What Is The Volume Of The Following Figure

Determining the volume of a figure isa fundamental skill in geometry, physics, engineering, and everyday problem‑solving. Whether you are calculating how much water a container can hold, estimating the material needed for a construction project, or simply satisfying curiosity about a shape’s size, knowing the volume of the following figure provides concrete, quantitative insight. This article walks you through the concept of volume, outlines a step‑by‑step method to find it, explains the underlying mathematics, offers worked examples, and answers common questions to solidify your understanding.

Introduction: Why Volume Matters

Volume measures the three‑dimensional space occupied by an object. Unlike area, which concerns only length and width, volume adds depth, giving a complete picture of how much “stuff” fits inside a shape. The phrase volume of the following figure appears frequently in textbooks, exams, and real‑world scenarios because it directs the reader to apply the appropriate formula to a specific diagram. Mastering this skill not only boosts academic performance but also enhances practical abilities such as packing, cooking, and fluid dynamics.

How to Find the Volume of a Figure: A Step‑by‑Step Guide

Finding the volume of any shape follows a logical sequence. By breaking the process into clear steps, you reduce errors and build confidence.

1. Identify the Shape

First, examine the diagram and name the solid. Common figures include cubes, rectangular prisms, cylinders, cones, spheres, pyramids, and composite shapes made from two or more of these solids. Correct identification is crucial because each solid has its own volume formula.

2. Note the Given Dimensions Write down every measurement provided in the figure—radius, height, length, width, side length, slant height, etc. If a dimension is missing but can be derived (for example, using the Pythagorean theorem), calculate it before proceeding.

3. Choose the Appropriate Formula

Match the identified shape to its standard volume equation. Below are the most frequently used formulas:

• Cube: ( V = s^{3} ) where s is the side length.
• Rectangular Prism: ( V = l \times w \times h ).
• Cylinder: ( V = \pi r^{2} h ).
• Cone: ( V = \frac{1}{3} \pi r^{2} h ).
• Sphere: ( V = \frac{4}{3} \pi r^{3} ). - Pyramid: ( V = \frac{1}{3} B h ) where B is the area of the base.

For composite figures, compute the volume of each part separately and then add or subtract as needed.

4. Plug in the Numbers and Calculate

Insert the measured values into the formula, respecting units. Perform the arithmetic carefully, keeping track of powers and constants like π (pi). If the problem asks for an approximate value, use 3.14 or 22/7 for π; otherwise, leave the answer in terms of π for exactness.

5. State the Answer with Proper Units Volume is expressed in cubic units (e.g., cm³, m³, in³). Always attach the unit to your final number to avoid ambiguity.

6. Verify Reasonableness

Quickly check whether the result makes sense. A tiny cube with 2‑cm sides should not yield a volume of 500 cm³. If the answer seems off, revisit each step for possible mistakes.

Scientific Explanation: The Mathematics Behind Volume Formulas

Understanding why these formulas work deepens comprehension and helps you adapt them to irregular shapes.

Derivation from Basic Principles

Volume can be thought of as the sum of infinitely thin slices. For a prism, stacking identical cross‑sections of area A along a height h yields ( V = A \times h ). A rectangular prism’s base area is ( l \times w ), giving the familiar product of three dimensions.

For solids with curved surfaces, calculus provides the derivation. Consider a cylinder: each horizontal slice is a circle of area ( \pi r^{2} ). Integrating this area from the bottom to the top (height h) results in ( V = \pi r^{2} h ). A cone is similar, except the radius of each slice shrinks linearly from r at the base to 0 at the tip, leading to the factor ( \frac{1}{3} ).

The sphere’s formula emerges from integrating circular slices whose radii vary according to the Pythagorean theorem: ( r_{\text{slice}} = \sqrt{R^{2} - y^{2}} ). Performing the integral yields ( V = \frac{4}{3} \pi R^{3} ).

Units and Dimensional Consistency

Because volume involves three length measurements, its unit is always a length cubed. This dimensional check helps catch errors: if you end up with a unit like cm², you likely omitted a dimension.

Composite and Irregular Shapes

When a figure combines multiple simple solids, the principle of additivity applies: the total volume equals the sum of the constituent volumes (if they do not overlap). If a shape contains a hollow part, subtract the volume of the void from the outer solid’s volume.

