Problem 10 - Basic Engineering Mathematics

Question

Calculate the volume and total surface area of a hemisphere of diameter \(5.0 \mathrm{~cm}\).

Step-by-Step Solution

Answer

Volume is approximately 32.73 cm³. Total surface area is approximately 58.90 cm².

Step 1: Calculate the Radius

The diameter of the hemisphere is given as 5.0 cm. The radius is half of the diameter, so we first calculate the radius as follows:\[\text{Radius} = \frac{\text{Diameter}}{2} = \frac{5.0}{2} = 2.5\, \text{cm}\]

Step 2: Calculate the Volume of the Hemisphere

The volume of a hemisphere is given by the formula:\[V = \frac{2}{3} \pi r^3\]Substituting the radius from Step 1:\[V = \frac{2}{3} \pi (2.5)^3 \approx \frac{2}{3} \pi \times 15.625 \approx 32.725\, \text{cm}^3\]

Step 3: Calculate the Curved Surface Area of the Hemisphere

The curved surface area of a hemisphere is given by the formula:\[A_c = 2 \pi r^2\]Substituting the radius from Step 1:\[A_c = 2 \pi (2.5)^2 = 2 \pi \times 6.25 \approx 39.27\, \text{cm}^2\]

Step 4: Calculate the Total Surface Area of the Hemisphere

The total surface area of a hemisphere includes the curved surface area plus the area of the base (which is a circle). The formula is:\[A_t = 2 \pi r^2 + \pi r^2 = 3 \pi r^2\]Substituting the radius from Step 1:\[A_t = 3 \pi (2.5)^2 = 3 \pi \times 6.25 \approx 58.90\, \text{cm}^2\]

Key Concepts

Volume Calculation

Understanding the volume calculation of a hemisphere is key to mastering basic geometry. Unlike a sphere, a hemisphere is essentially half a sphere. To calculate its volume, we use a formula specific to hemispheres, which accounts for only half the volume of a complete sphere.

The formula to find the volume of a hemisphere is: \[ V = \frac{2}{3} \pi r^3 \] Here, \( \pi \) is a constant (approximately 3.14159), and \( r \) represents the radius of the hemisphere.

In practice, to solve this, you first need to determine the radius if it's not given directly. The radius is half of the diameter. For instance, if a hemisphere has a diameter of 5 cm, the radius would be 2.5 cm. You substitute the radius into the volume formula to calculate the total volume.

• Step-by-step: Take the cube of the radius (e.g., \( r^3 = (2.5)^3 \)).
• Multiply by \( \pi \) and \( \frac{2}{3} \).
• This results in the volume, in this example, approximately equal to 32.725 cm³.

Surface Area Calculation

Calculating the surface area of a hemisphere involves understanding that we're dealing with two sections: the curved surface and the flat base circle. The curved surface area represents the outer dome-like part of the hemisphere.

The formula for calculating the curved surface area of a hemisphere is: \[ A_c = 2 \pi r^2 \] This formula, much like other surface area calculations, highlights the role of the radius and the constant \( \pi \). By substituting the radius (e.g., 2.5 cm) into the formula, we can determine the curved surface area. In this case, it calculates to approximately 39.27 cm².

For the total surface area, we also add the area of the flat base circle of the hemisphere. Adding these two areas uses the formula: \[ A_t = 3 \pi r^2 \] This accounts for both the curved surface and the base. Using the radius (again, 2.5 cm), we calculate a total surface area of approximately 58.90 cm².

Hemisphere Formulas

Mastering geometry often requires a good grasp of different shapes and their formulas. Hemispheres, derived from spheres, have distinct formulas for their volume and surface area. Remember, a hemisphere is simply half a sphere.

When you encounter problems regarding hemispheres, keep the following key formulas in mind:

• Volume of a Hemisphere: \( V = \frac{2}{3} \pi r^3 \) – This formula is used when you need the inside capacity or the amount of space the hemisphere occupies.
• Curved Surface Area: \( A_c = 2 \pi r^2 \) – This guides you to find the area of the dome-like part of the hemisphere.
• Total Surface Area: \( A_t = 3 \pi r^2 \) – This calculates both the curved surface and the circular base.

Understanding how to apply and rearrange these formulas is crucial for solving various geometry problems effectively. With practice, identifying when and how to use these equations becomes second nature, providing a foundation for more complex geometrical studies.