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# 13. SURFACE AREAS AND VOLUMES

## 13.1 INTRODUCTION

Wherever we look, usually we see solids. So far, in all our study, we have been dealing with figures that can be easily drawn on our notebooks or blackboards. These are called *plane figures*. We have understood what rectangles, squares and circles are, what we mean by their perimeters and areas, and how we can find them. We have learnt these in earlier classes. It would be interesting to see what happens if we cut out many of these plane figures of the same shape and size from cardboard sheet and stack them up in a vertical pile. By this process, we shall obtain some *solid figures* (briefly called *solids*) such as a cuboid, a cylinder, etc. In the earlier classes, you have also learnt to find the surface areas and volumes of cuboids, cubes and cylinders. We shall now learn to find the surface areas and volumes of cuboids and cylinders in details and extend this study to some other solids such as cones and spheres.

## 13.2 SURFACE AREA OF A CUBOID AND A CUBE

Have you looked at a bundle of many sheets of paper? How does it look? Does it look Like what you see in Fig. 13.1?

**Fig. 13.1**

That makes up a cuboid. How much of brown paper would you need, if you want to cover this cuboid? Let us see:

First we would need a rectangular piece to cover the bottom of the bundle. That would be as shown in Fig. 13.2 (a)

Then we would need two long rectangular pieces to cover the two side ends. Now, it would look like Fig. 13.2 (b).

Now to cover the front and back ends, we would need two more rectangular pieces of a different size. With them, we would now have a figure as shown in Fig. 13.2(c).

This figure, when opened out, would look like Fig. 13.2 (d).

Finally, to cover the top of the bundle, we would require another rectangular piece exactly like the one at the bottom, which if we attach on the right side, it would look like Fig. 13.2(e).

So we have used six rectangular pieces to cover the complete outer surface of the cuboid.

**Fig. 13.2**

This shows us that the outer surface of a cuboid is made up of six rectangles (in fact, rectangular regions, called the faces of the cuboid), whose areas can be found by multiplying length by breadth for each of them separately and then adding the six areas together.

Now, if we take the length of the cuboid as *l*, breadth as *b *and the height as *h*, then the figure with these dimensions would be like the shape you see in Fig. 13.2(f).

So, the sum of the areas of the six rectangles is:

**Area of rectangle 1 (= l × h) + Area of rectangle 2 (= l × b) + Area of rectangle 3 (= l × h ) + Area of rectangle 4 (= l × b) + Area of rectangle 5 (= b × h) + Area of rectangle 6 (= b × h ) **

**= 2( l × b) + 2(b × h) + 2(l × h) = 2(lb + bh + hl)**

This gives us:

where *l*, *b *and *h *are respectively the three edges of the cuboid.

**Note: **The unit of area is taken as the square unit, because we measure the magnitude of a region by filling it with squares of side of unit length.

For example, if we have a cuboid whose length, breadth and height are 15 cm, 10 cm and 20 cm respectively, then its surface area would be:

2[(15 × 10) + (10 × 20) + (20 × 15)] cm^{2}

= 2(150 + 200 + 300) cm^{2}

= 2 × 650 cm^{2}

= 1300 cm^{2}

Recall that a cuboid, whose length, breadth and height are all equal, is called a *cube*. If each edge of the cube is *a*, then the surface area of this cube would be 2(*a *× *a *+ *a *× *a *+ *a *× *a*) i.e., 6*a*^{2} (see Fig. 13.3), giving us

where *a *is the edge of the cube.

**Fig. 13.3**

Suppose, out of the six faces of a cuboid, we only find the area of the four faces, leaving the bottom and top faces. In such a case, the area of these four faces is called the **lateral surface area **of the cuboid. So, *lateral surface area of a cuboid of length l, breadth b and height h is equal to 2lh + 2bh or 2*(*l + b*)*h*. Similarly, *lateral surface area of a cube of side a is equal to 4a ^{2}*.

