Well, I've done several excercises about calculating the volume of a solid of revolution some time ago and I don't remember how I got the results of some of them.

The first says "Find the volume of the solid generated by rotating the region bounded by $y=x^2 -4x$ and $y=0$ about $x$ axis".

What I've computed here is $V=\pi \int_{a}^{b}R^2 -r^2 dx\ $, where $a=0$, $b=4$, $r=0$ and $R=x^2 -4x$, so I got $\frac{512}{15} \pi$.

But then, there's another problem that says "Find the volume of the solid generated by rotating the region bounded by $y=x^3,$ and $y=4x$ in the third quadrant about $y=8$".

Here, I used the same formula but I wrote $r=8-x^3$, $R=8-4x$, $a=-2$ and $b=0$. And that is what I don't understand. Why $r=8-x^3$ instead of $r=8+x^3$? Because there are 8 units, and then you have to go to $x^3$. And the same happens with $R=8-4x$ instead of $R=8+4x$. Or, at least, could I write $R=16-4x$ and $r=16-x^3$? Because there are 16 units minus the functions.

What's the difference with the first activity?

  • 1
    $\begingroup$ Note that $x^3$ is negative in that quadrant, so $8-x^3$ is adding the distance to the graph of $x^3$. $\endgroup$ Jul 30 '19 at 23:55
  • $\begingroup$ @GerryMyerson so what I did in the first activity is wrong? $\endgroup$
    – AaronTBM
    Jul 31 '19 at 1:04
  • $\begingroup$ $x^2-4x$ is negative for $0<x<4$, so, yes, technically, what you did was wrong; you should have used $R=4x-x^2$. But since $R$ only comes into the volume formula as $R^2$, it made no difference to the answer you get. $\endgroup$ Jul 31 '19 at 1:57
  • $\begingroup$ Is this a trick question? The graphs of both $y=x^3$ and $y=4x$ lie entirely in the first and third quadrants. I don’t see any area they enclose in the fourth quadrant. $\endgroup$
    – David K
    Jul 31 '19 at 11:03
  • $\begingroup$ @David, good point. I took it to mean 3rd quadrant. $\endgroup$ Jul 31 '19 at 12:58

Your course seems to have an unusual definition of fourth quadrant.

But let’s consider a concrete example. Let $x=-1.$ This is midway between your $a$ and $b$ so it’s certainly relevant.

For the curve $y=x^3,$ at $x=-1$ you have $y=(-1)^3=-1.$ So you are correct when you say you want to measure the inner radius by going from $8$ to $0$ and then the additional distance from $0$ to $x^3,$ because $x^3=-1.$ In this particular case the radius is $9.$

OK, so let’s try $8+x^3$ as the “additional distance” intuition might suggest. Since $x^3=-1,$ we find that $8+x^3=8+(-1)=7.$ But we already determined graphically that the correct radius is $9.$ So “additional distance $\implies$ use addition” is a faulty intuition.

On the other hand, $8-x^3=8-(-1)=9.$ So subtraction gives the correct answer after all, even when the $y$ values are on opposite sides of the $x$ axis.

The key takeaway for me is: Subtraction always gives the distance. That’s because $p-q$ is precisely how much we have to add to $q$ in order to arrive at $p.$ To be a little more rigorous we should say $p-q$ always gives the distance or the negative of the distance between $p$ and $q,$ because whether you get a positive or negative negative number depends on whether you list the greater number first. But if you’re only going to use the square of the distance then the positive/negative distinction is erased by the squaring.


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