Monday 4 December and Wednesday 6 December classes

(Wednesday after Test 4)

Topics:

• Finding the area bounded by two curves. Please make sure that you see this post about the WeBWorK on this topic.

When we find the area bounded between two curves, we are looking for a genuine area, that is, all areas are positive areas (unlike the “area” defined by the definite integral, which can be negative.) For this reason it is important to look at the graph and see which curve lies above and which below, and how many times they intersect each other. Only then can we set up the integral or integrals we need to compute. After that, it should be straightforward integration.

Another choice that has to be made is whether to integrate with respect to x or with respect to y. (Sometimes either one will do.)

One easy check: If both your curves are giving y as a function of x, you can integrate with respect to x. If one or both are giving x as a function of y (rather than vice-versa), it may be better to integrate with respect to y.

PatrickJMT has some videos on finding areas between curves which you may find useful: see this list

 

• Volumes of rotation: two methods (introduction)

There are two basic methods of computing the volume we get when rotating a region in the plane around some axis. They go by a variety of names.

The first I discussed is usually called the disc method or the washer method, but sometimes it is called the ring method as well.

The second I discussed is called the shell method or the cylinder method.

There may be some other names for them – I hope not – but be aware that different names may possibly be used in different sources!

 

•••I worked out an example of finding the volume you get by taking the region between x=0 and the curve $y = x^{2}$, below y=9, and rotating around the y- axis. . This region is shown in my two Desmos graphs, which also show a sample way of making rectangles for a Riemann sum for each of the two approaches.

• For the disc method, here is the graph: notice that the long sides of the rectangles are perpendicular to the axis we are rotating around. When we rotate any one of those rectangles around the y-axis, it gives a disc whose thickness is \Delta y. (Ignore the fact that the thickness according to the graph would be 0.5: remember that the length of this short side is going to go to 0, it is not fixed.)

To find the volume of the disc that goes with that rectangle, remember that the area of a circle is \pi r^{2}. For this disc, we can see that its radius is the long side of the rectangle, which is x (whatever x goes with this rectangle). So the volume of the disc is \pi x^{2} times \Delta y, but we need this to be all in terms of y, since we are going to have to integrate with respect to y.

From y=x^{2} we can substitute y for x^{2} and so the volume of the disc is \pi y \Delta y.

This tells us that we have to find the integral \displaystyle\int_{0}^{9}\pi y\,\textrm{d}y, which I worked out in class; it gives the volume $latex \frac{81\pi}{2}

 

• For the shell method, the graph is here: notice that the long sides of the rectangles are parallel to the axis we are rotating around. When we rotate any one of those rectangles around the y-axis, it gives a cylindrical shel whose thickness is \Delta x.

To find the volume of the shell, notice that the height of the cylinder (the long side of the rectangle) is 9 - x^{2} for whatever x goes with that rectangle. Also the “average” radius of the cylindrical shell is x, so its average circumference is 2\pi x. The area of the cylinder of radius x and height 9 - x^{2} is 2\pi x\left(9 - x^{2}\right) and we multiply this by \Delta x to get the volume of the shell:

2\pi x\left(9 - x^{2}\right)\Delta x

This tells us that we have to find the integral \displaystyle\int_{0}^{3}2\pi x\left(9 - x^{2}\right)\,\textrm{d}x

I leave it as an exercise for you to compute this integral and verify that it gives the same result as the disc method did. Clearly this method is more tedious for this example. You will begin to see that sometimes one method is distinctly easier than the other, but this means sometimes the washer/disc method is easier and sometimes the shell method is easier. We need both.

 

Here are some very nice videos that show what is going on in 3 dimensions better than I can by drawing on the board. I hope you find them enlightening!

Disc method (has very nice graphics)

washer (disc) method: Mathispower4u

shell method:   Mathispower4u

 

Homework:

• We have already started and most of us made substantial progress on the area between curves WeBWorK, so finish it for Sunday evening (not waiting to the last minute, of course!) – look at Patrick’s videos linked above if you find videos helpful.

• Compute the integral I got for the shell method on my problem and verify that it gives the same answer as the disc method did, namely \frac{81\pi}{2}

• You do not have to work the Shells and Washers 1 WeBWorK yet: please look at the videos if you have a chance, just trying to see what’s  going on. If you want to get a start on the WeBWorK, a good place to begin is with problem 2 and maybe problem 5, which are similar to what I did in class.

• Please take a look at the Final Exam Review sheet. For next time, please work on problems 1, 2, 8, and 9. I will ask for volunteers to put some parts of these on the board. (You can certainly also start working on the other problems: I recommend starting with 5, 6, and 7, which are older material. We will do as many as we can of those and the rest on Wednesday!)

 

Don’t forget, if you get stuck on a problem, you can post a question on Piazza. Make sure to give your question a good subject line and tell us the problem itself – we need this information in order to answer your question. And please only put one problem per posted question!

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