Keeping in view of the above, the surface area of a cuboid (or a cube) is sometimes also referred to as the **total surface area**. Let us now solve some examples.

Mary wants to decorate her Christmas tree. She wants to place the tree on a wooden box covered with coloured paper with picture of Santa Claus on it (see Fig. 13.4). She must know the exact quantity of paper to buy for this purpose. If the box has length, breadth and height as 80 cm, 40 cm and 20 cm respectively how many square sheets of paper of side 40 cm would she require?

Since Mary wants to paste the paper on the outer surface of the box; the quantity of paper required would be equal to the surface area of the box which is of the shape of a cuboid. The dimensions of the box are:

**Fig. 13.4**

Length =80 cm, Breadth = 40 cm, Height = 20 cm.

The surface area of the box = 2(*lb *+ *bh *+ *hl*)

= 2[(80 × 40) + (40 × 20) + (20 × 80)] cm^{2}

= 2[3200 + 800 + 1600] cm^{2}

= 2 × 5600 cm2 = 11200 cm^{2}

The area of each sheet of the paper = 40 × 40 cm^{2} = 1600 cm^{2}

Therefore, number of sheets required = ^{Surface area of box}/_{Area of one sheet of paper} = 7

So, she would require 7 sheets.

Hameed has built a cubical water tank with lid for his house, with each outer edge 1.5 m long. He gets the outer surface of the tank excluding the base, covered with square tiles of side 25 cm (see Fig. 13.5). Find how much he would spend for the tiles, if the cost of the tiles is Rs 360 per dozen.

Since Hameed is getting the five outer faces of the tank covered with tiles, he would need to know the surface area of the tank, to decide on the number of tiles required.

Edge of the cubical tank = 1.5 m = 150 cm (= *a*)

So, surface area of the tank = 5 × 150 × 150 cm^{2}

Area of each square tile = side × side = 25 × 25 cm^{2}

So, the number of tiles required = ^{Surface area of tank}/_{Area of each tile} = ^{5 x 150 x 150}/_{25 x 25 }=180

Cost of 1 dozen tiles, i.e., cost of 12 tiles = Rs 360

Therefore, cost of one tile = Rs ^{360}/_{12}= Rs 30

So, the cost of 180 tiles = 180 × Rs 30 = Rs 5400

**Fig. 13.5**

## 13.3 SURFACE AREA OF A RIGHT CIRCULAR CYLINDER

If we take a number of circular sheets of paper and stack them up as we stacked up rectangular sheets earlier, what would we get (see Fig. 13.6)?

**Fig. 13.6**

Here, if the stack is kept vertically up, we get what is called a *right circular cylinder*, since it has been kept at right angles to the base, and the base is circular. Letus see what kind of cylinder is *not *a right circular cylinder.

In Fig 13.7 (a), you see a cylinder, which is certainly circular, but it is not at right angles to the base. So, we can *not *say this a *right* circular cylinder.

Of course, if we have a cylinder with a non circular base, as you see in Fig. 13.7 (b), then we also cannot call it a right circular cylinder.

**Fig. 13.7**

**Remark : **Here, we will be dealing with only right circular cylinders. So, unless stated otherwise, the word cylinder would mean a right circular cylinder.

Now, if a cylinder is to be covered with coloured paper, how will we do it with the minimum amount of paper? First take a rectangular sheet of paper, whose length is just enough to go round the cylinder and whose breadth is equal to the height of the cylinder as shown in Fig. 13.8.

**Fig. 13.8**

The area of the sheet gives us the curved surface area of the cylinder. Note that the length of the sheet is equal to the circumference of the circular base which is equal to 2π*r*.

So, curved surface area of the cylinder = area of the rectangular sheet = length × breadth

*h*

*r*×

*h*

Therefore,

where *r *is the radius of the base of the cylinder and *h *is the height of the cylinder.

**Remark : **In the case of a cylinder, unless stated otherwise, 'radius of a cylinder' shall mean' base radius of the cylinder'.

