0
$\begingroup$

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?

$\endgroup$
5
  • 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
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.