Chapter 9: Problem 37
In Exercises \(37-40,\) describe the solid satisfying the condition. \(x^{2}+y^{2}+z^{2} \leq 36\)
Short Answer
Expert verified
The solid which satisfies the condition is a sphere with a radius of 6, centered at the origin (0,0,0), including its interior.
Step by step solution
01
Identify the main equation
The condition given is \(x^{2}+y^{2}+z^{2} \leq 36\). This equation is similar to the standard equation for a sphere, which is \(x^{2}+y^{2}+z^{2} = r^{2}\).
02
Identify the radius of the sphere
For the standard equation, \(r^{2}\) stands for the square of the radius of the sphere. In our equation, this is equivalent to 36. Therefore, you can say that the radius of the sphere, \(r\), is the square root of 36, which is 6.
03
Identify the solid satisfying the condition
Since our equation includes the \( \leq \) notation, it doesn't just represent the surface of the sphere. It represents the volume of the sphere, which includes all the points within and on the boundary of the sphere. Therefore, the solid satisfying this equation is a sphere with radius 6 at the origin (0,0,0), including all points inside the sphere.
04
Final description of the solid
In conclusion, the solid that satisfies the condition \(x^{2}+y^{2}+z^{2} \leq 36\) is a sphere with radius 6 centered at the origin (0,0,0), including its interior.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Solid Geometry and the Sphere
Solid geometry is the branch of mathematics that deals with three-dimensional shapes and their properties. A sphere, a classic example of a solid in geometry, is a round, symmetrical object where every point on its surface is equidistant from the center. The equation given in the exercise, \(x^{2}+y^{2}+z^{2} \leq 36\), represents not just the sphere's surface, but also the solid volume inside. In solid geometry, understanding equations like this one is crucial to visualizing and manipulating the shapes in three-dimensional space.
To better comprehend this, imagine the sphere as a collection of infinitesimal points that are all at a distance of 6 or less from a central point (the origin). This conceptualization helps in grasping the inclusiveness of the \(\leq\) notation in the equation, hinting towards the entirety of the space inside the sphere's boundary. Such understanding is fundamental in solid geometry as it lays the groundwork for further exploration into volumes, surface areas, and intersections with other solids.
To better comprehend this, imagine the sphere as a collection of infinitesimal points that are all at a distance of 6 or less from a central point (the origin). This conceptualization helps in grasping the inclusiveness of the \(\leq\) notation in the equation, hinting towards the entirety of the space inside the sphere's boundary. Such understanding is fundamental in solid geometry as it lays the groundwork for further exploration into volumes, surface areas, and intersections with other solids.
Spherical Coordinates
Spherical coordinates offer a different approach to describe points in three-dimensional space, which is especially useful when dealing with spherical objects. This coordinate system uses three parameters: the radius (\(r\)), the polar angle (\(\theta\)), and the azimuthal angle (\(\phi\)). Here, the radius extends from the origin to the point, the polar angle is measured from the positive z-axis, and the azimuthal angle is taken in the xy-plane from the positive x-axis.
For example, the equation from our exercise can be expressed in spherical coordinates as \(r \leq 6\), which simplifies the understanding of points within or on the sphere. This is because every point within the sphere will have a radius that is less than or equal to 6, irrespective of the angles \(\theta\) and \(\phi\). Spherical coordinates make calculations involving spheres more straightforward and are a powerful tool in fields like physics and engineering, where spherical symmetry often occurs.
For example, the equation from our exercise can be expressed in spherical coordinates as \(r \leq 6\), which simplifies the understanding of points within or on the sphere. This is because every point within the sphere will have a radius that is less than or equal to 6, irrespective of the angles \(\theta\) and \(\phi\). Spherical coordinates make calculations involving spheres more straightforward and are a powerful tool in fields like physics and engineering, where spherical symmetry often occurs.
Volume of a Sphere
The volume of a sphere is a fundamental concept in geometry, and it can be derived from the sphere's radius. The formula for the volume of a sphere is given by \(\frac{4}{3}\pi r^{3}\), where \(r\) is the radius of the sphere. Applying this to our exercise, with a given radius of 6, we find the volume by computing \(\frac{4}{3}\pi \times 6^{3}\), which equals \(\frac{4}{3}\pi \times 216\) or \(288\pi\) cubic units.
Understanding how to derive and apply the volume formula is crucial in many practical situations, such as calculating the capacity of spherical tanks or measuring the displacement of a sphere submerged in a fluid. These applications are not only significant in academic settings but also in various real-world scenarios across different industries and fields of research.
Understanding how to derive and apply the volume formula is crucial in many practical situations, such as calculating the capacity of spherical tanks or measuring the displacement of a sphere submerged in a fluid. These applications are not only significant in academic settings but also in various real-world scenarios across different industries and fields of research.