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I saw this one question that intrigued me, it had you revolve $\sqrt x$ around the axis from 0 to 9, and then revolve that again around the y axis. The only thing I could think of doing it was using the theorem of Pappus. I figured since we were revolving around the x axis I should find the volume of the first revolution which is simply $\pi\int_{0}^{9}(\sqrt{x})^2\text{d}x=\frac{81\pi}{2}$ then to revolve it again using the theorem of Pappus we need the center of mass of this first solid of revolution. I figured the solid of revolution will have the same abscissa but a different ordinate due to the circular path the ordinate takes. Using the formula $\bar{x}=\frac{\int_{a}^{b}xf(x)}{M}$, luckily we already know M and the top integral is simply $\frac{486}{5}$. Using this we get that $\bar{x}=\frac{18}{5\pi}$ now using our theorem of pappus formula we get $2\pi\bar{x}A=2\pi\cdot\frac{18}{5\pi}\cdot\frac{81\pi}{2}=\frac{1458\pi}{5}$. My first question, did I do this correctly? Also, is there a general way of doing this (solids of revolution around axes) without using multi-variable calculus? If not then please share the calc 3 method because I took the class and could not figure out how to apply it here. Thanks in advance.

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  • $\begingroup$ Three remarks: 1) If you have the half parabolic curve with equation $y=\sqrt{x}$, you have to revolve it only once around $oy$ to get a volume ? Why do you speak about a second revolution ? (2) What are you interested in ? A new method for computing a volume ? 3) Structure your presentation by changing lines frequently $\endgroup$
    – Jean Marie
    Mar 14, 2017 at 20:54
  • $\begingroup$ @JeanMarie 1) it's a double revolution because you go once around the x axis and the another time about the y axis. So you revolve to get a solid and then revolve again to get another solid. 2) I'm interested in finding the volume of this solid of revolution $\endgroup$
    – Teh Rod
    Mar 14, 2017 at 20:59
  • $\begingroup$ By "turning around the x-axis" do you mean 1) generating a curve (like gluing the two curves of $y=\sqrt{x}$ and $y=-\sqrt{x}$) and this curve, considered as a template, generates the volume by revolving it around y-axis, or 2) do you at once generate a paraboloid ? $\endgroup$
    – Jean Marie
    Mar 14, 2017 at 21:07
  • $\begingroup$ The big problem with your method is that Pappus works with rotating an area around an axis. Your calculation uses the volume. You plug in the value $\frac{81\pi}{2}$ (that you explicitly calculated as the volume of the object you get after the first revolution) for the area in the Pappus formula. $\endgroup$
    – Ingix
    Mar 15, 2017 at 14:22
  • $\begingroup$ Then what method would you recommend? @Ingix $\endgroup$
    – Teh Rod
    Mar 15, 2017 at 14:23

1 Answer 1

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If you revolve the area $\mathscr A$ under the curve $f(x)=\sqrt{x}$ from $0 \le x \le 9$ around the $x$-axis, you get a paraboloid $\mathscr P$. Each point of $\mathscr A$ will create a circle around the $x$-axis. For a point $(x_0,y_0) \in \mathscr A$ the circle will be the set of points $(x_0,y,z)$ with $y^2+z^2=y_0^2$. Since $y_0^2 \le x_0$, we have:

$$ (x,y,z) \in \mathscr P \Rightarrow 0 \le x \le 9, y^2+z^2 \le x.$$

On the other hand, if we have any point $(x,y,z)$ with $0 \le x \le 9, y^2+z^2 \le x$, then it is part of the circle created by the point $(x,\sqrt{y^2+z^2}) \in \mathscr A$ when rotating $\mathscr A$ around the $x$-axis. This means we have a complete description of the result of the first rotation:

$$ \mathscr P =\{(x,y,z) \in {\mathbb R}^3: 0 \le x \le 9, y^2+z^2 \le x\}.$$

If we do the second rotation of $ \mathscr P$ around the $y$-axis, we get a volume $ \mathscr M$. Clearly, $ \mathscr M$ can also be obtained by rotating some area $ \mathscr B$ that lies in the $x$-$y$-plane around the $y$ axis. If we knew $\mathscr B$, we could then apply the Pappus formula to get the volume of $ \mathscr M$. Since $\mathscr M$ is obtained by rotating $ \mathscr P$ around the $y$-axis, this unknown area $\mathscr B$ is the intersection of the $x$-$y$-plane and $\mathscr M$.

