Question

Let \(C\) denote the set \(\left\\{(x, y, z): x^{2}+y^{2}+z^{2} \leq 1\right\\}\). Using spherical coordinates, evaluate $$ Q(C)=\iiint_{C} \sqrt{x^{2}+y^{2}+z^{2}} d x d y d z $$

Short answer

The answer to this problem requires performing multiple integration operations, therefore a quick short answer can't be provided without going through all the steps.

Step-by-step solution

Convert to spherical coordinates

In spherical coordinates, \(x^{2}+y^{2}+z^{2} = r^{2}\) and \(dx\,dy\,dz = r^{2}\sin(\phi)\,dr\,d\theta\,d\phi\). Thus\[Q(C)=\iiint_{C} \sqrt{x^{2}+y^{2}+z^{2}} \,dx\,dy\,dz = \iiint_{C} r\,r^{2}\sin(\phi)\,dr\,d\theta\,d\phi\].

Set the limits of integration

For a sphere with radius 1, the limits are: \(r\) ranges from 0 to 1, \(\theta\) ranges from 0 to \(2\pi\), and \(\phi\) ranges from 0 to \(\pi\). Substitute those to the converted integral.

Compute the triple integral

The integral becomes separable, hence can be written as a product of three simpler integrals. Compute each of those simpler integrals. After the computations, multiply the three results together to get the final answer.

Simplify the answer

Simplify the answer, which should be a numerical value.

Key concepts

Spherical Coordinates

Understanding spherical coordinates is crucial for solving many problems in mathematics, particularly those involving three-dimensional spaces. Unlike Cartesian coordinates, which use x, y, and z to locate a point in space, spherical coordinates describe points with three different values: the radius r, the polar angle \(\theta\), and the azimuthal angle \(\theta\).

• r represents the distance from the origin to the point.
• \(\phi\), ranges between 0 and \(\pi\), denotes the angle between the positive z-axis and the line connecting the origin to the point.
• \(\theta\), which goes from 0 to \(2\pi\), is the angle between the positive x-axis and the projection of the point onto the xy-plane.

With spherical coordinates, computing the volume of complex shapes becomes much simpler. To visualize, think of Earth with its longitudinal and latitudinal lines; any point can be pinpointed with a form of spherical coordinate system.

Volume of a Sphere

The volume of a sphere is a fundamental concept in geometry, frequently illustrated using spherical coordinates. The formula for the volume of a sphere is given by \(V = \frac{4}{3}\pi r^3\), where r is the radius of the sphere.

When we express this in the language of calculus, we compute the volume by integrating over a three-dimensional region. In spherical coordinates, this involves an integral with respect to r, \(\theta\), and \(\phi\) over the sphere's dimensions.

It's crucial to recognize that, in particular scenarios, the function we integrate is simple: just a geometrical shape's density or other properties that depend on the point's location in space. For a sphere of uniform density, the integrand can be considered as the radius itself, which simplifies the computation significantly.

Multiple Integration

Multiple integration is an extension of single-variable integration to two or more variables. When dealing with the volume of three-dimensional objects or calculating properties like mass, charge, or density that are distributed throughout a space, we use triple integrals.

Triple integrals, especially in spherical coordinates, break down into three consecutive integrations, each over one variable while keeping the others constant. This stratification allows for a step-by-step approach to solving complex problems.

Improving the Triple Integral Solution

When solving triple integrals, it's essential to:

• Accurately convert Cartesian coordinates to spherical coordinates.
• Determine the appropriate limits of integration, which rely on the geometry of the object.
• Recognize symmetries to simplify the integral, if possible.
• Do the integrals sequentially and simplify at each step.
• Interpret the final solution within the context of the problem, confirming that it makes physical or geometrical sense.

This disciplined approach ensures that the student fully comprehends the solution to the triple integration problem, making the concept easier to grasp and apply to new problems.