SURFACE AREAS AND VOLUMES-Notes
Maths - Notes
Surface Area of a Right Circular Cone
Right circular cone
A right circular cone is a three-dimensional solid with:
- a circular base,
- a curved surface, and
- a vertex (tip) directly above the center of the base.
To understand its surface area, imagine a cone as a circular sheet rolled into a pointed shape. This gives the cone two measurable parts:
- The curved surface
- The circular base
Important Terms
- Radius \((r)\):
The radius of the circular base. - Height \((h)\):
The perpendicular distance from the center of the base to the vertex. - Slant Height \((l)\):
The distance from the vertex to any point on the edge of the base.
In a right circular cone, the slant height is found using the Pythagorean Theorem: \[l=\sqrt{h^2+r^2}\]
Curved Surface Area (CSA)
When the cone is cut along the slant height and opened out flat, it forms a sector of a circle.
The formula for the curved surface area comes from this idea.
The Curved Surface Area (CSA) of a right circular cone is:
\[
\boxed{
\text{CSA} = \pi rl}
\]
Total Surface Area (TSA)
A cone has two surfaces:
- the curved surface, and
- the circular base.
Key Points to Remember
- A cone’s surface area has two parts: curved and circular base.
- Slant height is essential for calculating surface area.
- Use CSA when only curved area is needed.
- Use TSA when the entire outer area is required.
Example-1
Find the curved surface area of a right circular cone whose slant height is 10 cm and base radius is 7 cm.
Solution:
Let the radius of the base of the cone be \(r = 7~\text{cm}\)
and the slant height be \(l = 10~\text{cm}\)
The formula for the curved surface area (CSA) of a right circular cone is \(\pi r l\)
Substituting the given values:
Example-2
The height of a cone is 16 cm and its base radius is 12 cm. Find the curved surface area and the total surface area of the cone (Use π = 3.14).
Solution:
Given:
Height of the cone, $$h = 16~\text{cm}$$ Base radius, $$r = 12~\text{cm}$$
First, calculate the slant height \(l\): $$ \begin{aligned} l &= \sqrt{h^2 + r^2} \\ &= \sqrt{16^2 + 12^2} \\ &= \sqrt{256 + 144} \\ &= \sqrt{400} \\ &= 20~\text{cm} \end{aligned} $$
Curved surface area \(A_{\text{CSA}}\) of the cone: $$ \begin{aligned} A_{\text{CSA}} &= \pi r l \\ &= 3.14 \times 12 \times 20 \\ &= 753.6~\text{cm}^2 \end{aligned} $$
Total surface area \(A_{\text{TSA}}\) of the cone: $$ \begin{aligned} A_{\text{TSA}} &= \pi r (l + r) \\ &= 3.14 \times 12 \times (20 + 12) \\ &= 3.14 \times 12 \times 32 \\ &= 1,206.72~\text{cm}^2 \end{aligned} $$
Answers:
Curved surface area = $$753.6~\text{cm}^2$$
Total surface area = $$1,206.72~\text{cm}^2$$
Example-3
A corn cob (see Fig. 11.5), shaped somewhat like a cone, has the radius of its broadest end as 2.1 cm and length (height) as 20 cm. If each 1 \(\mathrm{cm^2}\) of the surface of the cob carries an average of four grains, find how many grains you would find on the entire cob.
Given: Radius of the broadest end of the corn cob \(r\) is
\(2.1~\text{cm}\);
height
\(h\) is \(20~\text{cm}\).
First, calculate the slant height \(l\): $$ \begin{aligned} l &= \sqrt{h^2 + r^2} \\ &= \sqrt{20^2 + (2.1)^2} \\ &= \sqrt{400 + 4.41} \\ &= \sqrt{404.41} \\ &\approx 20.1~\text{cm} \end{aligned} $$
Curved surface area \(A_{\text{CSA}}\) of the cone-shaped cob: $$ \begin{aligned} A_{\text{CSA}} &= \pi r l \\ &\approx 3.14 \times 2.1 \times 20.1 \\ &\approx 3.14 \times 42.21 \\ &\approx 132.54~\text{cm}^2 \end{aligned} $$
Each \(1~\text{cm}^2\) of the surface carries 4 grains. $$\begin{aligned} \text{Total number of grains} &= A_{\text{CSA}} \times 4 \\&= 132.54 \times 4 \\&\approx 531 \end{aligned}$$
Answer: The corn cob contains approximately 531 grains.