If the top and the bottom of the cylinder are also to be covered, then we need two circles (infact, circular regions) to do that, each of radius *r*, and thus having an area of π*r*^{2} each (see Fig. 13.9), giving us the total surface area as 2π*rh *+ 2π*r*^{2} = 2π*r*(*r *+ *h*).

**Fig. 13.9**

So,

where *h *is the height of the cylinder and *r *its radius.

**Remark : **You may recall from Chapter 1 that π is an irrational number. So, the value of π is a non-terminating, non-repeating decimal. But when we use its value in our calculations, we usually take its value as approximately equal to ^{22}/_{7} or 3.14.

Savitri had to make a model of a cylindrical kaleidoscope for her science project. She wanted to use chart paper to make the curved surface of the kaleidoscope (see Fig 13.10). What would be the area of chart paper required by her, if she wanted to make a kaleidoscope of length 25 cm with a 3.5 cm radius? You may take π = ^{22}/_{7}

**Fig. 13.4**

Radius of the base of the cylindrical kaleidoscope (*r*) = 3.5 cm.

Height (length) of kaleidoscope (*h*) = 25 cm.

Area of chart paper required = curved surface area of the kaleidoscope

= 2π*rh*

=2×^{22 }/_{7}×3.5× 25 cm ^{2}

= 550 cm^{2}

**Fig. 13.10**

## 13.4 SURFACE AREA OF A RIGHT CIRCULAR CONE

So far, we have been generating solids by stacking up congruent figures. Incidentally, such figures are called *prisms*. Now let us look at another kind of solid which is not a prism. (These kinds of solids are called *pyramids*). Let us see how we can generate them.

This is called a *right circular cone*. In Fig. 13.13(c) of the right circular cone, the point A is called the vertex, AB is called the height, BC is called the *radius* and AC is called the slant height of the cone. Here B will be the centre of circular base of the cone. The height, radius and slant height of the cone are usually denoted by *h*, *r *and *l *respectively. Once again, let us see what kind of cone we can *not *call a right circular cone. Here, you are (see Fig. 13.14)! What you see in these figures are not right circular cones; because in (a), the line joining its vertex to the centre of its base is not at right angle to the base, and in (b) the base is not circular.

**Fig. 13.14**

As in the case of cylinder, since we will be studying only about right circular cones, remember that by 'cone' in this chapter, we shall mean a 'right circular cone.'

Find the curved surface area of a right circular cone whose slant height is 10 cm and base radius is 7 cm.

Curved surface area = π*rl*

=^{22}/_{7} × 7 × 10 cm^{2}

= 220 cm^{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).

Here, *h *= 16 cm and *r *= 12 cm.

So, from *l*^{2} = *h*^{2} + *r*^{2}, we have

*l *= √16^{2} + 12^{2} cm = 20 cm

So, curved surface area = π*rl*

= 3.14 × 12 × 20 cm^{2} = 753.6 cm^{2}

Further, total surface area = π*rl *+ π*r*^{2}

= (753.6 + 3.14 × 12 × 12) cm^{2} = (753.6 + 452.16) cm^{2}
= 1205.76 cm^{2}

A corn cob (see Fig. 13.17), 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 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.

Since the grains of corn are found only on the curved surface of the corn cob, we would need to know the curved surface area of the corn cob to find the total number of grains on it. In this question, we are given the height of the cone, so we need to find its slant height.

**Fig. 13.17**

Here, *l *= √r^{2} + h^{2} = √(2.1)^{2} + 20^{2} cm = 20.11 cm

Therefore, the curved surface area of the corn cob = π*rl* = ^{22}/_{7} × 2.1 × 20.11 cm^{2} = 132.726 cm ^{2} = 132.73 cm^{2} (approx.)

Number of grains of corn on 1 cm^{2} of the surface of the corn cob = 4

Therefore, number of grains on the entire curved surface of the cob = 132.73 × 4 = 530.92 = 531 (approx.)

