# How are surface and volume connected?

### Compound solids: volume

You already know composite and hollowed-out bodies from class 8. Many objects are composed of geometric bodies.

Example: This spire is composed of a cylinder and a cone.

Image: iStockphoto.com (Andrei Nekrassov)

Volume of body 1 + volume of body 2 = volume of total body

In the case of composite and hollowed-out bodies, you first determine the individual bodies. Then you calculate the volume of the individual bodies and you create a formula for the entire body.

You can choose whether you want to calculate the bodies individually or the whole body.

### 1st way

Mathematically, the spire consists of a cylinder and a cone.

1. Volume cylinder:

\$\$ V_1 = G * h_K \$\$

\$\$ V_1 = π * r ^ 2 * h_K \$\$

\$\$ V_1 = π * (1.5 \ m) ^ 2 * 2 \ m \$\$

\$\$ V_1 = 14.14 \ cm ^ 3 \$\$

2. Volume cone:

\$\$ V_2 = 1/3 G * h_K \$\$

\$\$ V_2 = 1/3 π * r ^ 2 * h_K \$\$

\$\$ V_2 = 1/3 π * (1.5 \ m) ^ 2 * 3.5 \ m \$\$

\$\$ V_2 = 8.25 \ m ^ 3 \$\$

3. Whole body:

\$\$ V = V_1 + V_2 \$\$

\$\$ V = 14.14 \ m ^ 3 + 8.25 \ m ^ 3 \$\$

\$\$ V = 22.39 \ m ^ 3 \$\$

### 2nd way

You can also write everything into an equation and plug in the values:

\$\$ V = V_1 + V_2 \$\$

\$\$ V = G * h_K + 1/3 * G * h_K \$\$

\$\$ V = π * r ^ 2 * h_K + 1/3 π * r ^ 2 * h_K \$\$

\$\$ V = π * (1.5 \ m) ^ 2 * 2 \ m + 1/3 π * (1.5 \ m) ^ 2 * 3.5 \ m \$\$

\$\$ V = 22.38 \ m ^ 3 \$\$

This value is more precise because no intermediate result has been rounded.

Image: iStockphoto.com (Andrei Nekrassov)

Circle: \$\$ G = π * r ^ 2 \$\$

Cylinder: \$\$ V = G * h_K \$\$

Cone: \$\$ V = 1/3 G * h_K \$\$

### Observatory

There are also compound bodies with spheres or hemispheres like this observatory.

You can also calculate the volume here:

### 1st way

Mathematically, the observatory consists of a cylinder and a hemisphere.

1st cylinder:

\$\$ V_1 = G * h_K \$\$

\$\$ V_1 = π * r ^ 2 * h_K \$\$

\$\$ V_1 = π * (2 \ m) ^ 2 * 2 \ m \$\$

\$\$ V_1 = 25.13 \ m ^ 3 \$\$

2nd hemisphere: \$\$ V_2 = (4 / 3π * r ^ 3): 2 \$\$

\$\$ V_2 = (4 / 3π * (2 \ m) ^ 3): 2 \$\$

\$\$ V_2 = 16.76 \ m ^ 3 \$\$

3. Whole body: \$\$ V = V_1 + V_2 \$\$

\$\$ V = 25.13 \ m ^ 3 + 16.76 \ m ^ 3 \$\$

\$\$ V = 41.89 \ cm ^ 3 \$\$

### 2nd way

You can also write everything into an equation and plug in the values:

\$\$ V = V_1 + V_2 \$\$

\$\$ V = π * r ^ 2 * h_K + (4 / 3π * r ^ 3): 2 \$\$

\$\$ V = π * (2 \ m) ^ 2 * 2 \ m + (4/3 π * (2 \ m) ^ 3): 2 \$\$

\$\$ V = 41.89 \ m ^ 3 \$\$

Image: Picture-Alliance GmbH (Hans Ringhofer)
This is the Kuffner observatory in Vienna.

Circle: \$\$ G = π * r ^ 2 \$\$

Cylinder: \$\$ V = G * h_K \$\$

Sphere: \$\$ V = 4 / 3π * r ^ 3 \$\$

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### The surface of composite bodies

Calculating the surface area is a little more difficult.

The surface area of ​​a composite body is all the surfaces that you can touch.

Therefore, you cannot simply add up the surface area of ​​the individual bodies. Some surfaces are adjacent to each other. You are not allowed to include them in the surface content.

Example:

In the picture you can see that the “top” of the cylinder and the “bottom” of the cone must not be included because they are on top of each other.

For the cylinder this means that you only calculate the circular area and the jacket once. With the cone, you only need the outer surface.

Image: iStockphoto.com (Andrei Nekrassov)

### 1st way

Divide compound bodies into individual bodies. Calculate the areas you need for the total surface.

1st cylinder: a base and the lateral surface \$\$ O = π * r ^ 2 + 2 * π * r * h_K \$\$

\$\$ O = π * (1.5 \ m) ^ 2 + 2 * π * 1.5 \ m * 2 \ m \$\$

\$\$ O = 25.92 \ m ^ 2 \$\$

2nd cone: lateral surface

\$\$ O = π * r * sqrt (r ^ 2 + h ^ 2) \$\$

\$\$ O = π * 1.5 \ m * sqrt ((1.5 \ m) ^ 2 + (3.5 \ m) ^ 2) \$\$

\$\$ O = 17.94 \ m ^ 2 \$\$

3. Whole body: \$\$ O = O_ (cylinder) + O_ (Ke g e l) \$\$

\$\$ O = 25.92 \ m ^ 2 + 17.94 \ m ^ 2 \$\$

\$\$ O = 43.86 \ m ^ 2 \$\$

Surface of composite bodies

### 2nd way

As with volume, you can also write everything into an equation and insert the values:

\$\$ O = O_ (cylinder) + O_ (cone) \$\$

\$\$ O = π * r ^ 2 + 2 * π * r * h_K + π * r * sqrt (r ^ 2 + h ^ 2) \$\$

\$\$ O = π * (1.5 \ m) ^ 2 + 2 * π * 1.5 \ m * 2 \ m + π * 1.5 \ m * sqrt ((1.5 \ m) ^ 2 + (3.5 \ m) ^ 2) \$\$

\$\$ O = 43.86 \ m ^ 2 \$\$

Image: iStockphoto.com (Andrei Nekrassov)

Surface cylinder:

\$\$ O = 2 * π * r ^ 2 + 2 * π * r * h_K \$\$

Surface cone:

\$\$ O = π * r ^ 2 + π * r * s \$\$

or if only \$\$ r \$\$ and \$\$ h_K \$\$ are given:

\$\$ O = π * r ^ 2 + π * r * sqrt (r ^ 2 + h_K ^ 2) \$\$

\$\$ O \$\$ does not stand for the entire surface of the cylinder, but for the area that you need from the cylinder for the entire body. You can choose the name as you like. The main thing is that it is understandable.