Surface Area of a Sphere
A sphere is one of the most perfectly shaped objects in geometry. It is completely round, smooth and has no edges, no vertices, and no flat faces. Everyday objects like a football, cricket ball, orange, globe, soap bubble, and even tiny water droplets are great examples of spherical shapes.
What is a Sphere?
- This fixed point is called the centre of the sphere.
- The constant distance from this centre to the surface is called the radius (r).
A sphere is a three-dimensional solid formed by all the points in space that are at the same distance from a fixed point.
Because the surface of a sphere is smooth and curved everywhere, we talk about its Curved Surface Area (CSA) or simply Total Surface Area (TSA).
Formula: Surface Area of a Sphere
\[\boxed{\text{Surface Area (TSA)}=4\pi r^2}\]Key Features of a Sphere
- Perfectly symmetrical in all directions
- Only one formula for surface area (no separate CSA or TSA)
- Surface area increases rapidly as radius increases (because of the square of radius)
Hemisphere
A hemisphere is exactly half of a sphere. If you cut a sphere into two equal parts along its centre, each
part
is called a hemisphere. Objects like a bowl, half coconut shell, half orange, dome-shaped buildings, and
some
lampshades look like hemispheres.
Just like the sphere, the hemisphere is a three-dimensional solid with curvature on one side and a flat
circular
face on the other.
Structure of a Hemisphere
- Curved Surface (outer curved part)
- Flat Circular Surface (base)
A hemisphere has two distinct surfaces:
Curved Surface Area (CSA) of a Hemisphere
Since it is half of a sphere, \[\scriptsize\begin{aligned}\text{CSA of hemisphere}&=\frac{1}{2}\text{Surface Area of sphere}\\\\ \text{CSA}&=\frac{1}{2}\times 4\pi r^2\\\\ \text{CSA}&=2\pi r^2 \end{aligned}\]
Total Surface Area (TSA) of a Hemisphere
TSA includes:
- Curved Surface Area
- Area of the circular base
Base area =\(\pi r^2\)
\[\begin{aligned}\text{TSA} &=2\pi r^2 + \pi r^2\\
\text{TSA}&=3\pi r^2\end{aligned}\]
Final Formulas for Hemisphere
Curved Surface Area (CSA) \[\boxed{\text{CSA}=2\pi r^2}\] Total Surface Area (TSA) \[\boxed{\text{TSA}=3\pi r^2}\]
Important Points to Remember
- For spheres, TSA = CSA (only one curved surface).
- Surface area depends only on radius, not on diameter directly.
- Larger the radius \(\Rightarrow\) more curved area it covers.
- To find radius from surface area, use \[r=\sqrt{\frac{\text{Surface Area}}{4\pi}}\]
Example-4
Find the surface area of a sphere of radius 7 cm.
Solution:
Given:
Radius of the sphere, \(r = 7~\text{cm}\).
The formula for the surface area of a sphere is $$A = 4\pi r^2$$
Substituting the given value: $$ \begin{aligned} A &= 4 \pi r^2 \\ &= 4 \times \frac{22}{7} \times 7 \times 7 \\ &= 4 \times 154 \\ &= 616~\text{cm}^2 \end{aligned} $$
Answer: The surface area of the sphere is $$616~\text{cm}^2$$
Example-5
Find
(i) the curved surface area and
(ii) the total surface area of a
hemisphere of radius 21 cm.
