/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 46 Evaluate surface integral \(\iin... [FREE SOLUTION] | 91Ó°ÊÓ

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Evaluate surface integral \(\iint_{S} z d \mathrm{~A}\), where \(\mathrm{S}\) is surface \(z=\sqrt{x^{2}+y^{2}}, 0 \leq z \leq 2\).

Short Answer

Expert verified
The surface integral evaluates to \( \frac{16\pi\sqrt{2}}{3} \).

Step by step solution

01

Parameterize the Surface

To evaluate the surface integral, we first need to parameterize the surface. The surface \( S \) is defined by \( z = \sqrt{x^2 + y^2} \), which represents a cone. Using cylindrical coordinates, we can parameterize the surface as: \[ x = r \cos(\theta), \quad y = r \sin(\theta), \quad z = r \] where \( 0 \leq r \leq 2 \) and \( 0 \leq \theta < 2\pi \).
02

Compute the Differential Area Element

The differential area element \( dA \) on the surface is given by the magnitude of the cross-product of the partial derivatives of the parameterization with respect to \( r \) and \( \theta \). The partial derivative of the parameterization with respect to \( r \) is \( (\cos(\theta), \sin(\theta), 1) \), and with respect to \( \theta \) is \( (-r \sin(\theta), r \cos(\theta), 0) \). The cross-product is:\[ \left| \begin{array}{ccc} \mathbf{i} & \mathbf{j} & \mathbf{k} \ \cos(\theta) & \sin(\theta) & 1 \ -r \sin(\theta) & r \cos(\theta) & 0 \end{array} \right| = (-r \cos(\theta), -r \sin(\theta), r) \]The magnitude of this vector, which gives us \( dA \), is \( \sqrt{r^2 + r^2} = \sqrt{2}r \). Thus, \( dA = \sqrt{2}r \, dr \, d\theta \).
03

Set Up the Integral

We now set up the integral to evaluate \( \iint_{S} z \, dA \). In our parameterization, \( z = r \), so the integral becomes:\[ \int_{0}^{2\pi} \int_{0}^{2} r \cdot \sqrt{2}r \, dr \, d\theta \]which simplifies to:\[ \sqrt{2} \int_{0}^{2\pi} \int_{0}^{2} r^2 \, dr \, d\theta \]
04

Evaluate the Integral

Evaluate the integral by first integrating with respect to \( r \):\[ \int_{0}^{2} r^2 \, dr = \left[ \frac{r^3}{3} \right]_{0}^{2} = \frac{8}{3} \]Next, integrate with respect to \( \theta \):\[ \int_{0}^{2\pi} \, d\theta = 2\pi \]Combining both results, the integral becomes:\[ \sqrt{2} \times \frac{8}{3} \times 2\pi = \frac{16\pi\sqrt{2}}{3} \]
05

Conclusion

The surface integral evaluates to \( \frac{16\pi\sqrt{2}}{3} \). This is the value of the integration of \( z \, dA \) over the given surface.

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Key Concepts

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

Understanding Surface Parameterization
Surface parameterization is a method used in mathematics to express a surface as a continuous mapping from a parameter domain, usually a two-dimensional region like a rectangle or a disk, to a surface in three-dimensional space. It is particularly helpful for evaluating surface integrals, where we need to smoothly transition over every point on a surface.

In our problem, the surface is a cone described by the equation \( z = \sqrt{x^2 + y^2} \). We can use this to parameterize the cone using cylindrical coordinates, which utilize the variables \( r \) and \( \theta \) to represent the radial distance and the angular position, respectively.

To parameterize the surface, the expressions become:
  • \( x = r \cos(\theta) \)
  • \( y = r \sin(\theta) \)
  • \( z = r \)
These expressions map out every point \((x, y, z)\) on the cone as \( r \) varies from 0 to 2 and \( \theta \) ranges from 0 to \( 2\pi \). Thus, parameterization provides an efficient way to handle the complex surface shape by converting it into a simpler set of variables.
Navigating with Cylindrical Coordinates
Cylindrical coordinates are an extension of polar coordinates into three dimensions. They are especially useful for problems involving symmetry around an axis, like cones, cylinders, and circular shapes.

