Cuboid water tank

I guess the fastest way to the solution: Notice that $\tan\theta = \dfrac{\text{base diagonal}}{80} = \dfrac{5}{8}$ . Also, we have:

$\begin{aligned} c' &= d\tan\theta \\ &= 50 \times \frac{5}{8} \\ &= \frac{125}{4}\end{aligned}$ .

Then, notice at the picture that we added the exact same volume of water, but this time we have a pretty straightforward formula ( $\text{base} \times \text{height}$ ): $\begin{aligned} 2V &= 30 \times 40 \times \dfrac{125}{4} \\ V &= 150 \times 125 \\ V &= \boxed{18,750}. \end{aligned}$

The volume and base area of the cuboid are

$V=30\cdot 40\cdot 80$
$a=30\cdot 40$

The short and long diagonals are

$d=\sqrt { { 30 }^{ 2 }+{ 40 }^{ 2 } }$
$D=\sqrt { { 30 }^{ 2 }+{ 40 }^{ 2 }+{ 80 }^{ 2 } }$

The angle of the tilt is

$\theta=ArcTan \left( \dfrac {d}{80} \right)$

The height of bottom vertex furthest from the floor is

$h=dSin(\theta)$

And the area of the surface of the water is

$A=aSec(\theta)$

From all this, we can find the volume of the water as follows

$v=\dfrac{1}{2}\left(V-\left(D-2h\right)A\right)\ = 18750$

A very nicely crafted problem with an integer result!

I found the length of a diagonal of a small face of the cuboid which is 50 cm (from the easy Pythagorean triple {30, 40, 50}). Then I imagined a coordinate plane through this diagonal and the opposite corner 80 cm edges at each end of this 50 cm diagonal such that the "space diagonal" was the segment from (0, 0) to (50, 80) in this plane. I then found the y-intercept of the line perpendicular to the space diagonal line with x-intercept of (50, 0). The y-intercept was 31.25 cm which is the highest point of the water on any of the 80 cm edges in the scenario presented. Now by symmetry , the volume of a cuboid 30 cm by 40 cm by 31.25 cm would contain twice the water volume of interest. Therefore, I calculated .5 times 30 times 40 times 31.25, giving me 18750.

In order to make the space diagonal vertical the cuboid (ABCDA'B'C'D') is rotated two times as shown below: If we look at the faces ABCD and A'B'C'D' respectively ( being directly above them), water level will rise upto A'E'FG Now the total volume of the water can be divided into two parts, the right traingular prism ABEA'B'E' and semi parallelopiped AEE'A'GF the base AEE'A' of which is a rectangle as shown in figure: Clearly volume of prism = 0.5 x 40 x 20 x 30 = 12000 (area of triangular base x height)

and volume of semi parallelopiped = 0.5 x 30 x 20 $\sqrt{5}$ x h (0.5 x area of rectangular base x height)

= 6750 ( h = 90/4 $\sqrt{5}$ )

Thus total volume = 12000 + 6750 = 18750

Construct a right handed coordinate system with the origin at the bottom vertex of the cuboid in such a way that the space diagonal is

$\begin{pmatrix} 40 \\ 30\\ 80 \end{pmatrix}$

The water surfaces are planes at right angles to this vector, and so they have equations

$40x+30y+80z=\text{constant}$

where increasing the constant increases the depth of water.

We want the surface passing through $(40,30,0)$ which fixes the constant. So the equation for the surface is

$40x+30y+80z=40^{2}+30^{2}=2500$

Now we want the volume of the part of the cuboid which lies beneath (i.e. closer to the origin) than this plane.

I found this volume with a sneaky computer assist by asking Wolfram Alpha (see note) to evaluate

Integrate[Boole[40x+30y+80z<2500],{x,0,40},{y,0,30},{z,0,80}]

and almost unbelievably $\boxed{18750}$ popped out just like that!

note.

In Wolfram Alpha, Boole[expression] returns 1 if the expression is true and 0 if it is false. So we are asking Wolfram Alpha to find the volume V by integrating $1 dx dy dz=dV$ over the region we want!

