• <blockquote id="a6oys"></blockquote>
  • <tt id="a6oys"></tt>
  • Paul's Online Notes
    Paul's Online Notes
    Home / Calculus III / Surface Integrals / Divergence Theorem
    Show General Notice Show Mobile Notice Show All Notes Hide All Notes
    General Notice

    I apologize for the outage on the site yesterday and today. Lamar University is in Beaumont Texas and Hurricane Laura came through here and caused a brief power outage at Lamar. Things should be up and running at this point and (hopefully) will stay that way, at least until the next hurricane comes through here which seems to happen about once every 10-15 years. Note that I wouldn't be too suprised if there are brief outages over the next couple of days as they work to get everything back up and running properly. I apologize for the inconvienence.

    Paul
    August 27, 2020

    Mobile Notice
    You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width.

    Section 6-6 : Divergence Theorem

    In this section we are going to relate surface integrals to triple integrals. We will do this with the Divergence Theorem.

    Divergence Theorem

    Let \(E\) be a simple solid region and \(S\) is the boundary surface of \(E\) with positive orientation. Let \(\vec F\) be a vector field whose components have continuous first order partial derivatives. Then,

    \[\iint\limits_{S}{{\vec F\centerdot d\vec S}} = \iiint\limits_{E}{{{\mathop{\rm div}\nolimits} \vec F\,dV}}\]

    Let’s see an example of how to use this theorem.

    Example 1 Use the divergence theorem to evaluate \(\displaystyle \iint\limits_{S}{{\vec F\centerdot d\vec S}}\) where \(\vec F = xy\,\vec i - \frac{1}{2}{y^2}\,\vec j + z\,\vec k\) and the surface consists of the three surfaces, \(z = 4 - 3{x^2} - 3{y^2}\), \(1 \le z \le 4\) on the top, \({x^2} + {y^2} = 1\), \(0 \le z \le 1\) on the sides and \(z = 0\) on the bottom.
    Show Solution

    Let’s start this off with a sketch of the surface.

    This is a graph with the standard 3D coordinate system.  The positive z-axis is straight up, the positive x-axis moves off to the left and slightly downward and positive y-axis moves off the right and slightly downward.  The walls of the solid in this graph are the cylinder given in the problem statement and whose “cap” is the elliptic paraboloid that starts at z=4 and opens along the z-axis in the negative z direction until it hits the cylinder at z=1.

    The region \(E\) for the triple integral is then the region enclosed by these surfaces. Note that cylindrical coordinates would be a perfect coordinate system for this region. If we do that here are the limits for the ranges.

    \[\begin{array}{c}0 \le z \le 4 - 3{r^2}\\ 0 \le r \le 1\\ 0 \le \theta \le 2\pi \end{array}\]

    We’ll also need the divergence of the vector field so let’s get that.

    \[{\mathop{\rm div}\nolimits} \vec F = y - y + 1 = 1\]

    The integral is then,

    \[\begin{align*}\iint\limits_{S}{{\vec F\centerdot d\vec S}} & = \iiint\limits_{E}{{{\mathop{\rm div}\nolimits} \vec F\,dV}}\\ & = \int_{{\,0}}^{{\,2\pi }}{{\int_{{\,0}}^{{\,1}}{{\int_{{\,0}}^{{4 - 3{r^2}}}{{r\,dz}}\,dr}}\,d\theta }}\\ & = \int_{{\,0}}^{{\,2\pi }}{{\int_{{\,0}}^{{\,1}}{{4r - 3{r^3}\,dr}}\,d\theta }}\\ & = \int_{{\,0}}^{{\,2\pi }}{{\left. {\left( {2{r^2} - \frac{3}{4}{r^4}} \right)} \right|_0^1\,d\theta }}\\ & = \int_{{\,0}}^{{\,2\pi }}{{\frac{5}{4}\,d\theta }}\\ & = \frac{5}{2}\pi \end{align*}\]
    亚洲欧美中文日韩视频 - 视频 - 在线观看 - 影视资讯 - av网 <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <文本链> <文本链> <文本链> <文本链> <文本链> <文本链>