Calculus II: Work Problem - Find Work in Pumping Water to Top of Tank

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"Work" (Calculus II)

This is a problem from the chapter called "Work":

A water tank in the form of an inverted right-circular cone is 29 ft across the top and 15 ft deep. If the surface of the water is 5 ft. below the top tank, find the work done in pumping the water to the top of the tank. Assume water weights W lbs. per cubic feet (which means your answer will be expressed in terms of W)


I have read it many times but I don't know what to do to solve it...please help, I really appreciate it.
 
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Work is weight times distance. At a given distance h below the top of the tank take a slice of the water of thickness dh. The weight is volume of the slice times W. The work to pump it to the top is h*(volume of the slice)*W. Now you need to integrate that over the range of h that includes all of the water in the tank. Is that enough to get started? It's a lot like finding the volume but with those extra factors of W and h.
 


thank you very much...i'm going to see if I can solve it using this :)
 
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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