Let $L,W,H$ be the dimensions of the cuboid.

Choose a coordinate system with the origin at the highest point of the tilted base and axes along the directions of the cuboid. The water surface is a parallelogram through $(0,0,0)$ and with its center at $(L/2,W/2,z)$ , where $z$ is the average height of the water above the base (viewed along the $z$ -axis). The parallelogram is also perpendicular to the body diagonal $(-L,-W,H)$ .

This leads to the vector dot product equation $\left(\begin{array}{c} L/2 \\ W/2 \\ z\end{array}\right) \cdot \left(\begin{array}{c} -L \\ -W \\ H\end{array}\right) = 0,$ i.e. $-L^2/2 - W^2/2 + Hz = 0\ \ \ \therefore\ \ \ z = \frac{L^2 + W^2}{2H}.$ The volume of the water may be rearranged as a cuboid with length $L$ , width $W$ , and height $z$ (see drawing below): $V = LWz = \frac{LW(L^2 + W^2)}{2H}.$ In this case, $V = \frac{30\cdot 40\cdot(30^2 + 40^2)}{2\cdot 80} = \frac{1200\cdot 2500}{160} = \boxed{18\,750}.$

By projection, it is quickly deductible that the surface that the water adopts is that of a parallelogram . First, we will name the vertices of our interest and the measurements of some segments. Now we work with the condition that the diagonal is perpendicular to the parallelogram that forms the plane. We will do this for vertices $A$ , $B$ , $D$ . We will call $P$ , $P_1$ and $P_2$ at the points of intersection of the perpendiculars drawn from $A$ , $B$ and $D$ to the parallelograms, respectively. $l$ , $l_1$ and $l_2$ are the lengths of segments $AP$ , $BP_1$ and $DP_2$ , respectively. Using the similarity of triangles, we will relate to the triangles $P_1FB$ , $P_2GD$ , $PEA$ and $PAC$ with the triangle that is formed with the diagonal of the container. It is immediate obtain the following expressions: $\begin{array}{lcl} \ c=\dfrac{50\times 80}{10\sqrt{89}}\, \end{array}$ $\begin{array}{lcl} \ l=\dfrac{50\times 50}{10\sqrt{89}} \, \end{array}$ $\begin{array}{lcl} \ b=\dfrac{50\times 50 \times 50}{80\times 10\sqrt{89}} \, \end{array}$ $\begin{array}{lcl} \ a+f=\dfrac{50\times 50}{80} \, \end{array}$ $\begin{array}{lcl} \ f=\dfrac{l_1\times \sqrt{89}}{8} \, \end{array}$ $\begin{array}{lcl} \ a=\dfrac{l_2\times \sqrt{89}}{8} \, \end{array}$ To find the volume, it is necessary to know how much the area of the $ABFE$ trapezoid is, as it moves along the $z$ axis, understanding as axis $z$ the direction of the vector $AD$ Calculating the integral: $\begin{array}{lcl} \ 30\displaystyle\int_{0}^{40} \dfrac{a}{2} + f - \dfrac{f\cdot z}{40} \, dz\, \end{array}$ $\begin{array}{lcl} \ V = 30\cdot (20a-40f+20f) = 30\cdot 20 \cdot (a+f) = 30\cdot 20 \cdot \dfrac{50\times 50}{80} \ \, \end{array}$ $\begin{array}{lcl} \ \boxed{V=18750}\, \end{array}$ Sorry for the bad english.

