How do I derive the volume of this cup using multivariable calculus?

Question

How do I derive the volume of this cup? It's been many years since I've taken calculus... So far I've started with the radius of the bottom and integrated that around the circle. Did I start it right? What's the next step?

Answer 1

That volume is radial symmetric around the $z$ axis. So you could describe it as stack of discs with volume $$ dV = A \, dz = \pi r(z)^2 \,dz $$ where $r(z)$ is the radius of such a disc at height $z$. Then integrate this from the lowest to the highest $z$ coordinate. $$ V = \pi \int\limits_{z_0}^{z_1} r(z)^2 dz $$

If the radius grows linear from $r_0$ to $r_1$ you can assume $$ r(z) = r_0 (1-t) + r_1 t = r_0 + (r_1-r_0)t \quad \mbox{with} \quad t = \frac{z-z_0}{z_1-z_0} $$ This results in \begin{align} V &= \pi \int\limits_{z_0}^{z_1}\!\! r(z)^2 dz \\ &= \pi (z_1-z_0)\int\limits_{0}^{1} \!\! \left( r_0^2 + 2r_0(r_1-r_0) t + (r_1-r_0)^2 t^2 \right) dt \\ &= \pi (z_1-z_0) \left[ r_0^2 t + r_0(r_1-r_0) t^2 + \frac{1}{3}(r_1-r_0)^2 t^3 \right]_{t=0}^{t=1} \\ &= \pi (z_1-z_0) \left( r_0^2 + r_0 r_1 - r_0^2 + \frac{1}{3}(r_1-r_0)^2 \right) \\ &= \frac{1}{3} \pi (z_1-z_0) \left( r_0^2 + r_0 r_1 + r_1^2 \right) \end{align}

Answer 2

The cup is in the shape of a Conical Frustum. If the cup has a radius of $r_1$ at the bottom and $r_2$ at the top and a height of $h$, then its volume is $V = \dfrac{1}{3}\pi h(r_1^2+r_1r_2+r_2^2)$.

You can derive this formula without any calculus if you know the formula for the volume of a cone.

In the diagram above, the volume of the frustum is the volume of the large cone minus the volume of the small cone, i.e. $V = \dfrac{1}{3}\pi h_1r_1^2 - \dfrac{1}{3}\pi h_2r_2^2$.

Using similar triangles, we have $\dfrac{h_1}{r_1} = \dfrac{h_2}{r_2} = \dfrac{h}{r_1-r_2}$. Hence, $h_1 = \dfrac{hr_1}{r_1-r_2}$ and $h_2 = \dfrac{hr_2}{r_1-r_2}$.

Thus, the volume of the frustum is: $V = \dfrac{1}{3}\pi h_1r_1^2 - \dfrac{1}{3}\pi h_2r_2^2$ $= \dfrac{1}{3}\pi \cdot \dfrac{hr_1}{r_1-r_2} \cdot r_1^2 - \dfrac{1}{3}\pi \cdot \dfrac{hr_2}{r_1-r_2} \cdot r_2^2$ $= \dfrac{1}{3}\pi h \dfrac{r_1^3-r_2^3}{r_1-r_2}$ $= \dfrac{1}{3}\pi h(r_1^2+r_1r_2+r_2^2)$.

EDIT: If you are required to use calculus, then ignore this. I'll just leave this answer here in case anyone who searches for the volume of a cup needs an explanation.

Answer 3

We can take another approach using Pappus' (second) centroid theorem. We can regard the conical frustum as a (right circular) cylinder with a radius $ \ r \ $ surrounded by a conical "skirt". The cross-section of this outer solid of revolution is then a right triangle of height $ \ h \ $ and base $ \ R - r \ $ , making its area $ \ \frac{1}{2} \ h \ (R - r ) \ $ . [We take this approach here to avoid the more complicated algebra of calculating the centroid of the trapezoid depicted -- more about this below.]

The aforementioned centroid theorem states that the volume of the solid is the area of the figure being revolved about an axis multiplied by the circumference of the circle that the centroid of the figure "traces" around that axis, thus, $ \ V \ = \ A \ \cdot \ 2 \pi \ \overline{x} \ $ .

In the diagram above, the centroid of the right triangle is marked by the asterisk. We may use the familiar geometrical theorem that the centroid of this triangle is one-third of the base length away from the vertical side, thus at a distance $ \ \frac{(R - r)}{3} \ $ from that edge. (This is also easily obtained using calculus.) This places the centroid at a distance from the vertical axis of revolution (in light blue) of

$$ \overline{x} \ = \ r \ + \ \frac{(R - r)}{3} \ = \ \frac{R + 2r}{3} \ \ . $$

By the centroid theorem then, the volume of the "skirt" is

$$ \left[ \frac{1}{2} \ (R - r) \ h \right] \ \cdot \ 2 \pi \ \cdot \ \left( \frac{R + 2r}{3} \right) \ = \ \frac{\pi}{3} \ h \ (R - r) \ (R + 2r) \ \ . $$

Adding this volume to that of the central cylinder (the cross-section of which is shaded in light blue), we find the volume of the frustum to be

$$ V \ = \ \pi \ r^2 \ h \ + \ \frac{\pi}{3} \ h \ (R - r) \ (R + 2r) \ = \ \frac{\pi}{3} \ h \ ( \ 3r^2 \ + \ R^2 \ + \ 2 \ R \ r \ - \ R \ r \ - \ 2r^2 \ ) $$

$$ = \ \frac{\pi}{3} \ h \ ( \ R^2 \ + \ R \ r \ + \ r^2 \ ) \ \ . $$

$ \ \ $

We could use Pappus' theorem in the other direction with our knowledge of the frustrum volume to find the distance of the trapezoid centroid from the axis of revolution as

$$ \overline{x_{trp}} \ = \ \frac{V_{fr}}{2 \pi \ A_{trp}} \ = \ \frac{\frac{\pi}{3} \ h \ ( \ R^2 \ + \ R \ r \ + \ r^2 \ ) }{2 \pi \ [\frac{R + r}{2}] \ h } \ = \ \frac{1}{3} \ \frac{ ( \ R^2 \ + \ R \ r \ + \ r^2 \ ) }{ (R + r) } \ \ . $$

This is in agreement with the earlier-evaded computation of the trapezoid centroid by "composition" of the rectangular and triangular regions:

$$ \overline{x_{trp}} \ = \ \frac{ (A_{r} \cdot \overline{x_r} \ + \ A_{tri} \cdot \overline{x_{tri}} ) }{A_{r} \ + \ A_{tri} \ } \ = \ \frac{ ( \ [\ r h \ ] \cdot [ \ \frac{r}{2} \ ] \ + \ [ \frac{(R - r) \ h}{2} \ ] \cdot [ \frac{R + 2r}{3} ] \ ) }{r h \ + \ \frac{(R - r) \ h}{2} \ } $$

$$ = \ \frac{1}{3} \ \cdot \ \frac{ ( \ 3r^2 \ + \ R^2 \ + \ 2 \ R \ r \ - \ R \ r \ - \ 2r^2 \ ) }{(R + r) } \ \ , $$

leading to the expression already shown.