Unlike Cartesian coordinates that use \( x, y, \) and \( z \), cylindrical coordinates use \( r, \theta, \) and \( z \). Here \( r \) represents the distance from a central axis (usually the z-axis), \( \theta \) is the angle between the radial line and a reference direction, and \( z \) is the height above the plane.
  • Radial Distance (\( r \)): Takes on values from the surface's center to its edge.
  • Angle (\( \theta \)): Rotates around the surface's axis, covering all points in a circular sweep.
  • Height (\( z \)): Varies along the direction perpendicular to the plane, especially in 3D shapes like cones.
This coordinate system simplifies formulating the surface integral into an iterated integral over a rectangular region in the \( (r, \theta) \) plane.
Calculating the Differential Area Element
The differential area element \( dA \) is a small piece of a surface, used to accumulate parts of the whole surface during integration. It's crucial when calculating surface integrals as it represents the 'tiny pieces' of the surface over which we integrate.

To find \( dA \), we need the cross-product of partial derivatives of the parameterization with respect to the variables. For our cone surface in cylindrical coordinates, the partial derivatives are:
  • With respect to \( r \): \( (\cos(\theta), \sin(\theta), 1) \)
  • With respect to \( \theta \): \( (-r \sin(\theta), r \cos(\theta), 0) \)
The cross-product of these vectors gives us a new vector \((-r \cos(\theta), -r \sin(\theta), r)\).

The magnitude of this vector determines the differential area element \( dA \), allowing us to express it as \( \sqrt{2} r \, dr \, d\theta \). This provides the measure of the small area embedded in the 3-dimensional space, thus completing our setup for evaluating the surface integral.

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Most popular questions from this chapter

Compute \(\iint_{S} \mathbf{F} \cdot \mathbf{N} d S\), where \(\mathbf{F}(x, y, z)=x y z \mathbf{i}+x y z \mathbf{j}+x y z \mathbf{k}\) and \(\mathrm{N}\) is an outward normal vector \(S\), where \(S\) is the surface of the five faces of the unit cube \(0 \leq x \leq 1,0 \leq y \leq 1,0 \leq z \leq 1\) missing \(z=0\).

For the following exercises, evaluate \(\iint_{S} \mathbf{F} \cdot \mathbf{N} d s\) for vector field \(F\), where \(\mathbf{N}\) is an outward normal vector to surface \(S\). \(\mathbf{F}(x, y, z)=x^{2} \mathbf{i}+y^{2} \mathbf{j}+z^{2} \mathbf{k}\), and \(S\) is the portion of plane \(z=y+1\) that lies inside cylinder \(x^{2}+y^{2}=1\)

For the following exercises, Fourier's law of heat transfer states that the heat flow vector \(\mathrm{F}\) at a point is proportional to the negative gradient of the temperature; that is, \(\mathbf{F}=-k \nabla T\), which means that heat energy flows hot regions to cold regions. The constant \(k>0\) is called the conductivity, which has metric units of joules per meter per second-kelvin or watts per meter-kelvin. A temperature function for region \(D\) is given. Use the divergence theorem to find net outward heat flux \(\iint_{S} \mathbf{F} \cdot \mathbf{N} d S=-k \iint_{S} \nabla T \cdot \mathbf{N} d S\) across the boundary \(S\) of \(D\), where \(k=1$$T(x, y, z)=100+x+2 y+z ; D=\\{(x, y, z): 0 \leq x \leq 1,0 \leq y \leq 1,0 \leq z \leq 1\\}\)

For the following exercises, use a CAS and the divergence theorem to compute the net outward flux for the vector fields across the boundary of the given regions \(D .\)Use the divergence theorem to evaluate \(\iint_{S} \mathbf{F} \cdot d \mathbf{S}\), where \(\mathbf{F}(x, y, z)=x \mathbf{i}+y \mathbf{j}+2 \mathbf{k}\) and \(S\) is the boundary of the solid enclosed by paraboloid \(y=x^{2}+z^{2}-2\), cylinder \(x^{2}+z^{2}=1\), and plane \(x+y=2\), and \(S\) is oriented outward.

Use a computer algebra system to find the curl of the given vector fields. $$ \mathbf{F}(x, y, z)=\sin (x-y) \mathbf{i}+\sin (y-z) \mathbf{j}+\sin (z-x) \mathbf{k} $$

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