So, there would be approximately 531 grains of corn on the cob.

## 13.5 SURFACE AREA OF A SPHERE

What is a sphere? Is it the same as a circle? Can you draw a circle on a paper? Yes, you can, because a circle is a plane closed figure whose every point lies at a constant distance (called **radius**) from a fixed point, which is called the **centre **of the circle. Now if you paste a string along a diameter of a circular disc and rotate it as you had rotated the triangle in the previous section, you see a new solid (see Fig 13.18). What does it resemble? A ball? Yes. It is called a **sphere**.

**Fig. 13.18**

Can you guess what happens to the centre of the circle, when it forms a sphere on rotation? Of course, it becomes the centre of the sphere. So, *a sphere is a three* *dimensional figure *(*solid figure*), *which is made up of all points in the space, * *which lie at a constant distance called the radius, from a fixed point called the* *centre of the sphere.*

**Note : **A sphere is like the surface of a ball. The word *solid sphere *is used for the solid whose surface is a sphere.

What have you achieved in all this?

The string, which had completely covered the surface area of the sphere, has been used to completely fill the regions of four circles, all of the same radius as of the sphere. So, what does that mean? This suggests that the surface area of a sphere of radius *r* = 4 times the area of a circle of radius *r *= 4 × (π *r*^{2})

So,

*r*is the radius of the sphere.

How many faces do you see in the surface of a sphere? There is only one, which is curved.

Now, let us take a solid sphere, and slice it exactly 'through the middle' with a plane that passes through its centre. What happens to the sphere?

Yes, it gets divided into two equal parts (see Fig. 13.20)!

What will each half be called? It is called a **hemisphere**.

(Because 'hemi' also means 'half')

And what about the surface of a hemisphere? How many faces does it have?

Two! There is a curved face and a flat face (base).

The curved surface area of a hemisphere is half the surface area of the sphere, which is ½ of 4π*r*^{2}.

**Fig. 13.20**

Therefore,

*r*is the radius of the sphere of which the hemisphere is a part.

Now taking the two faces of a hemisphere, its surface area 2π*r*^{2} + π*r*^{2}

So,

Find the surface area of a sphere of radius 7 cm.

The surface area of a sphere of radius 7 cm would be

4π*r*^{2} = 4 × ^{22}/_{7} × 7 × 7 cm ^{2} = 616 cm^{2}

Find (i) the curved surface area and (ii) the total surface area of a hemisphere of radius 21 cm.

The curved surface area of a hemisphere of radius 21 cm would be

= 2π*r*^{2} = 2 × ^{22}/_{7} × 21 × 21 cm^{2} = 2772 cm^{2}

(ii) the total surface area of the hemisphere would be

3π*r*^{2} = 3 × ^{22}/_{7} × 21 × 21 cm^{2} = 4158 cm^{2}

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.

Diameter of the sphere = 7 m. Therefore, radius is 3.5 m. So, the riding space available for the motorcyclist is the surface area of the 'sphere' which is given by

4π*r*^{2} = 4 × ^{22}/_{7} × 3.5 × 3.5 m^{2}

= 154 m^{2}

A hemispherical dome of a building needs to be painted (see Fig. 13.21). 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 Rs 5 per 100 cm^{2}.

Since only the rounded surface of the dome is to be painted, we would need to find the curved surface area of the hemisphere to know the extent of painting that needs to be done. Now, circumference of the dome = 17.6 m. Therefore, 17.6 = 2π*r*.

So, the radius of the dome = 17.6 × ^{7}/_{2 x 22} m = 2.8 m

The curved surface area of the dome = 2π*r*^{2}

= 2 × ^{22}/_{7} × 2.8 × 2.8 m^{2} = 49.28 m^{2}

Now, cost of painting 100 cm ^{2} is Rs 5.