The question becomes now: When rotating $\mathscr P$ around the $y$-axis, what points on the $x$-$y$-plane can be reached? Obviously, the points of $\mathscr P$ that already are in the $x$-$y$-plane are among them, but there are more. For example, we have $(9,0,3) \in \mathscr P$ and if we rotate that point around the $y$-axis, we intersect the $x$-$y$-plane in $(3\sqrt{10},0,0)$, a point with an $x$ coordinate bigger than 9, which does not exist in $\mathscr P$!

More generally, if we rotate a point $(x_0,y_0,z_0) \in \mathscr P$ around the $y$-axis, we get all points $(x,y_0,z)$ with $x_0^2+z_0^2=x^2+z^2$. Since we are only interested in those points that are in the $x$-$y$-plane, we have $z=0$ and this means the points $\left(\pm\sqrt{x_0^2+z_0^2},y_0,0\right)$. There are 2 solutions as $\mathscr B$ is symmetric about the $y$-axis. We will only consider the positive sign from now on.

So we now need to solve the following problem: How can we describe the set

$$ S = \left\{\left(\sqrt{x^2+z^2},y,0\right),0 \le x \le 9, y^2+z^2 \le x\right\}?$$

Let $(r,s,0) \in S$. We know that $y^2 \le x \le \sqrt{x^2+z^2}$, so $s^2 \le r$ is a necessary condition. If, on the other hand, $0 \le r \le 9, s^2 \le r$, then we can choose $x=r,y=s,z=0$ which means that $(x,y,z) \in \mathscr P$ and $\left(\sqrt{x^2+z^2},y,0\right) = (r,s,0)$.

So in the interval $ 0 \le r \le 9$, the answer is

$$ S \cap [0,9] \times {\mathbb R} \times {\mathbb R} = \{(r,s,0): s^2 \le r\}.$$

Again, let $(r,s,0) \in S$, then we have $z^2 \le x$ and it follows $r = \sqrt{x^2+z^2} \le \sqrt{x^2+x} \le \sqrt{9^2+9} = 3\sqrt{10}$, because $x \le 9$. So the remaining open case is the interval $9 < r \le 3\sqrt{10}$.

Now we have $r=\sqrt{x^2+z^2} \le \sqrt{81+z^2}$, from which $z^2 \ge r^2-81$ follows. We have $s^2 = y^2 \le x - z^2 \le 9 -z^2 \le 9-(r^2-81) = 90-r^2$.

Now, on the other hand, let there be some $9 < r < 3\sqrt{10}, s^2 \le 90 - r^2.$ We choose $x=9,y=s,z=\sqrt{r^2-81}$ and find that $(x,y,z) \in \mathscr P$ and that $\left(\sqrt{x^2+z^2},y,0\right) = (r,s,0)$. That completes our determination of S:

$$ S \cap (9,\infty) \times {\mathbb R} \times {\mathbb R} = \{(r,s,0): 9 < r < 3\sqrt{10}, s^2 \le 90-r^2\}.$$

So the set S of points in the $x$-$y$-plane, that when rotated around the $y$ axis produces $\mathscr M$, is the set of points $(x,y)$ with $0 \le x \le 3\sqrt{10}$ between $f(x)$ and $-f(x)$, where

$$ f(x)= \sqrt{x}, 0 \le x \le 9$$ $$ f(x) =\sqrt{90-x^2}, 9 < x \le 3\sqrt{10}$$

You can now use Pappus formula for this area.

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  • $\begingroup$ I am thoroughly impressed $\endgroup$
    – Teh Rod
    Mar 15, 2017 at 16:13

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