Question

Find the mass of each thin sheet. \(\mathcal{S}\) is the unit sphere, with \(\delta(x, y, z)=x+y+z+10\).

Short answer

Mass of the sheet is 40\(\pi\) after solving the integrals.

Step-by-step solution

Understand the Problem

We are tasked with finding the mass of a thin sheet represented by the unit sphere \(\mathcal{S}\) and density function \(\delta(x, y, z) = x + y + z + 10\). The unit sphere means we are working within the space where \(x^2 + y^2 + z^2 = 1\).

Set Up the Integral for Mass

The mass of the sheet can be found by integrating the density function over the surface of the sphere. The mass is given by \(M = \iint_{\mathcal{S}} \delta(x, y, z) \, dS\), where \(dS\) is the surface area element.

Choose a Parameterization for the Surface

A common parameterization for a unit sphere is using spherical coordinates: \(x = \sin \theta \cos \phi\), \(y = \sin \theta \sin \phi\), \(z = \cos \theta\). Here, \(0 \leq \phi < 2\pi\) and \(0 \leq \theta \leq \pi\).

Compute the Surface Element \(dS\)

For a parameterized surface \(\mathbf{r}(u,v)\), \(dS = \|\mathbf{r}_u \times \mathbf{r}_v\| \, dudv\). For the unit sphere in spherical coordinates, \(dS = \sin \theta \, d\theta \, d\phi\).

Rewrite the Density in Terms of Parameters

Substitute \(x = \sin \theta \cos \phi\), \(y = \sin \theta \sin \phi\), \(z = \cos \theta\) into \(\delta(x, y, z)\) to get \(\delta(\theta, \phi) = \sin \theta \cos \phi + \sin \theta \sin \phi + \cos \theta + 10\).

Evaluate the Integral for Mass

The mass \(M\) is found by evaluating \[M = \int_0^{2\pi} \int_0^{\pi} (\sin \theta \cos \phi + \sin \theta \sin \phi + \cos \theta + 10) \sin \theta \, d\theta \, d\phi.\] Simplify and calculate this integral accordingly.

Compute the Double Integral Applying Calculus

Breaking it down, we have a few parts to compute:\[\int_0^{2\pi} \int_0^{\pi} (\sin \theta \cos \phi \sin \theta + \sin \theta \sin \phi \sin \theta + \cos \theta \sin \theta + 10\sin \theta) \, d\theta \, d\phi\]. Due to symmetry, the integral over terms with \(\cos \phi\) and \(\sin \phi\) vanish. Compute separately and add contributions.

Simplify and Solve the Integral

Integrate each part separately, specifically focusing on: \[\int_0^{2\pi} d\phi = 2\pi\]\[\int_0^{\pi} \cos \theta \sin \theta \, d\theta = \frac{1}{2}\]\[\int_0^{\pi} 10 \sin \theta \, d\theta = 20\]. Sum these results to get the mass.

Key concepts

Unit Sphere

A unit sphere is a familiar concept in geometry and is defined as the set of points equidistant from a fixed central point with a radius of one unit. In mathematical terms, a unit sphere in three-dimensional space is described by the equation:

• \[ x^2 + y^2 + z^2 = 1 \]

This equation implies that any point \( (x, y, z) \) on the sphere must satisfy this condition, meaning it lies exactly one unit away from the origin.
Unit spheres are often used in mathematical exercises because of their symmetrical properties, which can simplify surface integrals, like the one needed in this exercise. Utilizing the concept of the unit sphere helps to build an intuitive understanding of spherical surfaces when parameters are introduced, as it offers a straightforward framework for converting complex three-dimensional shapes into simpler, more manageable forms.

Density Function

The density function in this problem, \( \delta(x, y, z) = x + y + z + 10 \), represents how the mass is spread across the surface of the unit sphere. Imagine laying a thin sheet on the sphere's surface, where every single point on this sheet might have a different mass concentration.
Density functions often help us understand how material or mass is distributed over a surface or throughout a volume. In this case, the function is linear in terms of the coordinates, which suggests a simple, consistent density variation across the surface. The constant term '+10' ensures that the density is consistently positive, reflecting a uniform minimum density over the entire sphere, with additional variations based on the \(x, y, z\) coordinates.
Using a density function in calculations enables us to quantify the mass when combined with surface area elements in an integral, thus forming the basis of the mass calculation process on unit spheres and other surfaces.

Spherical Coordinates

Spherical coordinates are a system of curvilinear coordinates that are naturally suited to problems involving spheres, thanks to their inherent symmetry. For a point represented as \( (x, y, z) \), the transformation into spherical coordinates is given as:

• \[ x = \sin \theta \cos \phi \]
• \[ y = \sin \theta \sin \phi \]
• \[ z = \cos \theta \]

Here, \( \theta \) and \( \phi \) are angles, with \( 0 \leq \theta \leq \pi \) and \( 0 \leq \phi < 2\pi \). These coordinates elegantly describe positions on a sphere by angle measures rather than Cartesian coordinates, minimizing the complexity involved in describing curved surfaces.
Applying spherical coordinates converts the calculation over the unit sphere into an integral over these parameters. The integration bounds follow directly from the definitions of \( \theta \) and \( \phi \,\), making them exact for a full sphere.
This system underscores how surface integrals, like the mass calculation above, are greatly simplified by using spherical coordinates, as they inherently align with the geometry of the sphere.

Mass Calculation

In calculating the mass of an object with variable density distributed over a surface like a unit sphere, we rely on the concept of surface integrals. Here, the total mass \( M \) is given as:

• \[ M = \iint_\mathcal{S} \delta(x, y, z) \, dS \]

This integral sums the contributions of the density function \( \delta(x, y, z) \) over each differential piece of the surface area \( dS \).
After substituting the density function into spherical coordinates, we arrive at a double integral over \( \theta \) and \( \phi \), namely:

• \[ M = \int_0^{2\pi} \int_0^{\pi} (\sin \theta \cos \phi + \sin \theta \sin \phi + \cos \theta + 10) \sin \theta \, d\theta \, d\phi \]

Breaking this integral down allows each part to be computed and simplified separately. Many terms involving trigonometric angles balance each other out due to the symmetrical nature of the sphere, simplifying calculations.
In the end, these adjustments and computations yield the mass of the sphere based on its given density distribution.