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So I've got this shape

enter image description here

How would I calculate the volume? I thought about splitting it up into a cone somehow but I don't have the rest of the information to do that, I think...What's to do?

This is not homework.

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You can use calculus to solve it pretty easily, but the tag you used was algebra-precalculus. So, are you looking for a non-calculus based solution? – JEET TRIVEDI Apr 30 '14 at 16:47
Calculus would be fine. I'm not sure how to do it, though. I'm guessing putting it up in integrals somehow. I'll change the tags. – user3200098 Apr 30 '14 at 16:48
It's not clear whether $100$ is the distance between the two centers of the circles, or the length of the segment generating the conical surface. – Christian Blatter Apr 30 '14 at 18:04
Looks like the difference between two cones. – Floris Apr 30 '14 at 19:44
up vote 8 down vote accepted

enter image description here

Since the radii in your cylinder are different from one another, it can be seen as a cone with its tip cut off.

Drawing the complete cone, we notice that the tip, that has been cut off, is similar to the complete cylinder.

Naming points on the cylinder according to the picture above, similarity gives us that

$$\frac{AB}{CD} = \frac{BC+CE}{CE} \Rightarrow CE+BC=\frac{CE \cdot AB}{CD} \Rightarrow CE(1-\frac{AB}{CD}) = -BC \Rightarrow CE = \frac{BC \cdot CD}{AB-CD}$$

Using the known lengths we get

$$CE = \frac{BC \cdot CD}{AB-CD} = \frac{100 \cdot 6}{10-6} = \frac{600}{4} = 150$$

The Pythagorean theorem gives us the lengths DE and AE:

$$DE = \sqrt{CE^2-CD^2} = \sqrt{150^2-6^2} = \sqrt{22464}$$

$$AE = \sqrt{BE^2-AB^2} = \sqrt{(BC+CE)^2-AB^2} = \sqrt{250^2-10^2} = \sqrt{62400}$$

Now, the volume we are searching for is the volume of the tip subtracted from the volume of the entire cone. Using the formula

$$V=\frac{\pi r^2 h}{3},$$

where $r$ is the bottom radius and $h$ the height, to calculate the volume of the entire cylinder and the tip, we get that the searched for volume is

$$\frac{\pi AB^2 AE}{3}-\frac{\pi CD^2 DE}{3} = \frac{\pi 10^2 \sqrt{62400}}{3}-\frac{\pi 6^2 \sqrt{22464}}{3} = \frac{\pi}{3}(4000\sqrt{39}-864\sqrt{39}) = \frac{3136\pi\sqrt{39}}{3}$$

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You know that the radius has shrunk by 4 meters after lenght 100 meters. It seems as though the rate of shrinking is linear. How long until $raduis = 0$? Then you can split the figure into cones.

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$$\frac{10}{100} = \frac{6}{x}$$ does this work? – user3200098 Apr 30 '14 at 16:57
linear function: $r(x) = ax + b$ where $b = r(0) = 10$ and $a$ is the rate of change per meter, which is what you are looking for. Plug in the known point $r(100) = a*100 + 10 = 6$, solve for $a$ and you got your linear function of $r$. You can then find out how long the big cone is by solving for $r=0$ – apetiss Apr 30 '14 at 17:11

This is a frustum, whose volume is $$ V = \frac{\pi}{3} h (R^2 + r^2 + rR) = \frac{\pi}{3}100(10^2+6^2+10\times 6) = \frac{19600\pi}{3} $$

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That is correct, except the fact that the height $h$ does not seem to be $100$. According to the picture it rather looks like it is the segment connecting the edges of the bases that is $100$. – TheR Apr 30 '14 at 18:51

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