Solution:
Given:
Radius of the hemisphere, $$r = 21~\text{cm}$$
(i) The curved surface area (CSA) of a hemisphere is given by: $$ \begin{aligned} \text{CSA} &= 2\pi r^2 \\ &= 2 \times \frac{22}{7} \times 21 \times 21 \\ &= 2 \times 22 \times 21 \times 3 \\ &= 2772~\text{cm}^2 \end{aligned} $$
(ii) The total surface area (TSA) of a hemisphere is given by: $$ \begin{aligned} \text{TSA} &= 3\pi r^2 \\ &= 3 \times \frac{22}{7} \times 21 \times 21 \\ &= 4158~\text{cm}^2 \end{aligned} $$
Answers:
Curved surface area = $$2772~\text{cm}^2$$
Total surface area = $$4158~\text{cm}^2$$
Example-6
The hollow sphere, in which the circus motorcyclist performs his stunts, has a diameter of 7 m. Find the area available to the motorcyclist for riding.
Solution:
The diameter of the hollow sphere is given as \(7~\text{m}\) This means the radius \(r\) can be calculated as: $$ r = \frac{7}{2} = 3.5~\text{m} $$
To find the area available for the motorcyclist to ride on, we use the formula for the surface area of a sphere: $$ \text{Surface area} = 4\pi r^2 $$
Substitute the value of \(r\) into the formula: $$ \begin{aligned} \text{Surface area} &= 4 \times \frac{22}{7} \times 3.5 \times 3.5 \\ &= 4 \times \frac{22}{7} \times 12.25 \\ &= 4 \times 38.5 \\ &= 154~\text{m}^2 \end{aligned} $$
Therefore, the motorcyclist has $$154~\text{m}^2$$ of area available for riding inside the hollow sphere.
Example-7
A hemispherical dome of a building needs to be painted (see Fig. 11.9). If the circumference of the base of the dome is 17.6 m, find the cost of painting it, given the cost of painting is ₹5 per 100 cm2.
Solution:
The circumference of the base of the hemispherical dome is given as $$17.6~\text{m}$$ The formula for the circumference of a circle is \(2\pi r\), so we set up the equation: $$ 2\pi r = 17.6 $$ To find the radius \(r\), rearrange and substitute $$\pi \approx \frac{22}{7}$$ $$\begin{aligned} r &= \frac{17.6}{2\pi} \\\\ &= \frac{17.6}{2 \times \frac{22}{7}} \\\\ &= \frac{17.6 \times 7}{44} \\\\ &= \frac{123.2}{44} \\\\&= 2.8~\text{m} \end{aligned}$$
Now, calculate the curved surface area (CSA) of the hemisphere: $$ \begin{aligned} \text{CSA} &= 2\pi r^2 \\\\ &= 2 \times \frac{22}{7} \times 2.8 \times 2.8 \\\\ &= 2 \times \frac{22}{7} \times 7.84 \\\\ &= 2 \times 24.64 \\\\ &= 49.28~\text{m}^2 \end{aligned} $$
Since the cost of painting is ₹5 per \(100~\text{cm}^2\), and
\(1~\text{m}^2 =
10,000~\text{cm}^2\):
Therefore, Cost per \(1 m^2\)
$$\begin{aligned}
&= \frac{10,000}{100} \times 5 \\\\&= 100 \times 5 \\\\&= ₹500
\end{aligned}$$
Total cost for painting the dome: $$ \begin{aligned} \text{Total cost} &= 49.28 \times 500 \\ &= ₹24,640 \end{aligned} $$
Therefore, the cost of painting the hemispherical dome is ₹24,640.
Volume of a Right Circular Cone
A right circular cone is a three-dimensional solid obtained when a right-angled triangle is revolved around one of its perpendicular sides. It has:
- A circular base
- A curved surface
- A pointed top called the vertex
Formula: Volume of a Right Circular Cone
\[\boxed{V=\frac{1}{3}\pi r^2h}\] Where:- \(r\) is radius
- \(h\) is height
Important Points
- Volume depends on both radius and height.
- If radius doubles, volume becomes four times.
- If height doubles, volume becomes double.
- A cone can hold only one-third the amount of a cylinder of same base and height.
Example-8
The height and the slant height of a cone are 21 cm and 28 cm respectively. Find the volume of the cone.
