Problem 58 - Calculus: Early Transcendental Functions

Question

Use the formulas of exercises 53 and 54 to evaluate the surface integral. \(\iint_{S}\left(y^{2}+z^{2}\right) d S,\) where \(S\) is the hemisphere \(x=\sqrt{4-y^{2}-z^{2}}\)

Step-by-Step Solution

Answer

Please solve the final integration and simplify for the value. This is straightforward following the integral rules and putting the limits.

Step 1: Conversion to spherical coordinates

Convert the expression and the limits of the integral from Cartesian to spherical coordinates. In this case, we have \[y = \rho \sin(\phi) \cos(\theta)\], \[z = \rho \sin(\phi) \sin(\theta)\] and \[x = \rho \cos(\phi)\], with \(\rho^2 = x^2 + y^2 + z^2\). For the given hemisphere \(x=\sqrt{4-y^{2}-z^{2}}\), it can converted to \(\rho\cos(\phi) = 2\), which correspond to a hemisphere of radius 2 with \(\phi\) in \([0,\pi/2]\).

Step 2: Substitution

Substitute the converted expression into the integral, replacing \(y^2 + z^2\) with \(\rho^2\sin^2(\phi)\) so the integral is now \(\iint_{S}\left(\rho^{2}\sin^2(\phi)\right) d S\).

Step 3: Set up the bounds

Set up the integral bounds. The \(\theta\) goes from 0 to \(2\pi\) and \(\phi\) goes from 0 to \(\pi/2\). So the surface integral over the hemisphere is now double integral of \(\rho^{2}\sin^2(\phi)\) where \(\theta\) goes from 0 to \(2\pi\) and \(\phi\) from 0 to \(\pi/2\).

Step 4: Solve the integral

Solve the integral, considering the Jacobian determinant for spherical to Cartesian coordinates. The Jacobian for spherical coordinates is \(\rho^2\sin(\phi)\), so the integral to solve is \(\iiint\rho^4\sin^3(\phi) d\rho d\phi d\theta\). Solve this using integral rules and limits.

Key Concepts

Spherical Coordinates

Spherical coordinates provide a way to describe a point in three-dimensional space through a radius and two angles:

• The radius, \( \rho \), measures the distance from the origin to the point.
• The azimuthal angle, \( \theta \), is the angle in the \( xy \)-plane from the positive \( x \)-axis.
• The polar angle, \( \phi \), is the angle from the positive \( z \)-axis.

Spherical coordinates are particularly useful for problems with symmetry about a sphere, like the ones involving a hemisphere. They simplify the math by making angles easier to handle than Cartesian coordinates.
In our problem, to convert the hemisphere equation and the limits of the integral, use \( y = \rho \sin(\phi) \cos(\theta) \), \( z = \rho \sin(\phi) \sin(\theta) \), and \( x = \rho \cos(\phi) \). This change forms a much more manageable expression.

Hemispheres

A hemisphere is one half of a sphere, typically cut by a plane. In the context of this exercise, the hemisphere is defined mathematically as the surface \( x = \sqrt{4-y^{2}-z^{2}} \). When dealing with a hemisphere in spherical coordinates:

• The radius is the same throughout, represented as \( \rho = 2 \) in this problem.
• Since we are only considering the "upper" hemisphere, the angle \( \phi \) ranges from 0 to \( \pi/2 \), covering the top half of the sphere.

Hemisphere problems often involve finding areas or volumes, and these are evaluated using integrals. This involves setting appropriate bounds that capture the hemisphere's limits in the angular parameters \( \phi \) and \( \theta \).

Jacobian Determinant

The Jacobian determinant is a factor that appears when you change variables in multiple integrals. It accounts for the change of variables in the coordinate system transformation, ensuring the integral's value remains accurate.
For spherical coordinates, the Jacobian determinant is \( \rho^{2}\sin(\phi) \), which corresponds to the scaling factor present when transforming from Cartesian coordinates.
Take notice that while setting up the integrals in spherical coordinates, you must include this Jacobian determinant. It adjusts the measure element \( dS \) in the surface integral. \[ \iiint \rho^{4}\sin^{3}(\phi) d\rho\, d\phi\, d\theta \]
In our problem, the integration part includes \( \rho^{4}\sin^{3}(\phi) \) since it involves both the function being integrated and the Jacobian.

Double Integrals

Double integrals calculate the volume under a surface over a plane region. They are often expressed in polar, cylindrical, or spherical forms in three dimensions. In our hemisphere surface integral exercise, the double integral is adapted to spherical coordinates, taking into account the Jacobian.

• The variables of integration are typically \( \rho \) and \( \phi \), or \( \theta \) in polar cases.
• The boundary regions are defined by the ranges of these variables, \( \theta = [0, 2\pi] \) and \( \phi = [0, \pi/2] \).

This setup makes integration over curved surfaces manageable. In our case, the double integral turns into a triple one due to the Jacobian, incorporating the third dimension adjustment as \[ \iiint_{0}^{2} \rho^{4}\sin^{3}(\phi) d\rho\, d\phi\, d\theta \]. Solve it sequentially to find the total surface integral over the hemisphere.