We can put the tank on a three dimensional Cartesian plane so that its $30$ cm side is on the $x$ -axis, its $40$ cm side is on the $y$ -axis, its $80$ cm side is on the $z$ -axis, and the point that the tank is balanced on is at the origin, as pictured below:

Then the surface of the water is on a plane that is perpendicular to the vector $(30, 40, 80)$ and through the point $(30, 40, 0)$ and therefore has an equation of $3x + 4y + 8z = 250$ . This plane can be extended to intersect the axes at $(\frac{250}{3}, 0, 0)$ , $(0, \frac{125}{2}, 0)$ , $(0, 0, \frac{125}{4})$ , which along with the origin makes a right angle tetrahedron at a volume of $V = \frac{1}{6} \cdot \frac{250}{3} \cdot \frac{125}{2} \cdot \frac{125}{4} = \frac{1953125}{72}$ . The plane also intersects the edges of the tank at $(30, 0, 20)$ and $(0, 40, \frac{45}{4})$ to make two more smaller tetrahedra with volumes of $V = \frac{1}{6} \cdot \frac{160}{3} \cdot 40 \cdot 20 = \frac{64000}{9}$ and $V = \frac{1}{6} \cdot \frac{45}{4} \cdot \frac{45}{2} \cdot 30 = \frac{10125}{8}$ . The volume of the water is then the volume of the large tetrahedron minus the volumes of the two smaller tetrahedra which is $V = \frac{1953125}{72} - \frac{64000}{9} - \frac{10125}{8} = \boxed{18750}$ cubic centimeters.

Tilt the water, not the tank!   Suppose the space diagonal has endpoints at A(30,40,80) and the origin O. Assuming gravity acts in the direction from A to O, the surface of the water would lie in the plane normal to the vector N=((3,4,8)) and contain the point P(30,40,0). Using the standard equation for a plane give us 3 (x - 30) + 4 (y - 40) = 0. Solving for z and integrating over [0,30] × [0,40] finds the volume.

Let the dimensions of the tank be $a=30 \text{ cm}$ , $b = 40 \text{ cm}$ , $c = 80 \text{ cm}$ , and let's attach a reference frame $Ox'y'z'$ to the tilted tank, with its origin at the lowest point, and with its axes aligned to the edges of the tank. That is, the $Ox'$ axis is aligned with the $a$ edge, and the $Oy'$ axis is aligned with the $b$ edge , and the $Oz'$ aligned with the $c$ edge. It follows that the highest point of the tilted base corresponds to $(x', y', z') = (a, b, 0)$ .

The relation between the coordinates in the $Ox' y' z'$ frame and the absolute frame $O x y z$ is given by

$r = R r'$

Where R is a rotation matrix. From the condition given of the space diagonal, we know that

$R (a, b, c) = \sqrt{a^2 + b^2 + c^2} (0, 0, 1)$

So that,

$(a, b, c) = \sqrt{a^2 + b^2 + c^2 } R^T (0, 0, 1) = \sqrt{a^2 + b^2 + c^2 } (R_{31}, R_{32}, R_{33} )$

The water level throughout is the water level at $(x', y', z') = (a, b, 0)$

Hence,

$(x, y, z^* ) = R(a, b, 0) = R (a, b, c) - c R (0, 0, 1)$

which implies, $z^* = \sqrt{a^2 + b^2 + c^2} - c R_{33} = \sqrt{a^2 + b^2 + c^2} - \dfrac{c^2} {\sqrt{a^2 + b^2 + c^2}} = \dfrac{ a^2 + b^2 }{\sqrt{a^2 + b^2 + c^2} }$

The equation of the water level surface in the absolute frame is given by

$k^T r = z^*$ , where $k = (0, 0, 1)$ . Substituting $r = R r'$ gives $k^T R r' = z^*$

Letting $r' = (a/2 , b/2, z_1 )$ be the point corresponding to center of the base, and plugging this in the above plane equation,

$k^T R (a/2, b/2, z_1) = z^*$

so that,

$a/2 R_{31} + b/2 R_{32} + z_1 R_{33} = z^*$

Substituting for $R_{31}, R_{32}, R_{33} \text{ and } z^*$ results in,

$\dfrac{1}{2} \dfrac{ a^2 + b^2 + 2 z_1 c}{\sqrt{a^2 + b^2 + c^2}} = \dfrac{ a^2 + b^2 }{ \sqrt{a^2 + b^2 + c^2 }}$

Hence,

$z_1 = \dfrac{ a^2 + b^2 }{2 c }$

Finally, the volume is given by $V = a b z_1 = \dfrac{ a b (a^2 + b^2 ) }{2 c} = 18750$