Worked Examples: Applying the Steps

Below are three illustrative problems that demonstrate the process for different types of figures.

Example 1: Simple Rectangular Prism

A box measures 5 cm in length, 3 cm in width, and 4 cm in height. Find its volume.

1. Shape: rectangular prism.
2. Dimensions: ( l = 5 \text{ cm}, w = 3 \text{ cm}, h = 4 \text{ cm} ).
3. Formula: ( V = l \times w \times h ).
4. Calculation: ( V = 5 \times 3 \times 4 = 60 ).
5. Answer: ( 60 \text{ cm}^{3} ).

The result is reasonable: a modest‑sized box holds sixty cubic centimeters.

Example 2: Cylinder with Given Diameter

A cylindrical can has a diameter of 10 cm and a height of 15 cm. Determine its volume.

1. Shape: cylinder.

Example 2 (Continued): Cylinder with Given Diameter

1. Shape: cylinder.
2. Dimensions: Diameter = 10 cm → Radius ( r = \frac{10}{2} = 5 ) cm, Height ( h = 15 ) cm.
3. Formula: ( V = \pi r^{2} h ).
4. Calculation: ( V = \pi \times (5)^{2} \times 15 = \pi \times 25 \times 15 = 375\pi ).
   Using ( \pi \approx 3.1416 ): ( V \approx 375 \times 3.1416 \approx 1178.1 ).
5. Answer: ( 375\pi ) cm³ (exact) or approximately ( 1178 \text{ cm}^{3} ).
   Check: Units are cm³. Volume seems plausible for a can of this size.

Example 3: Composite Shape (Cylinder with Conical Depression)

A solid consists of a cylinder with a conical hole bored through its center. The cylinder has a radius of 4 cm and a height of 10 cm. The conical hole has the same radius (4 cm) and a depth of 6 cm. Find the volume of the remaining solid.

1. Shape: Composite (cylinder minus cone).
2. Dimensions:
   • Cylinder: ( r_c = 4 ) cm, ( h_c = 10 ) cm.
   • Cone (depression): ( r_k = 4 ) cm, ( h_k = 6 ) cm.
3. Formulas:
   • Cylinder Volume: ( V_{\text{cyl}} = \pi r_{c}^{2} h_{c} ).
   • Cone Volume: ( V_{\text{cone}} = \frac{1}{3} \pi r_{k}^{2} h_{k} ).
   • Solid Volume: ( V_{\text{solid}} = V_{\text{cyl}} - V_{\text{cone}} ).
4. Calculations:
   • ( V_{\text{cyl}} = \pi \times (4)^{2} \times 10 = \pi \times 16 \times 10 = 160\pi ) cm³.
   • ( V_{\text{cone}} = \frac{1}{3} \pi \times (4)^{2} \times 6 = \frac{1}{3} \pi \times 16 \times 6 = \frac{1}{3} \times 96\pi = 32\pi ) cm³.
   • ( V_{\text{solid}} = 160\pi - 32\pi = 128\pi ) cm³.
5. Answer: ( 128\pi ) cm³ (exact) or approximately ( 402.1 \text{ cm}^{3} ) (using ( \pi \approx 3.1416 )).
   Check: Units are cm³. The result is less than the cylinder's volume (160π ≈ 502.7 cm³), which makes sense due to the removed cone.

Conclusion

Mastering volume calculations hinges on a systematic approach: accurately identify the shape's geometry, select the appropriate formula based on its defining dimensions, perform the calculation meticulously, and always verify the result for dimensional consistency and practical reasonableness. Understanding the underlying mathematical principles—such as the summation of infinitesimal slices for prisms and the integration of varying cross-sections for curved solids—provides a robust foundation for applying these formulas confidently. Whether dealing with simple geometric forms or complex composite structures, the core strategy remains consistent: decompose, calculate, and combine. This methodology not only solves textbook problems but also equips learners with

the analytical tools to tackle real-world challenges in engineering, architecture, and beyond. By internalizing these steps and practicing with diverse examples, one develops both computational fluency and spatial reasoning, transforming abstract formulas into powerful instruments for quantifying three-dimensional space. The journey from recognizing a shape to confidently stating its volume encapsulates the essence of mathematical problem-solving: a blend of logical reasoning, precise execution, and critical verification that ensures accuracy and deepens understanding.