Problem 4 - Thomas Calculus

Question

Integrate the given function over the given surface. \begin{equation}\begin{array}{l}{\text { Hemisphere } G(x, y, z)=z^{2}, \text { over the hemisphere } x^{2}+y^{2}+} \\ {z^{2}=a^{2}, z \geq 0}\end{array}\end{equation}

Step-by-Step Solution

Answer

The integral evaluates to \( \frac{2}{3} \pi a^4 \).

Step 1: Define the Hemisphere Surface

The surface is defined by the hemisphere: \( x^2 + y^2 + z^2 = a^2, z \geq 0 \). This is the upper half of the sphere with radius \( a \) centered at the origin.

Step 2: Set Up the Surface Integral

The surface integral of a function \( G(x, y, z) \) over a surface \( S \) is given by \( \iint_{S} G(x, y, z) \, dS \). For this problem, \( G(x, y, z) = z^2 \), and the surface \( S \) is the hemisphere.

Step 3: Convert to Spherical Coordinates

For a sphere, converting to spherical coordinates simplifies the calculation. Use the transformations \( x = a \sin \theta \cos \phi \), \( y = a \sin \theta \sin \phi \), \( z = a \cos \theta \), where \( \theta \) ranges from \( 0 \) to \( \frac{\pi}{2} \) and \( \phi \) from \( 0 \) to \( 2\pi \).

Step 4: Calculate the Jacobian

In spherical coordinates, the area element \( dS \) on a sphere of radius \( a \) is given by \( a^2 \sin \theta \, d\theta \, d\phi \). This accounts for the change in area when converting coordinates.

Step 5: Substitute and Integrate

Substitute \( z = a \cos \theta \) and \( dS = a^2 \sin \theta \, d\theta \, d\phi \) into the integral. The integral becomes \( \iint_{S} (a \cos \theta)^2 \cdot a^2 \sin \theta \, d\theta \, d\phi \). Simplify it to \( a^4 \iint_{S} \cos^2 \theta \sin \theta \, d\theta \, d\phi \).

Step 6: Separate the Variables

The integral separates into the product of two integrals: \( a^4 \left( \int_0^{2\pi} \, d\phi \right) \left( \int_0^{\pi/2} \cos^2 \theta \sin \theta \, d\theta \right) \). The first integral, \( \int_0^{2\pi} \, d\phi \), is simply \( 2\pi \).

Step 7: Evaluate the Theta Integral

Evaluate \( \int_0^{\pi/2} \cos^2 \theta \sin \theta \, d\theta \) by substitution. Let \( u = \cos \theta \), then \( du = -\sin \theta \, d\theta \). Change in limits: \( u(\pi/2) = 0 \) and \( u(0) = 1 \). The integral becomes \( -\int_1^0 u^2 \, du = \int_0^1 u^2 \, du = \frac{1}{3} \).

Step 8: Combine the Results

Combine the results of the integrals: \( a^4 (2\pi)(\frac{1}{3}) = \frac{2}{3} \pi a^4 \). Thus, the surface integral of \( z^2 \) over the hemisphere is \( \frac{2}{3} \pi a^4 \).

Key Concepts

Spherical Coordinates

When dealing with surfaces like a sphere or hemisphere, spherical coordinates make calculations easier and more intuitive. In spherical coordinates, any point in space is expressed using three parameters:

• \( \rho \): the radial distance from the origin to the point. However, when dealing specifically with surfaces like a sphere or hemisphere, \( \rho \) becomes constant, equivalent to the radius \( a \).
• \( \theta \): the polar angle measured from the positive z-axis. It controls the vertical distance from the xy-plane. For the upper hemisphere, \( \theta \) ranges from 0 to \( \frac{\pi}{2} \) because \( z \geq 0 \).
• \( \phi \): the azimuthal angle in the xy-plane measured from the positive x-axis. \( \phi \) ranges from 0 to \( 2\pi \), allowing for full rotation around the z-axis.

In this context, the conversions used are:

• \( x = a \sin \theta \cos \phi \)
• \( y = a \sin \theta \sin \phi \)
• \( z = a \cos \theta \)

These transformations simplify integrating over spherical surfaces greatly.

Hemisphere Integration

Integrating a function over a hemisphere involves understanding the top half of a sphere. The challenge lies in setting up the integral correctly over this curved surface.The hemisphere in this problem is defined by the equation \( x^2 + y^2 + z^2 = a^2 \) with \( z \geq 0 \). This is the portion of the sphere where all points have non-negative z-coordinates, representing its upper half. When setting up the integral for a function like \( z^2 \), use spherical coordinates. This simplifies the elements of the integral into manageable components. Specifically:

• The integral begins by replacing \( z \) with \( a \cos \theta \).
• The differential area element \( dS \) on the hemisphere becomes \( a^2 \sin \theta \, d\theta \, d\phi \).
• The integral over the hemisphere becomes a product of multiple integrals that account for the entire surface area, considering both \( \theta \) and \( \phi \).

This ensures that the full range of the hemisphere is covered. It captures the surface's curvature accurately by tracing out each differential area piece.

Jacobian Determinant

When changing variables in integration, particularly converting from Cartesian to spherical coordinates, the Jacobian determinant plays a crucial role. It adjusts for the fact that the shape and size of area elements change under the transformation.In spherical coordinates for a sphere of radius \( a \), the area element \( dS \) doesn’t directly correspond to the simple dx, dy of Cartesian coordinates. Instead, it becomes \( a^2 \sin \theta \, d\theta \, d\phi \).This transformation involves these steps:

• Identify how each Cartesian differential (\( dx, dy, dz \)) is expressed in spherical terms.
• Recognize that the spherical coordinates introduce a scaling factor, \( \sin \theta \), which forms part of the determinant.
• While a full Jacobian determinant calculation would involve a 3x3 matrix for volume, for surface integrals, simplifying it to our specific usage results in the expression used

The Jacobian is, hence, key to moving seamlessly between coordinate systems, ensuring accurate representation and integration of quantities over curved surfaces.