Because the radius is changing as the rocket moves away, hence why the attraction force changes.

 

I've got more than

 

I'm doing #3. I got #1 right.

The answer to #1 ≈ 7.34 × 106 ft-lb

So for #3 I type in:

integrate (pi(12^2+(12-y)^2)(62.4)(23-y)) from 0 to 20 and I get 9.76518x10^6

the answer is ≈ 8.99 × 10^6 ft-lb

I know it has to do with how I'm calculating the radius. To calculate the radius I'm doing a triangle where x^2=r^2 which is 12^2-(12-y)^2=x^2

 

What book is this from?

 

my fukn math professors head

 

For #3 the surface area changes with the depth. Calculate the area as a function of y, then integrate A(y)*(y+3)*density*dy.

I always encourage my students to draw a picture, showing a representative slice of thickness Δy, the mass of that slice, how far that slice has to be moved (pumped), and so on.

 

Why y+3 and not 23-y?

 

I was assuming that y is measured vertically down from the top of the tank. It makes the calculation of the area of the slice a bit easier. In that case you will pump the water to the top of the tank (y) and an additional three feet.

As I wrote above: draw a picture. Define y in a way that makes it easiest for you to do the calculations.

 

Here's the picture I drew:





And I made the triangle where the radius was the hypotenuse and the "height" of the triangle is 12-y and the "width" of the triangle is x^2 which=r^2

 

Here's mine: