How much water? (Mathathon Problem 1)

I LOVE THE DREAM SMP!
There are many ways to look at this problem. Three good ways would be:

1. Directly calculate if you know the formula for frustum volume.

The formula for frustum volume, given bottom radius $R$ , top radius $r$ and height $h$ , is $V=\frac{1}{3}\pi h(R^2+r^2+Rr).$ Therefore we can plug in $\pi\approx \frac{22}{7}, ~~h=70,~~R=40,~~r=70$ to find the volume (must be converted to Liters): $V=\frac{1}{3}\times \frac{22}{7}\times 70(40^2+70^2+40\times 70=682000 (\text{cm}^3)=\color{#D61F06}682 \text{L}.$

2. Calculate difference of cones not knowing the formula for frustum volume.

Extending the cone by imagination and using similarity, we get the pic below: Let the height of the increased cone be $x$ . By similarity, we get $\frac{40}{x}=\frac{70}{70+x}$ . Solve this to get $x=\frac{280}{3}$ .
So the volume of the frustum is the volume of the larger cone minus the volume of the smaller cone (the formula for cone volume is $\frac{1}{3}\pi r^2h$ where $r$ stands for bottom radius and $h$ stands for height): $V_{\text{frustum}}=V_{\text{big cone}}-V_{\text{small cone}}=\frac{1}{3}\pi \times 70^2\times (70+\frac{280}{3})-\frac{1}{3}\pi \times 40^2\times \dfrac{280}{3}=682000(\text{cm}^3)~~~\color{#D61F06}\pi~\text{is replaced with}~\frac{1}{3}~\text{here}$

3. Use the ‘Baumkuchen’ integral.

$\scriptsize \color{#D61F06} \text{If under 18, do NOT do at home without adult supervision to avoid brain explosion.}$
This is mentioned in a book I read which finds volumes of 3D-shapes created by rotating the shape bounded by functions $f(x)$ and $g(x)$ in region $a\le x\le b$ : $V=2\pi \int_a^b|x||f(x)-g(x)| dx.$ The absolute value brackets are there to make sure $V$ is positive. THIS WORKS ONLY IF THE ROTATED BOUND IS ON THE SAME SIDE OF THE Y-AXIS.
Why is it called Baumkuchen?
It is because Baumkuchen is ‘log-like dessert’ in German, and the integral is like one!
Baumkuchen has log-like stripes Explanation:
We can split the integral into rings formed by the original bound split and rotated around the y-axis individually. Here I made a little mistake when labelling :P $x_i$ should be replaced with $dx$ :)
For simplicity, here I let $g(x)=0$ . Of course this can be generalised to other functions as well, given above.
If we cut a single ‘ring’ open, we get its volume by seeing it as a cuboid: I forgot the absolute value brackets :P

Here the volume of the cuboid is $2\pi |x||f(x)|$ because $g(x)=0$ and the $-g(x)$ term is therefore neglected. Summing infinite cuboids gives $V=2\pi \int_a^b|x||f(x)-g(x)| dx.$ So we can see the frustum as a cone chopped off from another as in example 2. This way, the volume of the big cone is the integral with $a=0,b=70$ , $f(x)=70$ and $g(x)=\frac{7}{3}x-\frac{280}{3}$ . Plug in these to get $V_1=2\pi\int_0^{70} |x||70-(\frac73x-\frac{280}{3})|=2\pi \int_0^{70} \frac{490}{3}x-\frac{7}{3}x^2\approx 2\times\frac{22}{7}(\left. \frac{245}{3}x^2-\frac79x^3\right|_0^{70})=75460009.$ Similarly we can apply the same to the small cone to get $V_2=\frac{1894000}{9}$ . $V_1-V_2=628000.$ Convert to liters: $628$ .

I used two different methods to obtain the answer.

The first one uses the volume of a cone in cm³.

I will try to remake it with bigger text, I couldn't use another definition because of the limit of the files...

The second one uses the volume of solids of revolution and is in dm³.

This method uses an integral to calculate the volume of a function in rotation around the axis.

I will explain why the volume of the frustum is approximately 682 litres using integrals. The main approach will be to think of it as being constructed out of many, many thin cylinders and then just take the sum of the volumes of these cylinders as their height $\text{d}h$ approaches zero.