Solution:
The height of the cone is given as $$h = 21~\text{cm}$$ and the slant height as $$l = 28~\text{cm}$$
To find the base radius \(r\), we use the Pythagorean relation: $$ l^2 = r^2 + h^2 $$ Substituting the values, $$\begin{aligned} r^2 &= l^2 - h^2 \\&= 28^2 - 21^2 \\&= (28+21)(28-21) \\&= 49 \times 7 \\&= 343 \end{aligned}$$ Thus, $$ r = \sqrt{343} = 7\sqrt{7}~\text{cm} $$
Now, calculate the volume of the cone using the formula $$V = \frac{1}{3}\pi r^2 h$$: $$ \begin{aligned} V &= \frac{1}{3} \pi r^2 h \\\\ &= \frac{1}{3} \times \frac{22}{7} \times (7\sqrt{7})^2 \times 21 \\\\ &= \frac{1}{3} \times \frac{22}{7} \times 49 \times 7 \times 21 \\\\ &= \frac{1}{3} \times \frac{22}{7} \times 343 \times 21 \\\\ \end{aligned} $$ Calculating further, $$ 343 \times 21 = 7203 $$ So, $$ V = \frac{1}{3} \times \frac{22}{7} \times 7203 $$ First, calculate $$\frac{22}{7} \times 7203$$ $$\begin{aligned} \frac{22}{7} \times 7203 &\approx 22 \times 1028.4286 \\&\approx 22,625.43 \end{aligned}$$ Now, divide by 3: $$ \frac{22,625.43}{3} \approx 7,541.81~\text{cm}^3 $$
Therefore, the volume of the cone is approximately $$7,542~\text{cm}^3$$ (rounded to the nearest integer).
Example-9
Monica has a piece of canvas whose area is 551 m2. She uses it to have a conical tent made, with a base radius of 7 m. Assuming that all the stitching margins and the wastage incurred while cutting, amounts to approximately 1 m2, find the volume of the tent that can be made with it.
Solution:
Monica starts with a piece of canvas having an area of $$551~\text{m}^2$$ Since stitching margins and cutting wastage amount to $$1~\text{m}^2$$ the effective area used for making the tent is: $$ \text{Net canvas area} = 551 - 1 = 550~\text{m}^2 $$
The tent will form a cone with base radius $$r = 7~\text{m}$$ The curved surface area (CSA) of the conical tent equals the canvas used for covering: $$ \pi r l = 550 $$ To find the slant height \(l\), rearrange the equation: $$ l = \frac{550}{\pi r} $$ Substitute $$\pi \approx \frac{22}{7}$$ and $$r = 7$$ $$\begin{aligned} l = \frac{550}{\frac{22}{7} \times 7} \\\\&= \frac{550}{22} \\\\&= 25~\text{m} \end{aligned}$$
Now, use the Pythagorean theorem to find the vertical height \(h\): $$\begin{aligned} l^2 &= r^2 + h^2 \\ h^2 &= l^2 - r^2 \\ h^2 &= 25^2 - 7^2 \\&= 625 - 49 \\&= 576\\ h &= \sqrt{576} \\&= 24~\text{m} \end{aligned}$$
The volume \(V\) of the tent (cone) is: $$ V = \frac{1}{3}\pi r^2 h $$ Substitute the known values: $$ \begin{aligned} V &= \frac{1}{3} \times \frac{22}{7} \times 7 \times 7 \times 24 \\ &= \frac{1}{3} \times \frac{22}{7} \times 49 \times 24 \\ \end{aligned} $$ First, calculate $$49 \times 24 = 1176$$ then $$1176 / 3 = 392$$ $$\begin{aligned} V &= \frac{22}{7} \times 392 \\\\&= 22 \times 56 \\&= 1232~\text{m}^3 \end{aligned}$$
Therefore, the maximum volume of the conical tent that Monica can have made is $$1232~\text{m}^3$$
Volume of a Sphere
A sphere is a perfectly round 3-D solid. All points on its surface are at the same distance from a fixed
point called the centre.
Objects like a football, cricket ball, orange, marble, globe, and soap bubble are shaped like
spheres.
To measure how much space is inside the sphere, we calculate its volume.