So, cost of painting 1 m ^{2} = Rs 500

Therefore, cost of painting the whole dome = Rs 500 × 49.28 = Rs 24640

**Fig. 13.21**

## 13.6 VOLUME OF A CUBOID

You have already learnt about volumes of certain figures (objects) in earlier classes. Recall that solid objects occupy space. The measure of this occupied space is called the **Volume **of the object.

** Note: **If an object is solid, then the space occupied by such an object is measured, and is termed the **Volume **of the object. On the other hand, if the object is hollow, then interior is empty, and can be filled with air, or some liquid that will take the shape of its container. In this case, the volume of the substance that can fill the interior is called the ** capacity of the container**. In short, the volume of an object is the measure of the space it occupies, and the capacity of an object is the volume of substance its interior can accommodate. Hence, the unit of measurement of either of the two is cubic unit.

So, if we were to talk of the volume of a cuboid, we would be considering the measure of the space occupied by the cuboid.

Further, the area or the volume is measured as the magnitude of a region. So, correctly speaking, we should be finding the area of a circular region, or volume of a cuboidal region, or volume of a spherical region, etc. But for the sake of simplicity, we say, find the area of a circle, volume of a cuboid or a sphere even though these mean only their boundaries.

**Fig. 13.23**

Observe Fig. 13.23. Suppose we say that the area of each rectangle is A, the height up to which the rectangles are stacked is *h * and the volume of the cuboid is V. Can you tell what would be the relationship between V, A and *h*?

The area of the plane region occupied by each rectangle × height= Measure of the space occupied by the cuboid

So, we get A × *h *= V

That is,

or *l *× *b *× *h*, where *l*, *b *and *h *are respectively the length, breadth and height of the cuboid.

**Note : **When we measure the magnitude of the region of a space, that is, the space occupied by a solid, we do so by counting the number of cubes of edge of unit length that can fit into it exactly. Therefore, the unit of measurement of volume is cubic unit.

Again

where ** a **is the edge of the cube (see Fig. 13.24).

So, if a cube has edge of 12 cm,

then volume of the cube = 12 × 12 × 12 cm^{3} = 1728 cm^{3}.

Recall that you have learnt these formulae in earlier classes. Now let us take some examples to illustrate the use of these formulae:

**Fig. 13.24**

A wall of length 10 m was to be built across an open ground. The height of the wall is 4 m and thickness of the wall is 24 cm. If this wall is to be built up with bricks whose dimensions are 24 cm × 12 cm × 8 cm, how many bricks would be required?

Since the wall with all its bricks makes up the space occupied by it, we need to find the volume of the wall, which is nothing but a cuboid.

Here, Length = 10 m = 1000 cm

Thickness = 24 cm

Height = 4 m = 400 cm

Therefore, Volume of the wall = length × thickness × height = 1000 × 24 × 400 cm^{3}

Now, each brick is a cuboid with length = 24 cm, breadth = 12 cm and height = 8 cm

So, volume of each brick = length × breadth × height = 24 × 12 × 8 cm^{3}

So, number of bricks required = ^{volume of the wall}/_{volume of each brick} = ^{1000 x 24 x 400}/_{24 x 12 x 8} = 4166.6

So, the wall requires 4167 bricks.

A child playing with building blocks, which are of the shape of cubes, has built a structure as shown in Fig. 13.25. If the edge of each cube is 3 cm, find the volume of the structure built by the child.

Volume of each cube = edge × edge × edge = 3 × 3 × 3 cm^{3} = 27 cm^{3}

Number of cubes in the structure = 15

Therefore, volume of the structure = 27 × 15 cm^{3} = 405 cm^{3}

**Fig. 13.25**

## 13.7 VOLUME OF A CYLINDER

Just as a cuboid is built up with rectangles of the same size, we have seen that a right circular cylinder can be built up using circles of the same size. So, using the same argument as for a cuboid, we can see that the volume of a cylinder can be obtained as : base area × height = area of circular base × height = π*r*^{2}*h*

So,

where *r *is the base radius and *h *is the height of the cylinder.

