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There are many ways to look at this problem. Three good ways would be:

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): $V=\frac{1}{3}\times \frac{22}{7}\times 70(40^2+70^2+40\times 70=682000 (\text{cm}^3)=\color{#D61F06}682 \text{L}.$

Extending the cone by imagination and using similarity, we get the pic below:
$x$. 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):
$V_{\text{frustum}}=V_{\text{big cone}}-V_{\text{small cone}}=\frac{1}{3}\pi \times 70^2\times (70+\frac{280}{3})-\frac{1}{3}\pi \times 40^2\times \dfrac{280}{3}=682000(\text{cm}^3)~~~\color{#D61F06}\pi~\text{is replaced with}~\frac{1}{3}~\text{here}$

$\scriptsize \color{#D61F06} \text{If under 18, do NOT do at home without adult supervision to avoid brain explosion.}$

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!**

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:
Enlarge to see clearer :)

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: $628$.

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## Comments

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TopNewestYou are 13 and u know integration? Now that's CoOoOOOoOlll!

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Oh uhh yeaaah :)

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U know what I’m actually making a calculus note to help unsubscribed users like me :)

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Everything awesome in brilliant is here :)

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I'm already subbed know! :D

## Great job !

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顺便说一句，因为我年纪大了，所以您可以问我任何与科学或Python（或心理学）有关的问题。我很乐意为您提供帮助:) Hope it made some sense :P

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Zakir Husain, he sometimes needs help with solving stuff with programs :)

Besides you might like to know sirLog in to reply

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You know what?

1. I joined brilliant one year ago, when I lacked at calculus :P (Besides brilliant doesn’t allow me to chose 12-year-old! I changed it back on my birthday)

2. You’re the only one to realise my age

3. I am the youngest brilliant user known for the time being

4. I feel like Sheldon

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1.Haha what a fun story ,the best part was u telling that u were not so good in calculus at the time when I didn't even know what the heck it is XP.

2.Are u serious?I mean it's mentioned in almost all the ques posted by u¯\

(ツ)/¯.3.I didn't wanna say it but.....

## Aww

.4. Haha ,weird flex but okay.. [face-palm emoji] ᕙ( • ‿ • )ᕗ

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