Formula: Volume of a Sphere
\[\boxed{V=\frac{4}{3}\pi r^3}\]Volume of a Hemisphere
Since a hemisphere is half of a sphere, its volume also half of sphere
\[\boxed{\scriptsize\text{Volume of a Hemisphere}=\frac{2}{3}\pi r^3}\]- Volume depends on \(r^3\), so even a small increase in radius greatly increases volume.
- A sphere holds more space than any other 3-D shape with the same surface area (it is the most space-efficient shape).
- Used in designing balls, bubbles, spherical tanks, and planets (planets and stars are nearly spherical).
Example-10
Find the volume of a sphere of radius 11.2 cm.
Solution:
The radius of the sphere is given as $$r = 11.2~\text{cm}$$
The formula for the volume of a sphere is: $$ V = \frac{4}{3} \pi r^3 $$
Substituting the given values: $$ \begin{aligned} V &= \frac{4}{3} \times \frac{22}{7} \times 11.2 \times 11.2 \times 11.2 \\\\ &= \frac{4}{3} \times \frac{22}{7} \times 1404.92 \\\\ \end{aligned} $$ First, calculate $$\frac{22}{7} \times 1404.92 \approx 4415.46$$ then multiply by $$\frac{4}{3}$$ $$\begin{aligned} V &= \frac{4}{3} \times 4415.46 \\&\approx 5887.28~\text{cm}^3 \end{aligned}$$
Therefore, the volume of the sphere is approximately $$5887~\text{cm}^3$$ (rounded to the nearest cubic centimeter).
Example-11
A shot-putt is a metallic sphere of radius 4.9 cm. If the density of the metal is 7.8 g per cm3, find the mass of the shot-putt.
Solution:
The shot-putt is a solid metallic sphere with a radius of $$r = 4.9~\text{cm}$$ The density of the metal is given as $$7.8~\text{g/cm}^3$$
The formula for the volume of a sphere is: $$ V = \frac{4}{3} \pi r^3 $$
Substitute the values to find the volume: $$ \begin{aligned} V &= \frac{4}{3} \times \frac{22}{7} \times 4.9 \times 4.9 \times 4.9 \\\\ &= \frac{4}{3} \times \frac{22}{7} \times 117.649 \\\\ &= \frac{4}{3} \times 370.52 \\\\ &= 494.03~\text{cm}^3 \end{aligned} $$
Now, use the given density to calculate the mass of the shot-putt: $$\begin{aligned} \text{Mass} &= \text{Density} \times \text{Volume} \\&= 7.8~\text{g/cm}^3 \times 494.03~\text{cm}^3 \\&\approx 3853~\text{g} \end{aligned}$$
When converted to kilograms: $$ 3853~\text{g} = 3.85~\text{kg} $$
Therefore, the mass of the shot-putt is approximately $$3.85~\text{kg}$$
Example-12
A hemispherical bowl has a radius of 3.5 cm. What would be the volume of water it would contain?
Solution:
The radius of the hemispherical bowl is $$r = 3.5~\text{cm}$$
To find the volume of water the bowl can contain, we use the formula for the volume of a hemisphere: $$ V = \frac{2}{3} \pi r^3 $$
Substituting the values: $$ \begin{aligned} V &= \frac{2}{3} \times \frac{22}{7} \times 3.5 \times 3.5 \times 3.5 \\\\ &= \frac{2}{3} \times \frac{22}{7} \times 42.875 \\\\ &= \frac{2}{3} \times 134.75 \\\\ &= 89.83~\text{cm}^3 \end{aligned} $$
Therefore, the maximum volume of water the hemispherical bowl can contain is $$89.83~\text{cm}^3$$
Summary
- Curved surface area of a cone = \(\pi rl\)
- Total surface area of a right circular cone = \(\pi rl+\pi r^2\)
- Surface area of a sphere of radius \(r=4\pi r^2\)
- Curved surface area of a hemisphere = \(2\pi r^2\)
- Total surface area of a hemisphere = \(3\pi r^2\)
- Volume of a cone = \(\frac{1}{3}\pi r^2h\)
- Volume of a sphere of radius \(r=\frac{4}{3}\pi r^3\)
- Volume of a hemisphere = \(\frac{2}{3}\pi r^3\)
