Show Posts

This section allows you to view all posts made by this member. Note that you can only see posts made in areas you currently have access to.


Messages - Retain

Pages: [1]
1
Math and Other Problems / Help on another problem (Sorry! :( )
« 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."



2
Math and Other Problems / Need help on 2 problems involving disk/shell
« on: December 22, 2006, 02:45:31 pm »
Hi again, so I decided to finish my math homework today and get it over with, but had problems with 2 problems on the hw :P.

Problem #1 (Using Shell method):

x+y^2=9, x=0, revolve about the x-axis  (Find the volume)

The graph of this (I think) would be a parabola on it's side (like a U on it's side would look like a C)  I used the interval [-3,3] because it crosses the y axis or x=0, at -3 and 3. So I integrated (y)(9-y^2) and all that, but end up with 0 as my answer. Any idea what I did wrong?


Problem #2 (Using disk, well it was in the disk section, not sure if you need to use it or not)

"The plane region by y = sqrt(x), y=0, x=0, and x=4 is revolved about the x-axis.  (a) find the value of x in the interval [0,4] that divides the solid into two parts of equal volume.  (b) find the values of x in the interval [0,4] that divide the solid into three parts of equal volume."

I have no idea what to do with this one.  Do you just integrate sqrt(x) and then substitute the x value with 0, 1, 2, 3, and 4?

3
Thank you so much!!!!!!!!  ;D

4
Hi there!!! I'm assuming I'm posting this in the right place, because it said post math questions here or something, if not sorry! :(.

Anyway I keep getting the wrong answer for this one question and would greatly appreciate it if someone can tell me what I'm doing wrong.

Directions:
For exercises 21-25, find the volume of the solid generated by revolving the region bounded by the graphs of the given equations about the line y=4

Question #21: y=x, y=3, x=0

The answer is 18pi but I get something else, dunno what I'm doing wrong!  Maybe I'm approaching it wrong or something.

What I did:
Well I need to find the outer and inner radius, so thats what I did!

Since it's revolving around y = 4, I made the outer radius (R) = 4, then squared 4 so it becomes 16 and integrated it using the interval [0, 3] and end up with 48pi.

For the inner radius (which I assume is the empty space), I made it = 4 - x
 -4 from the outer subtracted by the line y = x at the interval [0,3]

I then square (4-x), integrate it with the interval [0,3], ending up with 21pi.

Obviously I did something wrong because 48-21 isn't 18 :(.

Any idea what I did wrong?


Pages: [1]