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Chapter 12: Problem 16

Compare the volumes of a hemisphere and a cone with congruent bases and equal heights.

Short Answer

Expert verified
The volume of the hemisphere is twice the volume of a cone when both have congruent bases and equal heights.

Step by step solution

01

Finding the volume of a hemisphere

To find the volume of the hemisphere, use the formula \(\frac{2}{3}\pi r^3\). Given that the radius is r, the volume of the hemisphere becomes, \(\frac{2}{3}\pi r^3\).
02

Finding the volume of the cone

The formula to find the volume of a cone is diverse, given by \(\frac{1}{3}\pi r^2h\). In the problem is given that these forms have equal heights, in a hemisphere the height 'h' would be the same as the radius 'r'. So in this case, the volume of the cone becomes \(\frac{1}{3}\pi r^3\).
03

Comparing the volumes

Comparing both volumes, it is easy to see that the volume of the hemisphere is 2 times greater than the cone. This can be confirmed by multiplying the volume of the cone, \(\frac{1}{3}\pi r^3\), by 2, it gives \(\frac{2}{3}\pi r^3\), which is the volume of the hemisphere.

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Key Concepts

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

Volume of a Hemisphere
When looking at the volume of a hemisphere, imagine slicing a perfect sphere in half. The formula to calculate this is \( \frac{2}{3} \pi r^3 \), where \( r \) represents the radius of the hemisphere. This formula shows that we're taking two-thirds of the sphere's volume. In simpler terms, the hemisphere is just half the sphere's space, adjusted by the formula factor.

The use of \( \pi \) ensures it maintains its circular nature, and \( r^3 \) emphasizes how volume is calculated in cubic units. Think of \( r \) as the distance from the center to the surface, and it's cubed because we're considering three dimensions: width, length, and height.
Volume of a Cone
The cone is another intriguing solid with its own formula for volume: \( \frac{1}{3} \pi r^2 h \). This involves multiplying the area of the base circle, \( \pi r^2 \), by the height, \( h \), and then taking a third of that product.

Here鈥檚 how it breaks down:
  • \( \pi r^2 \) finds the area of the base of the cone.
  • The \( h \) represents how tall the cone is from base to tip.
  • Dividing by three reflects how the cone's point reduces its space compared to a cylinder (which doesn鈥檛 come to a point). This factor is what sets cones apart in calculations of their volume.
If both the height and radius are equal, the calculation simplification nicely ties back to our scenario.
Comparative Analysis of Volumes
Now, let's examine how the volumes of these two shapes relate when they have the same radius and height. A quick comparison shows us:
  • The volume of the hemisphere is \( \frac{2}{3} \pi r^3 \).
  • The volume of a cone is \( \frac{1}{3} \pi r^3 \).
This means the hemisphere's volume is double that of the cone. Why? Because the factor of two in the hemisphere's formula.

Analyzing sizes helps illuminate how geometry shapes our understanding. Hemisphere fills more space because it bulks out, encompassing more volume than the tapering cone. Visualizing these calculations can make a striking difference in recognizing spatial relationships.

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Most popular questions from this chapter

The radii of two spheres are in a ratio of \(2: 5 .\) a Find the ratio of their volumes. b Find the ratio of their surface areas.

PABCD is a regular square pyramid. a If each side of the base has a length of 14 and the altitude (PQ) is \(24,\) find the pyramid's lateral area and total area. b If each slant height is 17 and the altitude is \(15,\) find the pyramid's lateral area and total area. (Figure can't copy)

ABCD is a parallelogram, with \(A=(3,6), B=(13,6)\) \(\mathrm{C}=(7,-2),\) and \(\mathrm{D}=(-3,-2)\) a Find the slopes of the diagonals, \(\overline{A C}\) and \(\overline{B D}\).be your answers to part a to identify \(\square A B C D\) by its most specific name.

A rubber ball is formed by a rubber shell filled with air. The shell's outer diameter is \(48 \mathrm{mm},\) and its inner diameter is \(42 \mathrm{mm} .\) Find, to the nearest cubic centimeter, the volume of rubber used to make the ball.

A cistern is to be built of cement. The walls and bottom will be 1 ft thick. The outer height will be 20 ft. The inner diameter will be \(10 \mathrm{ft}\). To the nearest cubic foot, how much cement will be needed for the job? (Figure can't copy)
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