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Chapter 12: Problem 15

A sphere has radius 2 and a hemisphere ("half" a sphere) has radius 4. Compare their volumes.

Short Answer

Expert verified
The volume of the hemisphere is four times larger than the volume of the sphere.

Step by step solution

01

Calculating the Volume of the Sphere

The volume V of the sphere is calculated using the formula \( V = \frac{4}{3}\pi r^3 \). For a sphere with radius 2, we have: \( V = \frac{4}{3}\pi (2)^3 \ = \frac{4}{3}\pi \times 8 \ = \frac{32}{3}\pi \) cubic units.
02

Calculating the Volume of the Hemisphere

The volume V of the hemisphere is calculated using the formula \( V = \frac{2}{3}\pi r^3 \). For a hemisphere with radius 4, we have: \( V = \frac{2}{3}\pi (4)^3 \ = \frac{2}{3}\pi \times 64 = \frac{128}{3}\pi \) cubic units.
03

Comparing the Volumes

Comparing the two volumes, \( \frac{32}{3}\pi \) (volume of the sphere) and \( \frac{128}{3}\pi \) (volume of the hemisphere), it can be seen that the volume of the hemisphere is four times the volume of the sphere.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Sphere Volume
Calculating the volume of a sphere involves using a specific formula that considers the radius of the sphere. The formula is given by \( V = \frac{4}{3}\pi r^3 \). Here, \( r \) represents the radius of the sphere, and \( \pi \) is a mathematical constant approximately equal to 3.14159.

To find the volume of a sphere, simply plug the radius into this formula. For instance, if a sphere has a radius of 2, substitute 2 into the formula:
  • First calculate \( 2^3 \), which is the same as multiplying 2 by itself twice to get 8.
  • Next, multiply 8 by \( \pi \) and \( \frac{4}{3} \).
  • This gives \( \frac{32}{3} \pi \) cubic units for the volume of the sphere.
Understanding this formula is crucial for working with spheres in geometry. Remember, each radius value will give a unique volume, showcasing how it directly affects the size of the sphere.
Hemisphere Volume
A hemisphere is essentially half of a sphere. To calculate its volume, the sphere volume formula is adjusted to account for the 'half'. This is done using the formula \( V = \frac{2}{3}\pi r^3 \). Here, \( r \) is still the radius of the sphere from which the hemisphere is made.

Let's use a hemisphere with a radius of 4 as an example:
  • First, calculate \( 4^3 \), which equals 64.
  • Then, multiply 64 by \( \frac{2}{3} \) and \( \pi \).
  • This results in a volume of \( \frac{128}{3}\pi \) cubic units.
It's worth noting that this formula effectively divides the sphere volume formula by two because a hemisphere is half of a sphere. This approach gives you a reliable way to measure the volume of any hemispherical shape, provided you know the radius.
Volume Comparison
Comparing the volumes of a sphere and a hemisphere might seem complex at first, but it just involves straightforward division. In our specific problem, the sphere has a radius of 2 and a volume of \( \frac{32}{3}\pi \) cubic units, while the hemisphere has a radius of 4 and a volume of \( \frac{128}{3}\pi \) cubic units.

To compare these volumes, consider their mathematical relationship:
  • The hemisphere's volume is obtained from a larger radius, which results in a different scale.
  • To see how much larger, compare \( \frac{128}{3}\pi \) to \( \frac{32}{3}\pi \).
  • Dividing these volumes, \( \frac{128}{32} = 4 \).
  • This means that the hemisphere's volume is four times that of the sphere.
This comparison highlights the importance of the radius in determining volume. Even though a hemisphere is a fraction of a full sphere, its larger radius in this context makes it significantly larger than the given sphere.

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Most popular questions from this chapter

Find the area of the circle formed when a plane passes \(2 \mathrm{cm}\) from the center of a sphere with radius \(5 \mathrm{cm}\).

Construction engineers know that the strength of a column is proportional to the area of its cross section. Suppose that the larger of two similar columns is three times as high as the smaller column. a. The larger column is \(__________\) times as strong as the smaller column. b. The larger column is \(______________\) times as heavy as the smaller column. c. Which can support more, per pound of column material, the larger or the smaller column?

Four solid metal balls fit snugly inside a cylindrical can. A geometry student claims that two extra balls of the same size can be put into the can. provided all six balls can be melted and the molten liquid poured into the can. Is the student correct? (Hint: Let the radius of each ball be \(r .)\)

Determine whether each conjecture is true or false. Give a counterexample for any false conjecture. \(\angle L\) and \(\angle M\) are complementary angles. \(\angle N\) and \(\angle P\) are complementary angles. If \(m \angle L=y-2, m \angle M=2 x+3, m \angle N=2 x-y,\) and \(m \angle P=x-1\) find the values of \(x, y, m \angle L, m \angle M, m \angle N,\) and \(m \angle P .\)

If the length and width of a rectangular solid are each decreased by \(20 \%,\) by what percent must the height be increased for the volume to remain unchanged? Give your answer to the nearest whole percent.
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