Usually when your calculating work, energy, and the like, the choice of coordinate system is arbitrary, although a strategic choice can result in drastically simpler solutions. I've attached the one that I used.
Ok so heres the problem: The inner surface of a dog bowl named Gabby is hemispherical, with a radius of 5 inches. If it started out filled to the rim, and I let Gabby drink until the bowl was empty, how much work did she do if she lapped it up with her tongue while keeping her mouth level with the rim of the bowl? (Water has a density of 62.4 pounds/ft^3)
Alright so this is calc. 2, which means the general formula for work is W=Force*Distance. Force is really volume*density. So W=(volume)(density)(distance).
Now, I believe the volume of a hemisphere is V=2/3*pie*r^3. I converted the 5 inches to feet, so thats .42 feet. Now I plugged this into the V formula.
V=2/3*pie*(.42)^3. This yielded .155. I then multiplied this by the density and got 9.67 lbs *(change in X). I am stumped as to what to do next because the y coordinates arn't given, so the distance cannot be retrieved. Please help, thanks.
Thank you so much for your reply and calculations . I now realize that the y coordinates are arbitrary. I also didn't realize that the radius we were given was that side rim. I also didn't realize that pythag's theorum could be used even when the rim is curved. What I didn't exactly understand is why the distance, which is h, is 5-x. Regardless, I understand most of it now, thanks a lot.
You're right to be confused about why . That's a mistake on my part. It would be true if the origin were placed at the bottom of the bowl, but in the system that I used it should be . Sorry!
The reason for the appearance of the "Pythagorean" theorem in this case is because any point on the the surface of the bowl is on the circle defined by . If the bowl had been parabolic or cylindrical, this part would have been different.