a tank in the form of a cylinder has one hemispherical and one flat end. given that the diameter of the cylinder is 3.2 m and the overall length of the tank is 16.4 m, find the volume of the tank.
+131 You would probably have to make a few assumptions to solve this but
Volume of a cylinder
V= pi*r^2*h
Volume of a hemisphere
V = (2/3)*pi*r
radius of cylinder/hemisphere = 1.6m
height of cylinder = 16.4-1.6=14.8m
Total volume = pi*1.6^2*14.8 + (2/3)*pi*1.6
Total volume = 122.37m^3
+26400 a tank in the form of a cylinder has one hemispherical and one flat end. given that the diameter of the cylinder is 3.2 m and the overall length of the tank is 16.4 m, find the volume of the tank.
\(V_{\text{tank}}\) = volume of the tank.
L = length of the tank
r = radius of cylinder and hemisphere
h = length of the cylinder
\(\boxed{~ \begin{array}{lcll} V_{\text{cylinder}} &=& \pi \cdot r^2 \cdot h \qquad & | \qquad h = L - r\\ &=& \pi \cdot r^2 \cdot ( L - r ) \\ \hline V_{\text{hemisphere}} &=& \frac23 \cdot \pi \cdot r^3 \\ \end{array} ~}\)
\(\begin{array}{lcll} V_{\text{tank}} &=& V_{\text{cylinder}} + V_{\text{hemisphere}} \\ &=& \pi \cdot r^2 \cdot ( L - r ) + \frac23 \cdot \pi \cdot r^3 \\ &=& \pi \cdot r^2 \cdot L - \pi \cdot r^3 + \frac23 \cdot \pi \cdot r^3 \\ &=& \pi \cdot r^2 \cdot L - \frac33 \cdot \pi \cdot r^3 + \frac23 \cdot \pi \cdot r^3 \\ &=& \pi \cdot r^2 \cdot L - \pi \cdot r^3 \cdot \left( \frac33 - \frac23 \right) \\ &=& \pi \cdot r^2 \cdot L - \pi \cdot r^3 \cdot \frac13 \\ &=& \pi \cdot r^2 \cdot \left( L - r \cdot \frac13 \right) \\ &=& \pi \cdot r^2 \cdot \left( L - \frac{r}{3} \right) \qquad & | \qquad r = \frac{3.2}{2}\ m = 1.6\ m \qquad L = 16.4\ m\\\\ V_{\text{tank}} &=& \pi \cdot (1.6\ m)^2 \cdot \left( 16.4\ m - \frac{1.6\ m}{3} \right) \\ V_{\text{tank}} &=& \pi \cdot 1.6^2\ m^2\cdot \left( 16.4 - \frac{1.6 }{3} \right) \ m\\ V_{\text{tank}} &=& \pi \cdot 1.6^2\cdot \left( 16.4 - \frac{1.6 }{3} \right) \ m^3\\ V_{\text{tank}} &=& \pi \cdot 1.6^2\cdot ( 16.4 - 0.53333333333 ) \ m^3\\ V_{\text{tank}} &=& \pi \cdot 1.6^2\cdot 15.8666666667\ \ m^3\\ V_{\text{tank}} &=& \pi \cdot 2.56 \cdot 15.8666666667\ \ m^3\\ V_{\text{tank}} &=& \pi \cdot 40.6186666667\ \ m^3\\ \mathbf{ V_{\text{tank}} } & \mathbf{=} & \mathbf{127.607304799\ \ m^3} \\ \end{array}\)
