Find $V(x)$ if $V(x)$ is the volume of water of depth $x$ contained in a conical tank with vertice downwards. The tank is $8$ meters high and its diameter in the highest part is $6$ meters.
Answer:
$V(x)=3\pi\dfrac{x^3}{64}$.
I think that a sketch of the situation is:
I tried to use the formula of the conical volume: $V=\dfrac{\pi r^2h}{3}$, where $r=\dfrac62=3$ and $h=8$, but then $V=\dfrac{\pi3^28}3=24\pi$, which 1) does not depend on the depth and 2) does not have the same coefficients of the answer.
What am I doing wrong?
Thanks!
Your issue is that you found the volume of the whole conical tank, not the water.
Imagine taking a vertical cross-section of the tank:
The volume of the water is given by
$$V = \frac 1 3 \pi r^2 x$$
What is $r$? It can be shown that the triangle formed by the water is similar in the geometric sense to the entire triangle. Then we can set up a proportion:
$$\frac{\text{radius of the tank}}{\text{height of the tank}} = \frac{\text{radius of the water}}{\text{height of the water}} \implies \frac{3}{8} = \frac r x \implies r = \frac 3 8 x$$
Thus,
$$V = \frac 1 3 \pi \left( \frac 3 8 x \right)^2 x = \frac 1 3 \cdot \pi \cdot \frac{9}{64} \cdot x^2 \cdot x = \frac{3\pi}{64}x^3$$
matching the answer.