$V = \int \text{d}V$ , where $\text{d}V$ is one of those cylinders. The general formula for the volume of a cylinder is $V = \pi r^2\cdot h$ . Therefore $\text{d}V = \pi r^2\text{d}h$ . However, $r$ is not constant, but grows linearly as the height increases. Using the general formula for a linear function and two known points, $r(h)=40+\frac{70-40}{70}h = 40+\frac{30}{70}h$ .

Pluggin the expression for $\text{d}V$ and $r$ into the integral, we find that $V = \int_{0}^{70} \pi r^2 \text{d}h = \pi\int_0^{70} 40^2 + \frac{2\cdot 30\cdot 40}{70}h + \frac{30^2}{70^2}h^2 \text{d}h .$ Which is of course $V = \pi\left[40^2\cdot 70 + \frac{30\cdot 40\cdot 70^2}{70} + \frac{30^2}{3\cdot 70^2}70^3\right]\approx 681725.61 cm^3 .$ Noting that we have unit $cm^3$ , we will divide by 1000 to simplify and yield a volume in litres, which is our final result and thus $V\approx 681.73 dm^3\approx 682 l$

You can easily derive the general formula using the same method. Just substitute $70=R,40=r,70=h$ et voilà!

The bucket is in the shape of a frustum as specified in the question.

Volume of a frustum $= 1/3 \times π \times h ( R^2 + r^2 + R \times r)$

where h is the height of the frustum, r is the bottom radius and R is the top radius.

Volume of the bucket $= 1/3 \times 22/7 \times 70 ( 70^2 + 40^2 + 70 \times 40)$ $=\dfrac{22 \times 70 \times (4900+1600+2800)}{21}$ $=\dfrac{22 \times 70 \times 9300}{21}$ $=682000 cm^3$

Since 1 L = 1000cm^3

Volume in litres $=\dfrac{682000}{1000}$

$\boxed{682L}$

check out this proof

Given that the volume of a cone is given by $\frac{1}{3}πa^2b$ where $a$ is the radius of the cone and $b$ is its perpendicular height. For a frustum of a cone as shown above, we can generalize its volume to be,

$\text{Volume of the larger cone- Volume of the smaller cone}$

$\dfrac{1}{3}πR^2t-\dfrac{1}{3}πr^2m$

Putting $t=h+m$

$\dfrac{1}{3}πR^2h + \dfrac{1}{3}πR^2m- \dfrac{1}{3}πr^2m$

$\dfrac{1}{3}πR^2h + \dfrac{1}{3}πm(R^2-r^2)$

Remember that in both the upper and the lower cone the triangles made by radii, respective perpendicular heights and slant heights (hypotenuses) must be similar triangles here, i.e,

$\dfrac{m}{r}=\dfrac{h+m}{R} \text{ provides } m= \dfrac{hr}{R-r}$

Putting the value of $m$ we just got in the equation above we have,

$\dfrac{1}{3}πR^2h + \dfrac{1}{3}π(\dfrac{hr}{R-r})(R^2-r^2)$

Since $R^2-r^2= (R+r)(R-r)$ , we get

$\dfrac{1}{3}πR^2h + \dfrac{1}{3}π(hr)(R+r)$

And finally the volume of a frustum of a cone,

$\boxed{\dfrac{1}{3}πh(R^2+r^2+Rr)}$

Here we have $R= 70cm, r=40cm, h=70cm$ , which gives volume from the formula,

$\dfrac{1}{3}π(70)((70)^2+(40)^2+(70×40)) ≈682000cm^3$

Since $1000cm^3= 1\text{ litre}$ we get, the volume of the frustum, $\boxed{682\text{ litres}}$

The shape of the bucket is frustum $\therefore \blue{Volume} \red{=} \dfrac{1}{3}\pi\times\red{height}(\red{bottom\space circle's\space radius}^2+\red{top\space circle's\space radius}^2+\red{top\space circle's\space radius \times bottom\space circle's\space radius})$ $=\dfrac{1}{3}\times\dfrac{22}{\cancel{7}}\times \cancel{7}0 cm (70^2cm^2+40^2cm^2+70*40cm^2)$ $=\dfrac{22}{3}\times 10 \times 10^2 \times (7^2+4^2+7\times 4) \red{cm^3}$ $=\dfrac{22}{3}\times [(7+4)^2 - 7\times 4]\times \red{10^3 cm^3}$ $=\dfrac{22}{3}\times [11^2 - 28]\times \red{L}$ $=\dfrac{22}{3}\times [121 - 28]\times {L}$ $=\dfrac{22}{3}\times \red{93} \times {L}$ $=\dfrac{22}{\cancel{3}}\times \cancel{\red{3}}\times\red{31} \times {L}$ $=22\times 31 L=\boxed{\red{682 L}}$

