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This might be a somewhat unorthodox question. I am wondering if there are any simple guidelines or tips/tricks to better understand volumes of solids of revolution.

E.g. the simple assignment and my own thoughts. Please correct me!

$\text{Let a and h be positive, let A be the area in the first quadrant bounded by }$$ax^2$$ \text{the y-axis and y=h.}$

The area $A$ is rotated about the $y$-axis. Find the volume of the solid created.

My initial thoughts would be to graph a rough sketch of the scenario. So I have a shape bounded by $ax^2$ on the right, by $x=0$ on the left, and $y=h$ at the top. The lower bound is not relevant, as it is 0 before it crosses the axis.

So, if I set $y=h \land y=ax^2$ to be equal, I can find the $x$-value where I need to integrate to, i.e. $ax^2=h \implies x=\sqrt{\dfrac{h}{a}}$.

My guess would be: $$\pi\int_0^{\sqrt{\tfrac{h}{a}}}(ax^2)^2 - (h)^2 dx \implies$$

$$\pi\int_0^{\sqrt{\tfrac{h}{a}}}(a^2x^4) - \dfrac{a^2h^2}{a^2}\,dx=\pi a^2\int_0^{\sqrt{\tfrac{h}{a}}}x^4-\dfrac{h^2}{a^2}\,dx = \pi a^2 \left( \int_0^{\sqrt{\tfrac{h}{a}}}x^4\,dx - \dfrac{h^2}{a^2}{\sqrt{\dfrac{h}{a}}}\right)$$

$$\pi a^2 \left(\dfrac{\sqrt{\tfrac{h}{a}}^5}{5} - \dfrac{h^2}{a^2}{\sqrt{\dfrac{h}{a}}}\right) = \dots$$

So, at this point I am both algebraically stuck and somewhat close to giving up. Any advice?

Cheers.

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Just to offer a closure to this question, this may benefit from the inclusion of diagrams. Erlend appears to be proposing the use of "disks", but it should be keep in mind that disk "slices" are always perpendicular to the rotation axis. Since that is "vertical" in this problem, the slices are "horizontal" and so will have "thicknesses" $ \ dy \ . $ So we will need to integrate in the $ \ y-$ direction, which requires expressing the disk radii as a function of $ \ y \ . $

enter image description here

The function inversion for the parabola is $ \ x = \sqrt{\frac{y}{a}} \ $ , and this is applicable over the entire interval in the $ \ y-$ direction, $ \ 0 \le y \le h \ , $ so the volume integral is

$$ \int_0^h \ \pi \ [x(y)]^2 \ \ dy \ = \ \pi \ \int_0^h \ \frac{y}{a} \ \ dy \ = \ \frac{\pi}{ a} \ \left( \ \frac{y^2}{2} \vert_0^h \ \right) \ = \ \frac{\pi \ h^2}{2 \ a} \ , $$

as already shown by Jean-Claude Arbaut.

[The apparent discrepancy in "dimensionality" is due to the fact that $ \ a \ $ has the "dimension" of inverse length. Our result is correctly a three-dimensional volume.]

Some of Erlend's confusion may be arising from the requirements for applying the "shell method." For that, the shell-wall "slices" are always parallel to the rotation axis.

enter image description here

The shell "thicknesses" are now $ \ dx \ , $ so the integration is carried out in the $ \ x-$ direction, which will use the single interval $ \ 0 \le x \le \sqrt{\frac{h}{a}} \ , $ as he was writing (but for the wrong method). It is the integrand which must be "split", as the "height" of the shells extends from the parabola "upward" to the "horizontal" line $ \ y = h \ . $ Here, we express the function for the parabola in terms of $ \ x \ , $ making the shell heights $ \ h - ax^2 \ . $

The radii of the shells again extend perpendicularly from the rotation axis, so these are given by $ \ r = x \ . $ The shell volume integration is thus

$$ \int \ 2 \pi \ r \ h \ \ dr \ \ \rightarrow \ \ 2 \pi \ \int_0^{\sqrt{\frac{h}{a}}} \ x \ \cdot \ (h - ax^2) \ \ dx $$

$$ = \ 2 \pi \ \int_0^{\sqrt{\frac{h}{a}}} \ hx - ax^3 \ \ dx \ = \ 2\pi \ \left( \ \frac{hx^2}{2} \ - \ \frac{ay^4}{4} \ \vert_0^{\sqrt{h/a}} \ \right) $$

$$ = \ 2 \pi \ \left( \ \frac{h}{2} \cdot \frac{h}{a} \ - \ \frac{a}{4} \cdot \frac{h^2}{a^2} \ \right) \ = \ 2 \pi \ \cdot \ \frac{h^2}{4a} \ = \ \frac{\pi \ h^2}{2 \ a} \ , $$

as found above.

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Hint: if a volume is defined by "stacked" plane surfaces of area $A(y)$ for $y$ from $0$ to $H$, then the volume is given by

$$V=\int_0^H A(y) \, \mathrm{d}y$$

Here, your plane surfaces are discs, whose radii are defined by the $y=ax^2$ curve.

That is, $R(y) = x=\sqrt{y/a}$, so the area is $A(y)=\pi R(y)^2=\pi y/a$, and

$$V=\int_0^{h} \pi \frac y a \mathrm{d}y$$

Can you conclude now?

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  • $\begingroup$ I can, sure. I do wonder how you see that solution though..? $\endgroup$ – Erlend Dec 18 '13 at 12:42
  • $\begingroup$ Imagine you draw your volume, and you "slice" it. Actually, I was looking for such a slicing, because I know this formula and it's one of the easiest to compute volumes. You have also Guldin theorems, but they would be uneasy here (you have to compute the position of the centroid of your piece of parabola). Another way is writing the volume "directly" as a double integral, but again it's not as simple. $\endgroup$ – Stop hurting Monica Dec 18 '13 at 12:45
  • $\begingroup$ So my approach isn't wrong, just a heck of a lot more complex than necessary? Bah! I still don't quite grasp this whole area of rotating shapes about axis'. $\endgroup$ – Erlend Dec 18 '13 at 12:48
  • $\begingroup$ Your idea isn't wrong, but I'm not sure I follow your computation ;-) $\endgroup$ – Stop hurting Monica Dec 18 '13 at 12:48
  • $\begingroup$ oh my, is it wrong? I was thinking splitting the integrals was a good idea. $\endgroup$ – Erlend Dec 18 '13 at 13:13

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