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I was calculating the volume of water in a bowl, where I obtained $64\pi \; \text{units}^{3}$, a different part of the question wanted the volume in centimeters cubed, how would one convert from $\text{units}^{3}$ to $\text{cm}^{3}$? I looked it up all over the internet, there was unfortunately nothing helpful, I even tried to find a way in the book I was working on, there was also nothing.

Dewton
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    I think one would need the type of "units" to solve this, like cubic inches, cubic feet, etc. – coffeemath Jul 20 '19 at 21:08
  • @coffeemath That is why I am stuck, the book does not mention the type, it just writes $\text{units}^{3}$ for problems regarding solids of revolution. – Dewton Jul 20 '19 at 21:10
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    Are the original dimensions, ie radius given in a defined unit like inches? – Gnumbertester Jul 20 '19 at 21:13
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    Your question lacks information. Could you provide the FULL original question? – PackSciences Jul 20 '19 at 21:16
  • @Gnumbertester In the first part of the question there were no units given, the question only asked to obtain the volume of revolution, a graph was given. The second part of the question however, gave the radius and depth of water in centimeters, which looked the same as the graph, does that mean I write my same answer but as $64 \pi \approx 201$? – Dewton Jul 20 '19 at 21:19

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I did not fully understand the context of the exercise, so I will make up my own example. Hope it clarifies.

Suppose e.g. you are asked to integrate to find the volume of the set bounded below by the paraboloid of revolution given by the parabola $y = x^2$, and bounded above by the plane $z = 5$. Then the height of this specific solid is $5 \text{ units}$, and its volume is $$V = \int_0^5 \pi (\sqrt y)^2 dy = \frac{25}{2} \pi \text{ units}^3. $$ Now say you are told that the height of the solid is $20 \text{ cm}$. What will its volume be in $\text{cm}^3$? Well, the unit conversion factor for linear dimensions is $$\left(\frac{20\text{ cm}}{5\text{ units}} \right) = \left(4\ \frac{\text{cm}}{\text{units}} \right); $$ in other words, $1 \text{ unit}$ is the same as $4 \text{ cm}$ (which could in principle be taken to be the definition of $\text{unit}$). From this we find that the unit conversion factor for volumes is $$ \left(4\ \frac{\text{cm}}{\text{units}} \right)^3 = \left(64\ \frac{\text{cm}^3}{\text{units}^3} \right). $$ Hence you may calculate $$V = \frac{25}{2} \pi \text{ units}^3 \left(64\ \frac{\text{cm}^3}{\text{units}^3} \right) = 800\pi \text{ cm}^3 \approx 2513.27 \text{ cm}^3. $$

giobrach
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