The pillars of a temple are cylindrically shaped (see Fig. 13.26). If each pillar has a circular base of radius 20 cm and height 10 m, how much concrete mixture would be required to build 14 such pillars?

Since the concrete mixture that is to be used to build up the pillars is going to occupy the entire space of the pillar, what we need to find here is the volume of the cylinders.

Radius of base of a cylinder = 20 cm

Height of the cylindrical pillar = 10 m = 1000 cm

So, volume of each cylinder = π*r*^{2}*h*

**Fig. 13.26**

^{22}/

_{7}×20×20×1000 cm

^{3}

^{8800000}/

_{7}cm

^{3}

^{8.8}/

_{7}=m

^{3}(Since 1000000 cm

^{3}= 1m

^{3})

Therefore, volume of 14 pillars = volume of each cylinder × 14 = ^{8.8}/_{7} ×14m^{3} = 17.6 m^{3}

So, 14 pillars would need 17.6 m^{3} of concrete mixture.

At a Ramzan Mela, a stall keeper in one of the food stalls has a large cylindrical vessel of base radius 15 cm filled up to a height of 32 cm with orange juice. The juice is filled in small cylindrical glasses (see Fig. 13.27) of radius 3 cm up to a height of 8 cm, and sold for Rs 3 each. How much money does the stall keeper receive by selling the juice completely?

**Fig. 13.27**

The volume of juice in the vessel = volume of the cylinderical vessel

= πR^{2}H (where R and H are taken as the radius and height respectively of the vessel)

= π × 15 × 15 × 32 cm^{3}

Similarly, the volume of juice each glass can hold = π*r*^{2}*h* (where *r *and *h *are taken as the radius and height respectively of each glass)

= π × 3 × 3 × 8 cm^{3}

So, number of glasses of juice that are sold = ^{volume of vessel}/_{volume of each glass} = ^{π x 15 x 15 x 32 }/_{π x 3 x 3 x 8} = 100

Therefore, amount received by the stall keeper = Rs 3 × 100 = Rs 300

## 13.8 VOLUME OF A RIGHT CIRCULAR CONE

In Fig 13.28, can you see that there is a right circular cylinder and a right circular cone of the same base radius and the same height?

**Fig. 13.28**

The height and the slant height of a cone are 21 cm and 28 cm respectively. Find the volume of the cone.

From *l*^{2} = *r*^{2} + *h*^{2}, we have

*r *= √ l^{2} - h^{2} = √ 28^{2} - 21 ^{2 } cm = 7 √7

So, volume of the cone = ⅓ π*r*^{2}*h *= ⅓ × ^{22}/_{7} ×7 √7×7 √7 × 21 cm ^{3} = 7546 cm^{3}

Monica has a piece of canvas whose area is 551 m^{2}. 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 m^{2}, find the volume of the tent that can be made with it.

Since the area of the canvas = 551 m^{2} and area of the canvas lost in wastage is 1 m^{2}, therefore the area of canvas available for making the tent is (551 – 1) m^{2} = 550 m^{2}.

Now, the surface area of the tent = 550 m^{2} and the required base radius of the conical tent = 7 m

Note that a tent has only a curved surface (the floor of a tent is not covered by canvas!!).

Therefore, curved surface area of tent = 550 m^{2}.

That is, π*rl *= 550

or, ^{22}/_{7} × 7 × *l *= 550

or, *l *=3 ^{550}/_{22}m= 25 m

Now, *l*^{2} = *r*^{2} + *h*^{2}

Therefore, *h *= √ l^{2} - h^{2}= √ 25^{2} - 7^{2} m= √ 625 - 49 m = √ 576 m = 24 m

So, the volume of the conical tent = ⅓ π*r ^{2} h *= ⅓ ×

^{22}/

_{7}×7× 7× 24m

^{3}= 1232 m

^{3}.

## 13.9 VOLUME OF A SPHERE

Now, let us see how to go about measuring the volume of a sphere. First, take two or three spheres of different radii, and a container big enough to be able to put each of the spheres into it, one at a time. Also, take a large trough in which you can place the container. Then, fill the container up to the brim with water [see Fig. 13.30(a)].

