4
$\begingroup$

Compute the volume and surface area of the solid $S$ obtained by revolving the region $R$ (pictured below) enclosed by the graphs $y=q(x)=-(2x^2-7x+3),x=1,x=2$ and $y=-1$, around the $x$-axis, which crosses the region.

enter image description here


Volume:

I can just consider the region $T$ enclosed by the graphs $y=q(x),x=1,x=2$ and the $x$-axis (instead of $y=-1$) because the overlap (Volume of revolution on an area crossing the axis), seen precisely in the inequality $|q| > |-1|$ for $x = 1$ to $2$, means that the solid $P$ from the region $T$ is identical to $S$. That is $\text{Vol}(P) = \text{Vol}(S)$ because $P=S$?

Update: Wolfram (Solid P and Solid S) says $\pi + \frac{107 \pi}{15} = \text{Vol}(P) \ne \text{Vol}(S) = \frac{107 \pi}{15}$.

  • I notice $\pi$ is the volume of the solid $M$ obtained by rotating the square $N$ enclosed by $y=0$ and $y=-1$ from $x=1$ to $2$.

  • Could it be that Wolfram interprets the calculation for the volume of Solid S as $'\text{Vol}(S)' = \text{Vol}(P) + \text{Vol}(M)$ ?

  • Perhaps not because Wolfram looks like it's computing $$\int_1^2 \pi |\color{red}{-1} + (-q)^2| dx = \int_1^2 \pi |-(-1)^2 + (q)^2| dx$$ So, the '$\color{red}{-1}$' is actually '$-(-1)^2$' rather than the original '$y=-1$'? Hopefully, Wolfram assumes the axis doesn't cut the interior of the region.


Surface area:

For surface area, do we have $\text{SA}(P) = \text{SA}(S)$ because $P=S$ also?

Update: Wolfram (Solid P and Solid S) says $\text{SA}(P) = \text{SA}(S)$, but actually I think the answer should be $\text{the SA}(P)\text{ that wolfram gives} - 13 \pi$. Where does this extra $13 \pi$ come from please?

$\endgroup$
2
$\begingroup$

It appears that in the interpretation of Wolfram Alpha, two regions on opposite sides of the axis cancel each other out rather than simply reaffirming that the points swept by both are part of the solid.

Hence as we see in the figures on the two pages you linked to, the figure that is bounded at the $x$-axis produces a solid with no holes, whereas the figure bounded by $y = -1$ produces a solid with a cylindrical hole of radius $1$ drilled through the axis of the solid.

The result is even more complicated if the larger region extends farther from the $x$ axis in the positive direction for some values of $x$ and farther in the negative direction for other values of $x,$ as in this example, bounded by the line $y=-3$.

I suppose one could argue whether this is a good interpretation. It could come down to this: what is the purpose of rotating a region that crosses the axis of rotation?

For the surface area, Wolfram Alpha apparently counts all of the surface area of the solid, including the area at the flat "ends" of the solid, not just the parts formed by the rotation of parts of the curves $y=2x^2-7x+3$ and $y=-1.$ For Solid P, which does not have a hole drilled through it, the total surface area includes the disk of radius $2$ (area $4\pi$) at $x = 1$ and the disk of radius $3$ (area $9\pi$) at $x=2,$ contributing $13\pi$ to the total surface area of the solid. For Solid S, the hole removes a disk of radius $1$ from the disk at each flat end of the solid, which reduces the total area by $2\pi,$ but it adds the lateral area of the cylindrical hole, which increases the total area by $2\pi,$ so the final result is the same. This happens only because the length and radius of the cylindrical hole are equal; try merely increasing the radius of the hole (while still staying inside Solid P), for example using the line $y=-2$, and you will see a smaller total surface area.

$\endgroup$
  • $\begingroup$ David K, may I please clarify directly, wolfram is doing the same or doing the opposite of what David Quinn might suggest here? I believe you mean opposite. Just checking. Thanks! $\endgroup$ – Mitjackson Mar 2 at 17:45
  • 1
    $\begingroup$ It is doing the opposite. I would probably do the same as David Quinn unless I was given a reason to do it differently. I don't know why Wolfram Alpha does it differently, but it seems to be very much intentional, so I assume someone had a reason. Their method is somewhat like the even-odd rule for the interior of a polygon. $\endgroup$ – David K Mar 2 at 20:29
  • $\begingroup$ Thanks David K! $\endgroup$ – Mitjackson Mar 3 at 3:16

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.