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Chapter 14: Problem 60

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

Short Answer

Expert verified
The result of the surface integral over the given hemisphere is \(\frac{128}{5}\pi\)

Step by step solution

01

Parametrization of the Hemisphere

Parametrize the hemisphere by using the spherical coordinates. The equation for the hemisphere in spherical coordinates is given as \(\rho = 2\). So, choose the parameters \(r\) and \(\theta\), where \(0 \leq r \leq 2\), and \(0 \leq \theta \leq 2\pi\). Then, we transform the coordinates: \(x = r \cdot \cos(\theta)\) and \(z = r \cdot \sin(\theta)\). The unit normal vector \(dS\) on the sphere can be written as \(\sqrt{1 + (-\frac{dz}{dx})^{2}} dx dz = \sqrt{1 + r^{2}} r dr d\theta\).
02

Evaluate the Function

Substitute the parametrized variables in the function to find \(f(r, \theta) = r^{2} \cdot \cos^{2}(\theta) + r^{2} \cdot \sin^{2}(\theta) = r^{2}\). Substitute \(f(r, \theta) = r^{2}\) and \(dS = \sqrt{1 + r^{2}} r dr d\theta\) to yield the surface integral in spherical coordinates as \(\int_{0}^{2 \pi} \int_{0}^{2} (r^{2}) (\sqrt{1 + r^{2}} r dr d\theta)\).
03

Integration

Perform the integral over \(r\) and \(\theta\). The integral for \(\theta\) will result in \(2\pi\), and we only have to integrate \((r^{2})(\sqrt{1 + r^{2}} r)\) dr between 0 and 2. Solve the integral by substituting \(u = 1 + r^{2}\), \(du = 2r dr\), then \(r dr = du/2\). The integration range changes from \(u = 1\) to \(u = 5\). Transform the integral to \(\frac{1}{2} \int_{1}^{5} u\sqrt{u} du\), solve this to get \(\frac{64}{5}\). Multiply this with the result of integral over \(\theta (2\pi)\) to get the final solution.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Parametrization
Parametrizing a surface, like the hemisphere in our exercise, is about expressing it in terms of simpler variables. This makes complex integrations more manageable. Instead of working directly with three-dimensional coordinates \(x, y, z\), we use two parameters 鈥� typically \(r\) and \(\theta\) 鈥� that can cover the surface area.
For the hemisphere with equation \(y = \sqrt{4-x^{2}-z^{2}}\), we parametrize using spherical coordinates. This method is ideal as it translates well-suited parameters into space-friendly coordinates.
  • Choose \(r\) and \(\theta\) as parameters.
  • Define boundaries: \(0 \leq r \leq 2\) and \(0 \leq \theta \leq 2\pi\).
  • Convert: \(x = r \cdot \cos(\theta)\) and \(z = r \cdot \sin(\theta)\).
Thus, this parametrization allows us to represent every point on the hemisphere with \(r\) and \(\theta\), preparing our integral calculations.
Spherical Coordinates
Spherical coordinates are like a translator for geometry. They are especially useful when dealing with spheres or hemispheres as in this problem. Where Cartesian coordinates use \(x, y, z\) to describe a point, spherical coordinates use \(\rho, \theta, \phi\).
For our hemisphere, a special condition arises, reducing the complexity: the radius \(\rho\) is constant at 2. That means the surface is part of a sphere with radius 2.
  • \(x = r \cdot \cos(\theta)\)
  • \(z = r \cdot \sin(\theta)\)
  • \(y = \sqrt{4-x^2-z^2} = \sqrt{4 - r^2}\)
Thus, spherical coordinates give us a smoother and more intuitive handle on complex problems, allowing us to deal with them like simpler, flatter surfaces.
Integration
Integration in this context is about combining all these tiny pieces of area to find a total measure over the surface. The goal is to compute a surface integral where we handle functions over a surface in 3D space.
Substitute the parametrized variables to simplify the integral. The function becomes \(f(r, \theta) = r^{2}\) and the area element (often written as \(dS\)) turns into \( \sqrt{1 + r^{2}} r \, dr \, d\theta\).
Now, the integration task translates to:
  • Integrate \(\int_{0}^{2\pi} \int_{0}^{2} r^2 \sqrt{1 + r^2} r \, dr \, d\theta\).
  • The \( \theta \) integral is simple: \(2\pi\).
  • The \( r \) integral is executed via substitution: \(u = 1 + r^{2}, \, du = 2r \, dr\), simplifying boundaries from \(u = 1\) to \(u = 5\).
  • Evaluate \(\frac{1}{2} \int_{1}^{5} u\sqrt{u} \, du = \frac{64}{5}\).
Combining these results gives us the total value of the surface integral.
Hemisphere
A hemisphere is simply half of a sphere. When doing calculations involving hemispheres, we often focus on one hemisphere of a full sphere, either the top or bottom half. Here, we're dealing with the equation \(y=\sqrt{4-x^{2}-z^{2}}\), representing the upper hemisphere.
This problem restricts \(y\) to non-negative values, reflecting only the upper part of the sphere with a radius of 2.
  • The hemisphere's equation limits the surface by carving out the space where \(x^{2} + z^{2} < 4\).
  • The boundary is at \(r = 2\) where \(r = \sqrt{x^{2} + z^{2}}\).
  • This surface section can be smoothly parametrized and integrated.
Understanding the hemisphere's nature is crucial for solving the integration problem, as it confirms the bounded area over which the integration takes place.

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