Calculus - Hard Volume Problem :\

calculusisfun
Messages
30
Reaction score
0

Homework Statement


A pool in the shape of a rectangle is ten (10) m wide and twenty five (25) m long. The depth of the pool water x meters from the shallow part/end of the pool is 1 + (x^2)/175 meters.

Write a definite integral that yields the volume of water in the rectangular pool exactly. And then evaluate this integral.

2. The attempt at a solution

So, to find one section's volume I take the following integral: [PLAIN]http://img801.imageshack.us/img801/5991/calc1.png

So, that gives me one of the 25 foot long section's volumes. Thus, I multiply that integral by ten to yield the following: [PLAIN]http://img80.imageshack.us/img80/1236/calc2.png

I'm not sure if I interpreted the question the right way. Any explanations/help would be greatly appreciated. :)
 
Last edited by a moderator:
Physics news on Phys.org
Looks fine to me.
 
Didn't you already ask 2 days ago? It was fine by then, and it remains fine now :smile:
 
Haha okay thanks. Yeah, sorry micromass, I just wanted to get a little more input as I severely doubted I would have gotten it on my first try. :p

Thanks a bunch guys. :)
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

Similar threads

Replies
3
Views
6K
Replies
10
Views
3K
Replies
13
Views
3K
Replies
1
Views
3K
Replies
6
Views
3K
Back
Top