Clan x86

General Forums => Academic / School => Math and Other Problems => Topic started by: Retain on January 10, 2007, 05:38:47 PM

Title: Help on another problem (Sorry! :( )
Post by: Retain on January 10, 2007, 05:38:47 PM
Hi again!  Anyway, how would I approach this question: (Involves using Disk method I guess)

"A tank on a water tower is a sphere of radius 50 feet.  Determine the depth of the water when the tank is filled to 21.6 percent of it's total capacity."


Title: Re: Help on another problem (Sorry! :( )
Post by: rabbit on January 10, 2007, 06:10:16 PM
Volume of tank * .216, find the radius.
Title: Re: Help on another problem (Sorry! :( )
Post by: Sidoh on January 10, 2007, 06:39:17 PM
Quote from: rabbit on January 10, 2007, 06:10:16 PM
Volume of tank * .216, find the radius.

I think "find the radius" is the part he's having trouble with.  You can't solve for it using volume of a sphere because that'd just give you the radius of a sphere that would have the volume of the water left.

Quote from: Retain on January 10, 2007, 05:38:47 PM
Hi again!  Anyway, how would I approach this question: (Involves using Disk method I guess)

"A tank on a water tower is a sphere of radius 50 feet.  Determine the depth of the water when the tank is filled to 21.6 percent of it's total capacity."

The disk method is used to calculate volume.  I can't think of a way to apply it to this problem, but maybe I'm overlooking something.
Title: Re: Help on another problem (Sorry! :( )
Post by: dark_drake on January 10, 2007, 08:40:20 PM
Anyways, here goes.  I'm not promising there aren't errors in my answer, but it made sense to me.  Ask if you need any explanations.

(http://sidoh.dark-wire.net/upload/viewitem.php?id=43)

Edit: When I said equation for area, I meant the equation of the line that when revolved around the y-axis will give you the sphere.
Title: Re: Help on another problem (Sorry! :( )
Post by: Nate on January 11, 2007, 03:41:48 AM
^That is wrong.  21.6% of the volume can not be 60% of the height in a uniform sphere because 50% of the volume = 50% of the height.

Find what the volume is, then take that and using the equation for a semicircle and the disk method solve for the downward shift.  Its kinda late now and I'm going abroad tomorrow but if you can wait, I'll actually solve it in a few weeks.  Also remember actual height is 25 minus downward shift.
Title: Re: Help on another problem (Sorry! :( )
Post by: dark_drake on January 11, 2007, 11:48:49 AM
Quote from: Nate on January 11, 2007, 03:41:48 AM
^That is wrong.  21.6% of the volume can not be 60% of the height in a uniform sphere because 50% of the volume = 50% of the height.

Find what the volume is, then take that and using the equation for a semicircle and the disk method solve for the downward shift.  Its kinda late now and I'm going abroad tomorrow but if you can wait, I'll actually solve it in a few weeks.  Also remember actual height is 25 minus downward shift.
After re-reading, I have no idea what the hell you're saying. Someone's very confused; I think it's you.  Test my integral by plugging 0 (the mid-point) in for z.  You will get half the volume. You can even plug in 50 and find the full volume.

I'm not trying to be rude, but your argument makes no sense. However, I'm feeling generous, so I present to you what you need to do to make this whole thing make sense: learn the difference between a radius and diameter.
Title: Re: Help on another problem (Sorry! :( )
Post by: Ender on January 11, 2007, 06:10:05 PM
Actually, (third edit) I got 20 ft. I integrated from a to r, and I got 30 for a.

here's my work (http://64.9.205.64/~andrew/math/depth.pdf)

And yeah, Nate, you confused radius with diameter.