Calculating Cross-Sectional Area Of A Sphere Intersected By A Plane
Introduction
In the realm of three-dimensional geometry, understanding the intersection of a plane and a sphere is a fundamental concept. This article delves into the scenario where a plane intersects a sphere, specifically focusing on the case where the plane passes through the center of the sphere. We will explore how to determine the area of the resulting cross-section, a concept crucial in various fields, including engineering, physics, and computer graphics. Our main focus will be on a sphere with a volume of approximately 33.5 cubic meters, and we aim to calculate the cross-sectional area to the nearest tenth. This exploration will involve understanding the relationship between the sphere's volume, its radius, and the geometry of the intersection.
Delving into the sphere's properties is crucial for solving this problem. The volume of a sphere is given by the formula V = (4/3)πr³, where 'V' is the volume and 'r' is the radius. Knowing the volume allows us to calculate the radius, a fundamental parameter for determining the cross-sectional area. When a plane intersects a sphere, the resulting cross-section is a circle. If the plane passes through the center of the sphere, this circle is a great circle, which has the same radius as the sphere itself. This geometric property simplifies our calculations significantly. The area of a circle is given by A = πr², where 'A' is the area and 'r' is the radius. By finding the sphere's radius from its volume, we can directly calculate the area of the cross-section. This article will guide you through the steps, ensuring a clear understanding of the underlying concepts and the practical application of the formulas involved. Understanding the cross-sectional area not only enhances our knowledge of geometry but also provides a foundation for solving more complex problems in related fields.
Calculating the Radius of the Sphere
To determine the area of the cross-section, we first need to calculate the radius of the sphere. Given that the volume of the sphere is approximately 33.5 cubic meters, we can use the formula for the volume of a sphere, V = (4/3)πr³, to find the radius. This involves algebraic manipulation and understanding the relationship between volume and radius. By isolating 'r' in the formula, we can express the radius in terms of the volume and the constant π. The accuracy of our final result depends on the precision with which we calculate the radius, highlighting the importance of careful calculation and the use of appropriate mathematical constants. This step is crucial because the radius is the foundation for calculating the cross-sectional area, and any error in this step will propagate through the rest of the solution. This section provides a detailed breakdown of the calculation process, ensuring clarity and accuracy in determining the sphere's radius.
Starting with the formula V = (4/3)πr³, we substitute the given volume, 33.5 cubic meters, into the equation. This gives us 33.5 = (4/3)πr³. To isolate r³, we multiply both sides of the equation by 3/4, resulting in (3/4) * 33.5 = πr³. Further simplification leads to 25.125 = πr³. Next, we divide both sides by π to get r³ = 25.125 / π. Using a calculator, we find that 25.125 / π is approximately 7.997. Therefore, r³ ≈ 7.997. To find the radius 'r', we need to take the cube root of 7.997. The cube root of 7.997 is approximately 1.99975. Rounding this to a more practical value, we get r ≈ 2 meters. It’s important to recognize the units; since the volume is in cubic meters, the radius is in meters. This radius value is crucial for the next step, where we calculate the cross-sectional area. The process of finding the cube root might seem complex, but it’s a fundamental mathematical operation when dealing with volumes and radii of spheres. This calculation demonstrates the inverse relationship between volume and radius, and how a seemingly small change in volume can affect the radius. Understanding this relationship is crucial for solving various problems in geometry and physics. The final radius value, approximately 2 meters, will be used in the subsequent calculation to find the cross-sectional area of the sphere.
Determining the Cross-Sectional Area
Once we have the radius of the sphere, we can calculate the area of the cross-section. Since the plane intersects the sphere through its center, the cross-section is a circle with the same radius as the sphere. The formula for the area of a circle is A = πr², where 'A' is the area and 'r' is the radius. By substituting the calculated radius into this formula, we can find the area of the cross-section. This step involves a straightforward application of the formula, but it is essential to ensure the correct units are used and the final answer is rounded to the nearest tenth as required. This section will detail the substitution and calculation process, providing a clear understanding of how the cross-sectional area is derived from the sphere's radius. The result will give us the area in square meters, reflecting the two-dimensional nature of the cross-section.
Now that we have the radius, r ≈ 2 meters, we can proceed to calculate the cross-sectional area. The formula for the area of a circle is A = πr². Substituting the value of the radius into this formula, we get A = π * (2)² = π * 4. Therefore, A = 4π square meters. To find the numerical value, we multiply 4 by π (approximately 3.14159), which gives us A ≈ 12.56636 square meters. The problem requires us to round the answer to the nearest tenth. Rounding 12.56636 to the nearest tenth gives us 12.6 square meters. Therefore, the area of the cross-section is approximately 12.6 square meters. It is important to include the units in the final answer to provide a complete and accurate solution. This calculation demonstrates how a simple geometric formula can be used to solve a practical problem involving three-dimensional shapes. The precision of the final answer depends on the accuracy of the initial radius calculation and the rounding process. This result provides a clear understanding of the size of the circular cross-section created when a plane intersects the sphere through its center. The cross-sectional area is a fundamental property that can be used in various applications, such as calculating flow rates through pipes or determining the amount of material needed to cover a spherical surface.
Conclusion
In conclusion, we have successfully determined the area of the cross-section formed when a plane intersects a sphere with a volume of approximately 33.5 cubic meters through its center. By first calculating the radius of the sphere using the volume formula and then applying the formula for the area of a circle, we found the cross-sectional area to be approximately 12.6 square meters. This exercise demonstrates the interconnectedness of geometric concepts and the importance of understanding fundamental formulas. The ability to calculate cross-sectional areas is crucial in various scientific and engineering applications, highlighting the practical relevance of this geometric problem. This article has provided a step-by-step guide, ensuring a clear understanding of the process and the underlying principles.
Throughout this article, we emphasized the importance of accuracy in calculations and the correct application of formulas. The process involved several steps, each building upon the previous one. First, we used the volume of the sphere to calculate its radius. This step required understanding the formula V = (4/3)πr³ and performing algebraic manipulations to isolate the radius 'r'. We then substituted the given volume into the formula and solved for 'r', which we found to be approximately 2 meters. Next, we used the calculated radius to find the cross-sectional area. Since the plane intersects the sphere through its center, the cross-section is a circle with the same radius as the sphere. We applied the formula for the area of a circle, A = πr², substituting the value of the radius to find the area. This calculation yielded an area of approximately 12.6 square meters when rounded to the nearest tenth. This entire process underscores the importance of meticulous calculations and a strong understanding of geometric principles. Understanding the relationships between different geometric properties, such as volume, radius, and area, is essential for solving a wide range of problems. The final result, 12.6 square meters, provides a concrete measure of the size of the cross-section, illustrating the practical application of these concepts. This exercise not only enhances our knowledge of geometry but also reinforces the problem-solving skills necessary in various fields.
