Problem 19 - Advanced Engineering Mathematics

Question

Evaluate the surface integral \(\iint_{S} G(x, y, z) d S\). \(G(x, y, z)=\left(x^{2}+y^{2}\right) z ; S\) that portion of the sphere \(x^{2}+y^{2}+z^{2}=36\) in the first octant

Step-by-Step Solution

Answer

The surface integral evaluates to \(162\pi\).

Step 1: Understand the Surface

First, let's identify the surface over which we are integrating. The given surface is part of the sphere defined by the equation \(x^2 + y^2 + z^2 = 36\) in the first octant. This means we will only consider the portion of the sphere where \(x \geq 0\), \(y \geq 0\), and \(z \geq 0\).

Step 2: Parameterize the Surface

In order to switch the integral to a more solvable form, convert the portion of the sphere in Cartesian coordinates to spherical coordinates. Use the parameterization:\[x = 6 \sin \phi \cos \theta,\quad y = 6 \sin \phi \sin \theta,\quad z = 6 \cos \phi,\]where \(0 \leq \theta \leq \frac{\pi}{2}\) and \(0 \leq \phi \leq \frac{\pi}{2}\).

Step 3: Write the Surface Integral in Spherical Coordinates

Transform \(dS\) to spherical coordinates using the formula for the sphere's metric: \(dS = \left|\mathbf{r}_\phi \times \mathbf{r}_\theta \right| d\phi d\theta\). This results in:\[dS = 36 \sin \phi \, d\phi \, d\theta.\]The function \(G(x, y, z)\) becomes \((6^2 \sin^2 \phi) (6 \cos \phi)\). Substitute these into the integral:\[\iint_{S} G(x, y, z) dS = \int_{0}^{\pi/2} \int_{0}^{\pi/2} (36 \sin^2 \phi \cdot 6 \cos \phi) (36 \sin \phi) \, d\phi \, d\theta.\]

Step 4: Simplify and Integrate

Simplify the integral:\[\int_{0}^{\pi/2} \int_{0}^{\pi/2} 216 \cdot 36 \sin^3 \phi \cos \phi \, d\phi \, d\theta.\]Solve the inner integral with respect to \(\phi\):\[\int_{0}^{\pi/2} \sin^3 \phi \cos \phi \, d\phi = \frac{1}{4}, \]using the substitution \(u = \sin \phi\), \(du = \cos \phi \, d\phi\). Then, the full integral becomes:\[216 \cdot 36 \cdot \frac{1}{4} \int_{0}^{\pi/2} d\theta = 324 \cdot \frac{\pi}{2}.\]

Step 5: Calculate the Final Result

Compute the final result for the integral:\[162 \pi.\]

Key Concepts

Spherical coordinates

Spherical coordinates are a three-dimensional coordinate system that extends polar coordinates by introducing a height component. In these coordinates, we represent a point in space with three values: radial distance \(r\), polar angle \(\theta\), and azimuthal angle \(\phi\).
Spherical coordinates are particularly useful for dealing with problems involving spheres or spherical geometry which can simplify the problem considerably.
This is because they align more naturally to the symmetry of a sphere as compared to Cartesian coordinates.

• Radial distance \(r\): The distance from the origin to the point.
• Polar angle \(\theta\): The angle made with the positive \(z\)-axis.
• Azimuthal angle \(\phi\): This measures rotation around the \(z\)-axis, much like the polar angle in two-dimensional polar coordinates.

For a sphere of radius \(6\), and given the symmetries of the problem, we use:\[x = 6 \sin \phi \cos \theta, \, y = 6 \sin \phi \sin \theta, \, z = 6 \cos \phi\]where \(0 \leq \theta \leq \pi/2\) and \(0 \leq \phi \leq \pi/2\). These equations come from transforming the standard coordinates to a spherical form, matching the sphere's geometry.

Sphere in the first octant

The first octant in three-dimensional space is where all three Cartesian coordinates \(x, y, z \) are non-negative. For our sphere \(x^2 + y^2 + z^2 = 36\), the section lies within this octant. It represents just one-eighth of the entire sphere.
The sphere’s constraints are naturally set by the intersection with the planes \(x=0, y=0, z=0\). Spheres in spherical coordinates are often more straightforward, letting us succinctly describe sections of it, such as this octant portion.
To understand the section of the sphere:

• Imagine the full sphere centered at the origin with radius \(r=6\).
• The first octant focuses only on positive coordinates: \(x \geq 0, y \geq 0, z \geq 0\).
• This restriction simplifies the problem by reducing the integration region.

Spheres in the octants allow you to work with manageable, simpler parts of three-dimensional objects, significantly easing the computational process.

Parameterization of surfaces

Parameterizing a surface is a method to express the positions on it as functions of two variables. It's akin to describing a curve by its \(x\) and \(y\) coordinates in terms of a single parameter, but for surfaces, you need two parameters.
For a sphere, like in our problem, spherical coordinates \(\phi\) and \(\theta\) effectively parameterize the x, y, and z positions on a spherical surface.
Through parameterization, complex surfaces can become easier to handle mathematically, especially when converting into integral form for calculations.
Consider:

• Using parameters, you describe the surface in terms of easily manageable functions.
• This kind of description benefits from the natural symmetry of spheres, often simplifying calculations and expressions for integrals.

In our surface, the parameters are \(\phi \) and \(\theta\), where \(\phi\) and \(\theta\) operate within defined limits to ensure computation over just the first octant.

Double integrals

Double integrals are a way to compute the volume under a surface in three-dimensional space. By integrating a function of two variables over a region, they extend the concept of regular integration from one dimension to two dimensions.
For surfaces like our sphere, double integrals offer a way to sum all infinitesimally small pieces of the surface area.
Here’s how it helps in evaluating a surface integral:

• It transforms integration into evaluating formulas over a defined parameter range.
• Double integrals take advantage of the symmetry and parameterization of surfaces, especially if converted to cylindrical or spherical coordinates.

In the context of our exercise, having converted into spherical coordinates, the sphere's surface integral is turned into a double integral over angles \(\theta\) and \(\phi\),
where boundaries are \( 0 \leq \theta, \phi \leq \pi/2\). This process simplifies evaluating integrals by tackling them in terms of these angles.