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Calculus III - Notes

Internet Explorer 10 & 11 Users : If you have been using Internet Explorer 10 or 11 to view the site (or did at one point anyway) then you know that the equations were not properly placed on the pages unless you put IE into "Compatibility Mode". I believe that I have partially figured out a way around that and have implimented the "fix" in the Algebra notes (not the practice/assignment problems yet). It's not perfect as some equations that are "inline" (i.e. equations that are in sentences as opposed to those on lines by themselves) are now shifted upwards or downwards slightly but it is better than it was.

If you wish to test this out please make sure the IE is not in Compatibility Mode and give it a test run in the Algebra notes. If you run into any problems please let me know. If things go well over the next week or two then I'll push the fix the full site. I'll also continue to see if I can get the inline equations to display properly.
 Applications of Partial Derivatives Previous Chapter Next Chapter Line Integrals Change of Variables Previous Section Next Section Area and Volume Revisited

## Surface Area

In this section we will look at the lone application (aside from the area and volume interpretations) of multiple integrals in this material.  This is not the first time that we’ve looked at surface area   We first saw surface area in Calculus II, however, in that setting we were looking at the surface area of a solid of revolution.  In other words we were looking at the surface area of a solid obtained by rotating a function about the x or y axis.  In this section we want to look at a much more general setting although you will note that the formula here is very similar to the formula we saw back in Calculus II.

Here we want to find the surface area of the surface given by  where  is a point from the region D in the xy-plane.  In this case the surface area is given by,

Let’s take a look at a couple of examples.

 Example 1  Find the surface area of the part of the plane  that lies in the first octant.   Solution Remember that the first octant is the portion of the xyz-axis system in which all three variables are positive.  Let’s first get a sketch of the part of the plane that we are interested in.       We’ll also need a sketch of the region D.     Remember that to get the region D we can pretend that we are standing directly over the plane and what we see is the region D.  We can get the equation for the hypotenuse of the triangle by realizing that this is nothing more than the line where the plane intersects the xy-plane and we also know that  on the xy-plane.  Plugging  into the equation of the plane will give us the equation for the hypotenuse.   Notice that in order to use the surface area formula we need to have the function in the form  and so solving for z and taking the partial derivatives gives,                                   The limits defining D are,                                              The surface area is then,

 Example 2  Determine the surface area of the part of  that lies in the cylinder given by .   Solution In this case we are looking for the surface area of the part of  where  comes from the disk of radius 1 centered at the origin since that is the region that will lie inside the given cylinder.   Here are the partial derivatives,                                                            The integral for the surface area is,                                                            Given that D is a disk it makes sense to do this integral in polar coordinates.
 Change of Variables Previous Section Next Section Area and Volume Revisited Applications of Partial Derivatives Previous Chapter Next Chapter Line Integrals

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