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Chapter 14: Problem 35

Set up a double integral and evaluate the surface integral \(\iint_{S} g(x, y, z) d S\) \(\iint_{S} x d S, S\) is the portion of \(x^{2}+y^{2}-z^{2}=1\) between \(z=0\) and \(z=1\)

Short Answer

Expert verified
The value of the surface integral is 0

Step by step solution

01

Parameterizing the surface S

A suitable parameterization for the surface S, adviced by the cylindrical coordinates, is \(x = r \cos \theta, y = r \sin \theta, z = r^2\), where \(r\) varies from \(0\) to \(1\) and \(\theta\) varies from \(0\) to \(2\pi\). Thus, the parameterization is \(X(r, \theta)=(r\cos\theta, r\sin\theta, r^2)\).
02

Calculating the magnitude of the cross product

Calculating the magnitude of cross product of the tangent vectors \(\frac{\partial X}{\partial r}\) and \(\frac{\partial X}{\partial \theta}\). This provides the differential of surface area, which represents the \(dS\) in the integral. \(\left|\frac{\partial X}{\partial r}\times\frac{\partial X}{\partial \theta}\right| = \sqrt{(2 r \cos \theta)^{2}+(2 r \sin \theta)^{2}+(-r)^{2}} = \sqrt{(r^2 + 1)}r\).
03

Setting up the surface integral

Substituting the differential dS by the expression obtained in the last step and also substituting \(g(x, y, z) = x\) by \(rcos\theta\), we obtain \(\iint_S g(x,y,z)dS = \int_0^1 \int_0^{2\pi} (r \cos\theta) \sqrt{r^2+1} r d\theta dr\).
04

Evaluating the double integral

Finally, we compute the double integral by first integrating with respect to \(\theta\) and then integrating with respect to \(r\), which results \(\int_0^1 \left[\sqrt{r^2+1}r^2\cos\theta\right]_0^{2\pi}dr=\int_0^1 r^2 \sqrt{r^2+1}dr - \int_0^1 r^2 \sqrt{r^2+1}dr = 0\). The result is zero because the function \(r^2 \sqrt{r^2+1}\cos\theta\) is an odd function and we are integrating this over a period of \(2\pi\).

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Key Concepts

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

Double Integral
A double integral is a way to integrate over a two-dimensional area. It is used extensively in surface integrals where you want to account for variables changing over an area rather than along a line.
In our surface integral scenario, we first parameterize the surface and then make use of a double integral to evaluate it. This involves:
  • Identifying the limits of integration which usually relate to the geometric dimensions of the surface.
  • Using the appropriate variable transformations to simplify the integration process.
  • Calculating the double integral over the specified domain, which involves two steps of integration: one for each variable.
Using double integrals, we combine both integrations into a single cohesive process that provides the integral over a given surface.
Cylindrical Coordinates
Cylindrical coordinates are an extension of polar coordinates to three dimensions by adding a height value. This system is especially useful for surfaces with circular symmetry.
In our exercise, we use the conversion:
  • r, the radius from the axis of rotation (cylindrical radius)
  • \(\theta\), the angle around the axis (azimuthal angle)
  • z, the height above (or below) the plane
By switching to cylindrical coordinates, the given surface equation \(x^2 + y^2 - z^2 = 1\) can be parameterized naturally. This makes the integration process easier and more intuitive for the given problem. The use of cylindrical coordinates simplifies the double integral, which would otherwise be more complex in Cartesian coordinates.
Parameterization
Parameterization involves expressing the coordinates of a geometry or a surface in terms of one or more parameters.
For the indicated surface, we use the parameterization given by:
  • \(x = r \cos \theta\)
  • \(y = r \sin \theta\)
  • \(z = r^2\)
This approach helps mediate between the expressions of the surface in their algebraic form to an easier format for applying a surface integral. Here, \(r\) and \(\theta\) are chosen due to the symmetry and the form of the surface, making the expression for the integrand simpler and more direct to handle. This transformation is essential for setting up the surface integral in terms of \(dS\).
Cross Product
The cross product is a vector operation that finds a vector orthogonal (perpendicular) to two input vectors.
In the context of this surface integral, we use the cross product to find the differential area element, \(dS\). This requires:
  • Calculating the partial derivatives \(\frac{\partial X}{\partial r}\) and \(\frac{\partial X}{\partial \theta}\) of our parameterized surface.
  • Taking the cross product of these derivatives to find a normal vector to the surface.
  • Finding the magnitude of this cross product vector, which gives \(|dS|\).
The applicable result here is \(\left| \frac{\partial X}{\partial r} \times \frac{\partial X}{\partial \theta} \right| = \sqrt{(r^2 + 1)}r\). This calculation is crucial as it allows the replacement of \(dS\) in the surface integral, transferring the problem to one of finding an integral over these derivatives.

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Most popular questions from this chapter

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