Quiz 6.5:
Question:
A conical tank of radius 4 ft and height 10 ft is filled with water of weight density, \( w = 50 \mbox{lb} / \mbox{ft}^3 \) has its top 8 ft. below the ground. Set up the integral that would find the work done in pumping all the water to ground level. Also give the units that the answer would have.
Solution:
Setting a coordinate system with the origin and the tip of the cone. The cone can be seen as a volume of revolution of the line \( y = \frac{5}{2}x \)
Since we will be integrating with respect to \( y \), we will write this as, \( x = \frac{2}{5}y \)
Taking a slice at \( y= c_k \) running \( \Delta y_k = y_k-y_{k-1} \), we approximate the work in lifting that slice as,
\( W_k \approx \pi \left ( \frac{2}{5} c_k \right )^2 \Delta y_k (50)(18-c_k) \)
The exact work then is \( W = \int \limits _ 0 ^{10} \frac{200\pi}{25}y^2(18-y)dy \)
With units ft-lb
Since we will be integrating with respect to \( y \), we will write this as, \( x = \frac{2}{5}y \)
Taking a slice at \( y= c_k \) running \( \Delta y_k = y_k-y_{k-1} \), we approximate the work in lifting that slice as,
\( W_k \approx \pi \left ( \frac{2}{5} c_k \right )^2 \Delta y_k (50)(18-c_k) \)
The exact work then is \( W = \int \limits _ 0 ^{10} \frac{200\pi}{25}y^2(18-y)dy \)
With units ft-lb
