What is the integral for finding the volume of a rectangular pool?

calculusisfun
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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. :)
 
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Yes, I think this is the correct way of doing it.
 
Well, that's good to hear. Any additional input? :)
 
Well, if you've seen multiple integrals. Then maybe you can also try to solve it with them. That would be a nice exercise :smile:
 
Thanks, any additional input from anyone?
 
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...

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