Chapter 18: Problem 4
If \(V\) is the scalar field \(V=x y z^{2}\), evaluate \(\int_{S} V \mathrm{~d} \mathbf{S}\) over the surface \(S\) defined by \(x^{2}+y^{2}=9\) between \(z=0\) and \(z=2\) in the first octant.
Short Answer
Expert verified
12
Step by step solution
01
Identify the surface
The surface is a quarter of the cylinder defined by the equation \(x^{2}+y^{2}=9\) with bounds 0 \(\leq z \leq 2\) in the first octant.
02
Express in cylindrical coordinates
Convert the scalar field and the surface into cylindrical coordinates, where \(x = r \cos \theta\), \(y = r \sin \theta\), and \(r = 3\). The scalar field thus becomes \(V = x y z^{2} = r \cos \theta r \sin \theta z^{2} = r^{2} \cos \theta \sin \theta z^{2}\).
03
Define the surface element
For a cylindrical surface where \(r = 3\) and \(0 \leq z \leq 2\), the surface element in cylindrical coordinates is \(\text{d}\textbf{S} = \textbf{k} \text{d}\theta \text{d}z\) where \(\textbf{k}\) is the unit vector in the z-direction.
04
Set up the surface integral
The integral becomes \[ \int_{S} V \text{dS} = \int_{0}^{2} \int_{0}^{\frac{\pi}{2}} 9 \cos \theta \sin \theta z^{2} \text{d}\theta \text{d}z \]
05
Integrate with respect to \(\theta\)
Evaluate the integral with respect to \(\theta\): \[ \int_{0}^{\frac{\pi}{2}} 9 \cos \theta \sin \theta \text{d} \theta = 9 \int_{0}^{\frac{\pi}{2}} \cos \theta \sin \theta \text{d} \theta = 9 \left[ \frac{\sin^{2} \theta}{2} \right]_{0}^{\frac{\pi}{2}} = 9 \left( \frac{1}{2} - 0 \right) = \frac{9}{2} \]
06
Integrate with respect to \(z\)
Evaluate the integral with respect to \(z\): \[ \frac{9}{2} \int_{0}^{2} z^{2} \text{d}z = \frac{9}{2} \left[ \frac{z^{3}}{3} \right]_{0}^{2} = \frac{9}{2} \left( \frac{8}{3} - 0 \right) = \frac{9}{2} \cdot \frac{8}{3} = 12 \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Scalar Field Integration
A 'scalar field' is a mathematical function that assigns a single value to every point in space. In this problem, the scalar field is given by the function \(V = xyz^2\).
'Integration of a scalar field' over a surface involves summing up the influences of the scalar field over that surface. This process is critical in solving various real-world engineering problems, such as determining the flux of a substance through a surface.
To integrate a scalar field, you multiply the scalar value at every point by a small surface element and then sum these products over the entire surface. This leads us to a 'surface integral'.
In simpler terms, you are calculating the total effect of the function over a specified surface.
'Integration of a scalar field' over a surface involves summing up the influences of the scalar field over that surface. This process is critical in solving various real-world engineering problems, such as determining the flux of a substance through a surface.
To integrate a scalar field, you multiply the scalar value at every point by a small surface element and then sum these products over the entire surface. This leads us to a 'surface integral'.
In simpler terms, you are calculating the total effect of the function over a specified surface.
Cylindrical Coordinates
Cylindrical coordinates are a way of describing points in three-dimensional space. They are especially useful when dealing with surfaces and volumes that have a circular symmetry, like cylinders and circles.
In cylindrical coordinates, a point is represented by three values:
The transformation rules are:
\(x = r \cos \theta\),
\(y = r \sin \theta\),
and the scalar field becomes \(V = r^2 \cos \theta \sin \theta z^2\).
Understanding these transformations is crucial for solving many complex engineering mathematics problems.
In cylindrical coordinates, a point is represented by three values:
- \(r\), the radial distance from the z-axis
- \(\theta\), the angle around the z-axis
- \(z\), the height along the z-axis
The transformation rules are:
\(x = r \cos \theta\),
\(y = r \sin \theta\),
and the scalar field becomes \(V = r^2 \cos \theta \sin \theta z^2\).
Understanding these transformations is crucial for solving many complex engineering mathematics problems.
Calculus in Engineering Mathematics
Calculus is a fundamental tool in engineering mathematics. It allows us to understand and describe how certain quantities change and accumulate.
In this problem, we use 'integral calculus' to evaluate a surface integral. Integral calculus helps us quantify the total accumulation of a quantity, and it is the backbone of many engineering disciplines like structural engineering, fluid dynamics, and electromagnetic theory.
The steps involve:
In this problem, we use 'integral calculus' to evaluate a surface integral. Integral calculus helps us quantify the total accumulation of a quantity, and it is the backbone of many engineering disciplines like structural engineering, fluid dynamics, and electromagnetic theory.
The steps involve:
- Identifying the problem and converting functions and domains to more workable forms (like cylindrical coordinates)
- Setting up the integral through proper limits and conversion
- Breaking down the integration into manageable parts, often in terms of standard functions or shapes
- Summing the effects over defined limits to get a complete solution
Surface Integrals
A 'surface integral' extends the idea of an integral to functions defined over surfaces. Whereas a simple integral sums up a function over an interval, a surface integral sums up a function over a two-dimensional surface.
In this problem, the surface is a quarter of a cylinder in the first octant. Surface integrals are especially useful in fields like physics and engineering to calculate quantities like flux through a surface and other multi-dimensional analyses.
In terms of surface elements in cylindrical coordinates, for the given surface, the element \(\text{d}\textbf{S}\) involves multiplying infinitesimal elements in the \(\theta\) and \(z\) directions.
The full setup for the integral in cylindrical coordinates becomes:
\( \int_{S} V \text{d}S = \int_{0}^{2} \int_{0}^{\frac{\pi}{2}} 9 \cos \theta \sin \theta z^{2} \text{d}\theta \text{d}z \)
After setting up the double integral, the integration is carried out one variable at a time, simplifying the process. Understanding the construction and evaluation of surface integrals gives engineers the tools to solve a vast array of practical problems.
In this problem, the surface is a quarter of a cylinder in the first octant. Surface integrals are especially useful in fields like physics and engineering to calculate quantities like flux through a surface and other multi-dimensional analyses.
In terms of surface elements in cylindrical coordinates, for the given surface, the element \(\text{d}\textbf{S}\) involves multiplying infinitesimal elements in the \(\theta\) and \(z\) directions.
The full setup for the integral in cylindrical coordinates becomes:
\( \int_{S} V \text{d}S = \int_{0}^{2} \int_{0}^{\frac{\pi}{2}} 9 \cos \theta \sin \theta z^{2} \text{d}\theta \text{d}z \)
After setting up the double integral, the integration is carried out one variable at a time, simplifying the process. Understanding the construction and evaluation of surface integrals gives engineers the tools to solve a vast array of practical problems.