Math Is Fun Forum
You are not logged in.
- Topics: Active | Unanswered
Pages: 1
#1 Re: Help Me ! » Calculus/Physics Work » 2008-03-13 07:49:54
Where did you get 72 * pi? If you need find the area of each cross section (of a circle) wouldn't it be pi * r^2 (144 * pi)?
Also, you don't need to factor in the 32 (gravity) when dealing with lbs. Pounds is weight and not mass like kg.
#2 Re: Help Me ! » Calculus/Physics Work » 2008-03-13 06:54:38
Thank you. I have a similar problem, but don't have access to the answer and want to make sure I'm doing it properly:
A circular swimming pool has a diameter of 24 ft, the sides are 5 ft high, and the depth of the water is 4ft. How much work is required to pump all of the water out over the side? (use the fact that water weights 62.5 lb/ft^3)
#3 Re: Help Me ! » Calculus/Physics Work » 2008-03-13 06:19:30
Actually if you used 1/2 and 0 as the limits then you would end up with 1225 J
#4 Help Me ! » Calculus/Physics Work » 2008-03-12 17:14:34
- mathminor88
- Replies: 9
An aquarium 2 m long, 1 m wide, and 1 m deep is full of water. Find the work needed to pump half of the water out of the aquarium. (Use the fact that the density of water is 1000 kg/m^3)
This was what I thought, but the answer is: 2450 J
#5 Help Me ! » integration problem (area between 2 functions) » 2008-02-16 11:13:22
- mathminor88
- Replies: 1
Alright, this is driving me nuts, I can't seem to figure out why I can't get a correct answer. I think it just has to do with my algebra at the end of the problem.
Find the area between x = y^2 -2 and x = e^y from y=-1 to y=1
Now, the answer is suppose to be
what gives?
#6 Re: Help Me ! » improper integral type 2 » 2008-02-15 15:33:10
Alright, I don't know what I was doing here...
#7 Help Me ! » improper integral type 2 » 2008-02-15 15:22:31
- mathminor88
- Replies: 3
Can something please explain to me how I can finish this. I thought that for t being 6, I would get 1/0 = infinity. Then it would be positive infinity * a negative number and I would get zero, yet somehow the answer is suppose to be infinity.
#8 Re: Help Me ! » finishing off integral problem » 2008-02-06 11:23:10
Thank you.
I'll just put this proof in here that you used for future reference by anyone:
#9 Help Me ! » finishing off integral problem » 2008-02-05 09:31:29
- mathminor88
- Replies: 2
This is where I'm not sure what to do next.. I don't know if I'm suppose to use a "look-up" table or if there is a simple algebraic way to solve this. I'm not sure if I did the partial fraction expansion part completely correct either. Any advice would be helpful.
#10 Re: Help Me ! » trig integral » 2007-10-01 12:38:49
Sure, it's easy when you know how to do it.
I didn't know to change the tan^2 to it's sec equivalent.
Thanks.
#11 Help Me ! » trig integral » 2007-09-30 13:59:32
- mathminor88
- Replies: 4
use u = sec x to evaluate
#12 Help Me ! » trig identity integral » 2007-09-24 13:51:13
- mathminor88
- Replies: 2
#13 Re: Help Me ! » another integration problem » 2007-09-18 14:59:21
I just realized that the answer my TI-89 was giving me is the same thing I was getting, just in a different form 
Therefore
Therefore
#14 Help Me ! » another integration problem » 2007-09-18 13:03:38
- mathminor88
- Replies: 3
i think the proper way to solve this is using integration by parts, i've tried letting u=x, and dv=1/e^x but no prevail
#15 Re: Help Me ! » integration by parts » 2007-09-17 13:18:19
thanks for the tip
#16 Re: Help Me ! » integration by parts » 2007-09-17 09:26:14
My problem here was that I didn't know how to do
Thanks again!
#17 Help Me ! » integration by parts » 2007-09-17 07:47:46
- mathminor88
- Replies: 4
I've been stuck on this for a while now, I'm suppose to solve this using integration by parts
Pages: 1