinder — neither face is exposed. So:

• Total surface area = CSA(cylinder) + CSA(cone) + circular base of cylinder (the floor of the tent).
• If only the canvas above the ground matters: just CSA(cylinder) + CSA(cone).
• Total volume = $\pi r^2 h_\text{cyl} + \tfrac{1}{3} \pi r^2 h_\text{cone}$.

Configuration 2: Hemisphere on top of a cylinder (a dome). The flat side of the hemisphere is on top of the cylinder — both circular faces are interior.

• Total surface area = CSA(cylinder) + curved SA(hemisphere) + circular base of cylinder.
• Total volume = $\pi r^2 h + \tfrac{2}{3} \pi r^3$.

Configuration 3: Cylinder with hemispheres on both ends (a capsule). Both flat hemisphere faces are inside.

• Total surface area = CSA(cylinder) + 2 × curved SA(hemisphere) = $2\pi r h + 4\pi r^2$.
• Total volume = $\pi r^2 h + 2 \cdot \tfrac{2}{3}\pi r^3 = \pi r^2 h + \tfrac{4}{3} \pi r^3$.

Configuration 4: Cone inside a cylinder (a conical hole). A solid cylinder has a cone-shaped depression at one end.

• Volume = cylinder volume - cone volume.
• Surface area depends on what is exposed — be careful.

Configuration 5: Cylinder around a sphere. A cylinder of radius $r$ and height $2r$ exactly encloses a sphere of radius $r$.

• Volume of sphere $= \tfrac{4}{3} \pi r^3$ = $\tfrac{2}{3}$ of cylinder volume.

Detailed worked example

A circus tent is in the shape of a cylinder topped by a cone. The cylinder has radius $20$ m and height $4$ m; the cone has radius $20$ m and height $15$ m. How much canvas is required (excluding the floor)?

Step 1. Compute the slant height of the cone: $\ell = \sqrt{20^2 + 15^2} = \sqrt{625} = 25$ m.

Step 2. CSA of cylinder: $2\pi r h = 2 \cdot \tfrac{22}{7} \cdot 20 \cdot 4 = \tfrac{3520}{7} \approx 502.86$ m$^2$.

Step 3. CSA of cone: $\pi r \ell = \tfrac{22}{7} \cdot 20 \cdot 25 = \tfrac{11000}{7} \approx 1571.43$ m$^2$.

Step 4. Total canvas $= \tfrac{3520 + 11000}{7} = \tfrac{14520}{7} \approx 2074.29$ m$^2$.

Another worked example

A solid is made of a cylinder of radius $7$ cm and height $14$ cm, with hemispherical ends. Find the total surface area and total volume.

Step 1. CSA of cylinder $= 2 \pi r h = 2 \cdot \tfrac{22}{7} \cdot 7 \cdot 14 = 616$ cm$^2$.

Step 2. CSA of two hemispheres $= 2 \cdot 2\pi r^2 = 4 \pi r^2 = 4 \cdot \tfrac{22}{7} \cdot 49 = 616$ cm$^2$.

Step 3. Total surface area $= 616 + 616 = 1232$ cm$^2$.

Step 4. Volume = cylinder + 2 × hemisphere = $\pi r^2 h + \tfrac{4}{3} \pi r^3 = \tfrac{22}{7} \cdot 49 \cdot 14 + \tfrac{4}{3} \cdot \tfrac{22}{7} \cdot 343 = 2156 + \tfrac{30184}{21} \approx 2156 + 1437.33 = 3593.33$ cm$^3$.

Worked examples

Example 1. A wooden toy is a cone mounted on a hemisphere of common radius $7$ cm. The cone's height is $24$ cm. Find the total surface area.

CSA of cone: $\pi r \ell$ where $\ell = \sqrt{49 + 576} = 25$. So CSA $= \tfrac{22}{7} \cdot 7 \cdot 25 = 550$. CSA of hemisphere: $2\pi r^2 = 308$. Total: $550 + 308 = 858$ cm$^2$.

Example 2. A cylindrical tank has a hemispherical top, both of radius $3$ m. The cylindrical part is $6$ m tall. Find the total volume.

Cylinder: $\pi r^2 h = \pi \cdot 9 \cdot 6 = 54\pi$. Hemisphere: $\tfrac{2}{3} \pi \cdot 27 = 18\pi$. Total: $72\pi$ m$^3$ $\approx 226.29$ m$^3$.

Example 3. A cuboidal block has a cylindrical hole drilled through it. Cuboid: $10 \times 8 \times 6$. Hole radius: $2$, depth: $6$ (drilled all the way through).

Volume of block $= 480$. Volume of hole $= \pi \cdot 4 \cdot 6 = 24\pi \approx 75.43$. Remaining $\approx 404.57$.

Example 4. An ice-cream cone is shaped like a cone (radius $3$ cm, height $9$ cm) with a hemispherical scoop (radius $3$ cm) on top. Find the total volume.

Cone: $\tfrac{1}{3} \pi \cdot 9 \cdot 9 = 27\pi$. Hemisphere: $\tfrac{2}{3}\pi \cdot 27 = 18\pi$. Total: $45\pi \approx 141.43$ cm$^3$.

Example 5. A grain silo has a cylindrical body of radius $5$ m and height $10$ m, with a conical top of slant height $6$ m on the same base. Find the total surface area excluding the bottom.

Cone height: $h = \sqrt{36 - 25} = \sqrt{11} \approx 3.32$. CSA of cone: $\pi r \ell = \pi \cdot 5 \cdot 6 = 30\pi$. CSA of cylinder: $2\pi r h = 2\pi \cdot 5 \cdot 10 = 100\pi$. Total: $130\pi \approx 408.41$ m$^2$.

Try it yourself

1. A circus tent has a cylinder of radius $10$ m and height $3$ m, with a cone of slant height $12$ m on top. Find the canvas required.
2. A capsule has cylindrical body of radius $0.5$ cm and length $1.4$ cm, with hemispherical ends. Find the volume.
3. A wooden article is a cylinder (radius $7$ cm, height $8$ cm) with a cone (same radius, height $4$ cm) on top. Find the total surface area.
4. A solid is half a sphere (radius $r$) on top of a cylinder (same radius, height $h$). Express its volume.
5. A cylindrical glass has a hemispherical bottom (open at top). It is filled with water. Compute the volume for radius $4$ and cylinder height $10$.
6. A cube of side $a$ has a cone of radius $\tfrac{a}{2}$ and height $a$ drilled from one face. Find the remaining volume.
7. A toy is a cone on a hemisphere, both of radius $3$ cm. The cone is $4$ cm tall. Find the volume.
8. A grain silo: cylinder ($r = 4, h = 8$) with cone on top ($r = 4, h = 6$). Find total volume.
9. A wooden block (cuboid, $20 \times 15 \times 10$ cm) has a cylindrical hole of radius $3$ drilled through the smallest face. Find the remaining volume.
10. An object is a sphere of radius $r$ inscribed in a cylinder of radius $r$ and height $2r$. Find the volume of the cylinder NOT occupied by the sphere.

Pitfalls / Insight

• Don't double-count interior surfaces. When two shapes meet, the shared face is not part of the exterior.
• Match dimensions. Ensure all radii and heights are in compatible units.
• Use the right value of $\pi$ as the problem requests.

Insight. Real-world objects almost always combine basic shapes. Once you can identify the parts and apply the formulas separately, no composite object will defeat you. The technique is exactly the same as Chapter 10's triangulation — break the unknown into knowns and add (or subtract).