Most Common Solution -

$\text{Volume of frustum} = \dfrac{\pi h ((r_{1})^{2} + (r_{2})^{2} + r_{1}r_{2})}{3}$

$\begin{aligned} &\Rightarrow \dfrac{22(70) ((40)^{2} + (70)^{2} + (40)(70))}{3(7)} \\ \\ &= \dfrac{(1540)(9300)}{21} \\ \\ &= \dfrac{14322000}{21} \\ \\ &= 682000 \text{ cm}^{3}\end{aligned}$

$\boxed{682000 \text{ cm}^{3} = 682 \text{ L}}$

Ingenuity points if you prove the formula :)

(Here,) Frustum is a part of a cone which we get if we cut a cone in a plane parallel to the base.

• Consider the $\color{#20A900}{green}$ $\color{#20A900}{frustum}$ :

Volume = Volume( $\color{#CEBB00}{yellow}$ $\color{#CEBB00}{cone}$ ) - Volume( $\color{#EC7300}{orange}$ $\color{#EC7300}{cone}$ )

Volume of Cone = $\dfrac{1}{3}$ * π (radius of base )( perpendicular height )

Volume of $\color{#20A900}{green}$ $\color{#20A900}{frustum}$

= $\dfrac{πR^2(h + k)}{3}$ - $\dfrac{πr^2k}{3}$

= $\dfrac{π}{3}$ * (70^2 * 70 + 70^2 k - 40^2 k)

= $\dfrac{π}{3}$ * (70^3 + k(70^2 - 40^2))

= $\dfrac{π}{3}$ * (70^3 + k(70 + 40)(70 - 40)) ..... [reason: a^2 - b^2 = (a+b)(a-b)]

= $\dfrac{π}{3}$ * (70^3 + k(110)(30)) ----(eqn1)

we have to find the value of 'k'

• Consider the $\color{#69047E}{purple}$ $\color{#69047E}{triangle}$ & $\color{#D61F06}{red}$ $\color{#D61F06}{triangle}$ :

r is parallel to R .....[definition of frustum]

Therefore by Basic Proportionality Theorem , and the common angle

$\color{#69047E}{purple}$ $\color{#69047E}{triangle}$ ~ $\color{#D61F06}{red}$ $\color{#D61F06}{triangle}$ .....[ S-A-S test of similarity]

$\dfrac{k}{r}$ = $\dfrac{k + h}{R}$

$\dfrac{k}{40}$ = $\dfrac{k + 70}{70}$

$\dfrac{k}{40}$ = $\dfrac{k}{70}$ + 1

$\dfrac{k}{40}$ - $\dfrac{k}{70}$ = 1

7k - 4k = 280

k = 280/3

---

• Substituting 'k' in (eqn1) :

Volume of $\color{#20A900}{green}$ $\color{#20A900}{frustum}$ = $\dfrac{π}{3}$ * (70^3 + $\dfrac{(280)(110)(30)}{3}$ )

Volume of $\color{#20A900}{green}$ $\color{#20A900}{frustum}$ = $\dfrac{22}{7*3}$ * 7000(49 + 44)

Volume of $\color{#20A900}{green}$ $\color{#20A900}{frustum}$ = $\dfrac{(22)(1000)(93)}{3}$

Volume of $\color{#20A900}{green}$ $\color{#20A900}{frustum}$ = 682000 cm^3

1000 cm^3 = 1 liter

the bucket can hold 682 liters of water

If we substitute the values with variables then we can get the formula

The volume of the bucket can be found by subtracting 2 cones.

These are the 2 cones

• First, by similarity (between the small and big cones) we get, $\frac{x}{x+70}$ = $\frac{40}{70}$
• By solving the equation we get x as $\frac{280}{3}$ cm

Volume of cone = $\frac{1}{3}$ $\pi$ r²h

Volume of bucket

• = Volume of big cone - Volume of small cone
• = $\frac{1}{3}$ $\pi$ (70)²(70+ $\frac{280}{3}$ ) - $\frac{1}{3}$ $\pi$ (40)²( $\frac{280}{3}$ )
• = 682000 cm³
• = 682 L

Sorry if I posted my solution to problem n. 3 is because Brilliant won't let me write my solution on that problem (and I have it correct):

Do I get notified if I post an answer and new answers come?