Let V be the volume of the container and W the volume of water. Taking the filled part of the container and putting it on the top making the bottom and the top areas coincide you get an equivalent oblique parallelepiped with base area A (surface of water) and height equal to length L of the space diagonal minus the length of the projection P of the bottom face diagonal onto the space one. Such projection has a length of (using decimeters as units)

$P =\frac{\begin{pmatrix} 3 \\ 4 \\ 8 \end{pmatrix}\cdot \begin{pmatrix} 3 \\ 4 \\ 0 \end{pmatrix}}{L}=\frac{25}{\sqrt{89}}$ .

The water surface area is then found $A=\frac{V}{L-P}=\frac{3}{2}\sqrt{89}$ . Returning to our container if we remove the filled part and an equivalent part at the top we are left with another oblique parallelepiped with base area A and height L-2P. So the water volume is $W=\frac{1}{2}AP=\frac{75}{4}=18.75$ .

Summarizing in a formula if d is the space diagonal vector and b the base diagonal vector of the container we have

$W=\frac{V}{2}\frac{\mathbf{d}\cdot \mathbf{b}}{\mathbf{d}\cdot \left( \mathbf{d} - \mathbf{b} \right)}$ .

In summary, this method uses vectors to save having to visualise. If P is on the water surface and $\hat A$ a unit vector in the new vertical direction then the dot product $P . \hat A$ that projects P on to $\hat A$ gives the water height at P above the origin. This is the same at each intersection point. This allows the fill heights along the original verticals to be determined for a given tilt $\hat A$ .

Suppose the water intercepts the original verticals at $P0, P1, P2$ and $P3$ .

$P0 = (40,30,h_0) \\ P1 = (40,0,h_1) \\ P2 = (0,0,h_2) \\ P3 = (0,30,h_3)$

where all are coplanar and $h_i \geq 0$ .

The volume of water $V$ is a cuboid with base 40cm x 30cm and corner heights $h_0, h_1, h_2, h_3$ .

The volume is 40cm multiplied by the average of the areas along the 30cm sides.

$V = \frac{1}{2} ( \frac{h_0 + h_1}{2} . 30 + \frac{h_2+h_3}{2} . 30) . 40 = \frac{1}{4} (h_0 + h_1 + h_2 + h_3) . 30 . 40$

Note the symmetry between the $h_i$ and 30cm and 40cm.

Now we need to find the $h_i$ .

The vector $A = (40,30,80)$ is the new vertical and will be normal (perpendicular) to the water surface.

The dot products $P_i . A$ are the projections of the $P_i$ on to the vector $A$ giving the height of the $P_i$ above the origin multiplied by $|A|$ .

So the following dot products are all equal.

$P0 . A = 80 h_0 + 2500 \\ P1 . A = 80 h_1 + 1600 \\ P2 . A = 80 h_2 \\ P3 . A = 80 h_3 + 900 \\$

For minimum filling $h_0 = 0$ and $h_1 = 90/8, h_2 = 250/8, h_3=160/8$ .

So $V = \frac{1}{4} . \frac{500}{8} . 30 . 40 = 18750$

This method will work for other vectors A assuming they aren't degenerate cases e.g. (0,1,0) or (1,0,0).

Alternatively given a sufficient volume of water it will give the fill heights $h_i$ .

I work with a 3d coordinate system with vector positions r =(x,y,z). The lower vertex (the one with more hydrostatic pressure) is at the oringin and his opposed vertex at the position v =(30,40,80). Then, the plane of the surface of the water is perpendicular to v and passes through the position p =(30,40,0) so its equation is:

v ( r - p )=0 or

3x+4y+8z=250 [1]

Then we can think about the simetry so we compute the intersection h=125/4 of the plane with the z-axis (by plugging x=y=0 to [1] and solving for z) so that our final answer is V=30·40h/2=18750. Else we can isolate z from [1] and integrate it with respect to x and y from (0,30) and (0,40) respectevely to get the same final answer.