Finding the volume of revolution in the case of self-intersections.

The issue arose when we looked at #6-2-49 in class. We are asked to "find the volume of the solid generated by revolving the region about the prescribed axis," where the region is "bounded by , and ," and the axis is the line .

Confusion arose because the region isn't so easy to interpret. Here are the curves in question, and the axis of revolution:

The region we decided upon is the following, which is clearly bounded by the three curves given to us.

Quick solution to the problem: the curves intersect at so using shells (easier than the washers we used in class) gives

This agrees with the text's answer, so it is surely what the author intended.

But as somebody pointed out, the following region is also bdd by the same three curves.

We decided that the authors must have meant the first region, because the second one contains the axis of revolution, which would lead to a problem.

But we LIKE problems, so the question became, What is the volume obtained when the second region is rotated? The difficulty lies with the fact that the region will intersect itself as it revolves, so care must be taken not to "count certain parts twice".

I prefer to handle it as follows. Sketch the graps of the curves obtained by rotating and through 180 degrees about the axis of rotation, . This is the same as reflecting them with respect to the line . The reflected curves are shown below in red, along with the original curves in black. (The curve reflects to itself and isn't emphasized.)

Note: the reflection of a curve about a line is the curve . Try to prove this.

Here is the same figure, along with the region we must rotate.

When we do rotate, we pick up more (the lavender regions):

So to obtain these in one rotation, without overlap, we can simply rotate the following regions.

That's the trick. The rest of the problem involves actually doing the integrals.

Notice that one of the points we need to consider is the intersection of one of the original curves () wirh one of the reflected curves (). In this case we get  a cubic equation with no "nice" solution. But that's life.

Use numerical techniques (I used Mathematica, which is handy) and you'll obtain for that intersection.

There is only one real solution, which you can see by examining the actual polynomial obtained when solving (how?). You should notice that Mathematica gives an exact form of the solution; it's the first value produced in the output of Solve, above. I'll refer to this value as .

I'll use shells for finding the volume. The "blue" region is easiest, namely  , where , and  .

The small beige region gives

and the large beige region yields

The volume of our solid is