Work - Pumping Water

[krazydog]

I have a problem in my book that i cant seem to figure out how they got the radius or weight.

Problem: An open tank has the shape of a right circular cone(see figure). The tank is 8 feet across the top and 6 feet high.
How much work is done in emptying the tank by pumping the water over the top?

I will attach the figure from the book.
I took half of the right side of the shape and drew a rectangle so it could be revolved around the y axis to make its shape.

Volume of disk \(\displaystyle \pi r^2\)

the solutions manual gives \(\displaystyle \pi( \frac{2}{3}y )^2 dy \)

How are they getting 2/3 y?
they are also getting a weight of disk 62.4 not sure where that came from since its not specified.

 

[tkhunny]

Draw a right triangle representing one half the cross sectional area of the cone. The vertical leg is 6. The horizontal leg is...?

 

[krazydog]

HORIZONTAL radius is 2 but the solutions manual shows 2/3y

 

[galactus]

HORIZONTAL radius is 2 but the solutions manual shows 2/3y

This is a work problem, not a solid of revolution.


That's because they used similar triangles to eliminate a variable.

\(\displaystyle \frac{4}{6}=\frac{r}{y}\)

\(\displaystyle r=\frac{2y}{3}\)

Each disc is a circle with area \(\displaystyle \pi (\frac{2y}{3})^{2}\)

The weight of each disc is then \(\displaystyle 62.4\pi (\frac{2}{3}y)^{2}dy\)

The distance it is pumped is 6-y.

\(\displaystyle \displaystyle W=\frac{4(62.4)\pi}{9}\int_{0}^{6}\underbrace{(6-y)}_{\text{distance}}\cdot \overbrace{y^{2}}^{\text{weight}}dy\)

Similar triangles are handy in many problems. Especially those involving cones.