Rotated strings volume

It's obvious that each cross section of the figure parallel to bases is a circle. Then, it's evident that we should integrate area of these circles with respect to the height of the figure: $V_{f} = \int_{0}^{H}r(h)^{2}\pi\,dh$ where $r(h)$ is a radius of the circle at the height $h$ . From the corresponding problem in basic section , we know that $H = |AB|= \sqrt{2}$ , where $AB$ is a chord which joins two, $90$ degrees apart, points on the base circle. Also, as $H$ and $AB$ form isosceles right triangle, we have $h(x) = x$ where $x$ is the distance from $B$ along $AB$ . Observe how moving along $H$ is equivalent to moving along $AB$ . That allow us to rewrite our formula and instead of integrating with respect to $h$ , we integrate with respect to $x$ : $V_{f} = \int_{0}^{|AB|}r(x)^{2}\pi\,dx$ Animation of behavior of $r$ with respect to $x$ By cosine theorem we have: $\begin{aligned} r^2 &= x^{2} + 1^{2} - 2x\cos{45^{\circ}} \\ &= x^{2} + 1 - 2x\frac{\sqrt{2}}{2} \\ & = x^{2} - \sqrt{2}x + 1\end{aligned}$ The graph resembles the curvature of the figure at the given image Now, we integrate: $\begin{aligned} V_{f} &= \int_{0}^{|AB|}r(x)^{2}\pi\,dx \\ &= \pi\int_{0}^{\sqrt{2}}x^{2}-\sqrt{2}x + 1\,dx \\ &= \pi\int_{0}^{\sqrt{2}}x^{2}\, dx -\pi\sqrt{2}\int_{0}^{\sqrt{2}}x \,dx + \pi\int_{0}^{\sqrt{2}}1 \,dx \\ &= \pi\left [ \dfrac{x^{3}}{3} \right ]_{0}^{\sqrt{2}} - \pi\sqrt{2}\left [ \dfrac{x^{2}}{2} \right ]_{0}^{\sqrt{2}} + \pi [x]_{0}^{\sqrt{2}} \\ &= \frac{2}{3}\pi\sqrt{2} - \pi\sqrt{2} + \pi\sqrt{2} \\ &= \frac{2}{3}\pi\sqrt{2} \approx \boxed{2.961922}\end{aligned}$

Consider a top view and side view of one of the strings. From the top, the string appears to have a length of $\sqrt{1^2 + 1^2} = \sqrt{2}$ (as a hypotenuse of a right isosceles triangle with unit legs). From the side, the height of the 3D figure would be the leg of the right triangle that has a hypotenuse of the actual string length of $2$ and a leg of the apparent top view string length of $\sqrt{2}$ , which would be $h = \sqrt{2^2 - (\sqrt{2})^2} = \sqrt{2}$ , making another right isosceles triangle.

Let $x$ be the vertical distance away from the center of the 3D figure. Then another right triangle is formed in the top view with $x$ and $\frac{\sqrt{2}}{2}$ as legs and the radius $r$ of each horizontal circle cross-section as a hypotenuse, and so $r^2 = x^2 + (\frac{\sqrt{2}}{2})^2 = x^2 + \frac{1}{2}$ .

The volume of the upper half of the 3D figure is then $V = \pi \int_{0}^{\frac{\sqrt{2}}{2}} x^2 + \frac{1}{2} dx = \frac{\pi \sqrt{2}}{3}$ , and since the 3D figure is symmetrical, both halves would have a volume of $V = 2 \cdot \frac{\pi \sqrt{2}}{3} \approx \boxed{2.96192195877}$ .

My solution is identical to David Vreken's but I thought this problem was fun so I made a Desmos graph to show the hyperbola.

Here it is.

The dotted circle is the top view with two points that are 90 degrees apart going around it.

The purple horizontal lines transfer the y-coordinates to the side views of the circles (that appear as black line segments.) Each purple intersects the opposite side.

Connecting these points of intersection is the blue line representing the strings.

I also added the hyperbola in blue so you can see it being traced out. You can also click #13 to toggle the shape of interest.

Desmos needs a way to have thicker and thinner lines and a wider color palette.

Had it not been the last problem of the advanced section I would have done it much Earlier; Anyways Here's what I have done. $$ Part 1 $$ Consider the bottom circle to be the reference and let $h$ be the height you rise from the bottom plane; From here onwards assuming that you know the distance between the two planes is $\sqrt{2}$ ; Now, why does the cylinder squeeze before all? It is because when you consider a point $P(x_0,y_0)$ on the rim of the bottom circle (say $C_0$ ) it is mapped to some point $Q(x_1,y_1)$ on the top circle say $C_1$ by some transformations which can be as purely as translational as $T: \begin{cases} x=x_0 \pm \lambda \\ y=y_0 \mp \lambda \end{cases}$ and; Now we see that by rotating the top circle by $\frac{\pi}{2}$ we applied the transformation $T$ on all points $(x,y)$ on the line PQ. I hope this suffices to elucidate why the cylinder squeezes on torsion. $$ Part 2 $$ Let $y$ be the distance a point $X(x,y)$ travels radially while moving a distance $h$ up the cylinder; Now proportionally, we can say that $y=h.\sqrt{2}$ because when $h=\sqrt{2}$ , $y=2$ . Now let $A$ be the area of the disc with $r=(1-y)$ as radius. So $A=\pi.(1-h.\sqrt{2})^2$ .Consider an elemental height $dh$ at a distance $h$ from the base of the cylinder. The volume of this element is $dv=\pi.(1-h.\sqrt{2})^2.dh$ . Integrating this element from $0$ to $\frac{\sqrt{2}}{2}$ and letting the total volume to be $V$ we get $\frac{V}{2}=\pi.\frac{\sqrt{2}}{3}$

this gives $V=2\pi\frac{\sqrt{2}}{3}=2.96192195877$ and the Last problem of Advanced section is down

Let's look at both circles on top view:

• The Red X would be the tip of the string at the lower circle and the Red dot would be the tip of the string at the upper circle.

