# Confusion related to the volume of a solid of revolution

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?

• Note that $x^3$ is negative in that quadrant, so $8-x^3$ is adding the distance to the graph of $x^3$. Jul 30 '19 at 23:55
• @GerryMyerson so what I did in the first activity is wrong? Jul 31 '19 at 1:04
• $x^2-4x$ 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. Jul 31 '19 at 1:57
• 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. Jul 31 '19 at 11:03
• @David, good point. I took it to mean 3rd quadrant. Jul 31 '19 at 12:58

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.