Now, carefully place one of the spheres in the container. Some of the water from the container will over flow into the trough in which it is kept [see Fig. 13.30(b)]. Carefully pour out the water from the trough into a measuring cylinder (i.e., a graduated cylindrical jar) and measure the water over flowed [see Fig. 13.30(c)]. Suppose the radius of the immersed sphere is *r *(you can find the radius by measuring the diameter of the sphere). Then evaluate ^{4}/_{3} π*r*^{3}. Do you find this value almost equal to the measure of the volume over flowed?

**Fig. 13.30**

Once again repeat the procedure done just now, with a different size of sphere. Find the radius R of this sphere and then calculate the value of ^{4}/_{3} πR^{3}. Once again this value is nearly equal to the measure of the volume of the water displaced (over flowed) by the sphere. What does this tell us? We know that the volume of the sphere is the same as the measure of the volume of the water displaced by it. By doing this experiment repeatedly with spheres of varying radii, we are getting the same result, namely, the volume of a sphere is equal to ^{4}/_{3}π times the cube of its radius. This gives us the idea that

*r*is the radius of the sphere.

Later, in higher classes it can be proved also. But at this stage, we will just take it as true.

Since a hemisphere is half of a sphere, can you guess what the volume of a hemisphere will be? Yes, it is ½of ^{4}/_{ 3}πr^{3} =π*r* ^{3}.

So,

*r*is the radius of the hemisphere.

Let us take some examples to illustrate the use of these formulae.

A shot-putt is a metallic sphere of radius 4.9 cm. If the density of the metal is 7.8 g per cm^{3}, find the mass of the shot-putt.

Required volume = ^{4}/_{3} πr^{3} = ^{4}/_{3} × ^{22}/_{7} ×11.2× 11.2× 11.2 cm^{3} = 5887.32 cm^{3}

Find the volume of a sphere of radius 11.2 cm.

Since the shot-putt is a solid sphere made of metal and its mass is equal to the product of its volume and density, we need to find the volume of the sphere.

Now, volume of the sphere = ^{4}/_{3} πr^{3} = ^{4}/_{3}× ^{22}/_{7} ×4.9× 4.9 ×4.9 cm^{3} = 493 cm^{3} (nearly)

Further, mass of 1 cm^{3} of metal is 7.8 g.

Therefore, mass of the shot-putt = 7.8 × 493 g = 3845.44 g = 3.85 kg (nearly)

A hemispherical bowl has a radius of 3.5 cm. What would be the volume of water it would contain?

The volume of water the bowl can contain = ^{2}/_{3}π*r*^{3} = ⅔ × ^{22}/_{7} ×3.5×3.5×3.5cm^{3} = 89.8 cm^{3}

In this chapter, you have studied the following points:

**Surface area of a cuboid = 2 (***lb*+*bh*+*hl*)**Surface area of a cube = 6***a*^{2}**Curved surface area of a cylinder = 2**π*rh***Total surface area of a cylinder = 2**π*r*(*r*+*h*)**Curved surface area of a cone =**π*rl***Total surface area of a right circular cone =**ππ*rl*+π*r*^{2}, i.e.,*r*(*l*+*r*)**Surface area of a sphere of radius**π*r*= 4*r*^{2}**Curved surface area of a hemisphere = 2**π*r*^{2}**Total surface area of a hemisphere = 3**π*r*^{2}**Volume of a cuboid =***l*×*b*×*h***Volume of a cube =***a*^{3}**Volume of a cylinder =**π*r*^{2}*h***Volume of a cone = ⅓ π***r*^{2}h**Volume of a sphere of radius***r*=^{4}/_{3}πr^{3}**Volume of a hemisphere = ⅔ πr**^{3}

[Here, letters *l*, *b*, *h*, *a*, *r*, etc. have been used in their usual meaning, depending on the context.]