Before Rotation:

After Rotation: It would seem to appear then that it moved along the X direction by a distance of √2.

Using the information that the string length was 2 and it moved along X by √2. We can calculate the height. Side View:

$H^{2}$ = $2^{2}$ - $√2^{2}$

$H^{2}$ = 4 - 2

$H$ = √2

As the top circle rotates to 90 degress, a smaller circle begins to emerge at the middle of the whole object.

Since the whole object was a prismatoid, we can use the Prismoidal Formula : V = $\frac{H(A_{1} + A_{m} +A_{2})}{6}$

V = $\frac{√2( π+4(0.5π)+π)}{6}$

V = $\frac{2√2π}{3}$ = $\boxed{2.96192195877}$

Setting the bottom circle on the xy plane with its centre at the origin, the z axis going through centre of the volume through the top circle's centre. We know the hieght is $\sqrt2$ from a previous basic question. The volume is V=\pi\int_0^\sqrt2{x^2+y^2}dz where $x$ , $y$ and $z$ follow one of the lines and we're simply integrating the area of the circles times $dz$ from bottom to top. One of the lines passes from $(1, 0, 0)$ to $(0,1,\sqrt2)$ . Its parametric equation is $(1-t,t,\sqrt2t)$ . Substituting $t$ into $V$ yields: $V=\pi\int_0^1{(1-t)^2+t^2}\sqrt2dt = \sqrt2\pi\int_0^11-2t+2t^2dt =\sqrt2\pi(t-t^2+\frac23t^3)|_0^1=\frac23\sqrt2\pi\approx2.961921958772244$

I solve it by using the 3-dimensional coordinate system.

We have x-axis and z-axis on the screen and y-axis toward into the screen. It is easy to see that the distance between two rings be $\sqrt2$ (You can see David's solution), then let an arbitrary red line have intersection $A$ and $B$ on the top ring and the bottom ring respectively.

Then, we can set $A(\cos\mu,\sin\mu,\frac1{\sqrt2})$ , $B(-\sin\mu, \cos\mu, -\frac1{\sqrt2})$ by the knowing that $-\frac{\pi}2\leq\mu\leq0$ . This is because we want to determine the function of the right-side curve in terms of $x$ and $z$ . But we will end up in parameter form, it is easier.

Let a point $P$ be the intersection point of line $AB$ and the plane $y=0$ , that is $P(x,0,z)$ , we have $\begin{aligned}\overrightarrow{AP}&=k\overrightarrow{AB}\\<x-\cos\mu,-\sin\mu,z-\frac1{\sqrt2}>&=k<-\sin\mu-\cos\mu,\cos\mu-\sin\mu,-\sqrt2>\end{aligned}$ We will get the expression of $k$ : $k=\frac{\sin\mu}{\sin\mu-\cos\mu}$

Substituting these into the expression of $x$ and $z$ get $x=\cos\mu-k(\sin\mu+\cos\mu)=\frac1{\cos\mu-\sin\mu}=\frac1{\sqrt2\cos(\mu+\frac\pi4)}$ and also $z=\frac1{\sqrt2}-\sqrt2k=\frac1{\sqrt2}\tan(\mu+\frac\pi4)$

For the sake of simplicity, we use $\mu+\frac\pi4=\theta$ and change the range to $-\frac\pi4\leq\theta\leq\frac\pi4$ , we get simpler parameter form of $x$ and $z$ : $x=\frac1{\sqrt2}\sec\theta,z=\frac1{\sqrt2}\tan\theta$ As you can see, this is the parameter form of hyperbola .

Do some integral to get the volume:

$\begin{aligned}\because z&=\frac{1}{\sqrt2}\tan\theta\\\therefore dz&=\frac{1}{\sqrt2}\sec^2\theta d\theta\\\therefore V&=\int\pi x^2dz\\&=\frac{\pi}{2\sqrt2}\int_{-\frac\pi4}^{\frac\pi4}\sec^4\theta d\theta\\&=\frac{\pi}{2\sqrt2}\int_{-\frac\pi4}^{\frac\pi4}(1+\tan^2\theta)d\tan\theta\\&=\frac{\pi}{2\sqrt2}\bigg(\tan\theta+\frac13\tan^3\theta\bigg)\bigg|_{-\frac\pi4}^{\frac\pi4}\\&=\boxed{\frac{4\pi}{3\sqrt2}}\end{aligned}$

I don't know if it was fair, but instead of reinventing the wheel I remembered the formula

$V=\frac{(2a^2+r^2)h\pi}{3}$

where $a=\frac{1}{\sqrt{2}}$ is the radius at the middle height.

It's easy to remember: if a=0, you get two cones, if a=r you get a cylinder.