Volume of a cone using integrals

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Homework Statement



A water tank is shaped like an inverted cone with height 6 m and base radius 1.5 m

If the tank is full, how much work is required to pump the water to the level of the top of the tank and out of the tank?


Homework Equations



Integral of ( density * g (acceleration due to gravity) * A(y) (area of a cross section ) * change in y )

The Attempt at a Solution



The integral is from zero to 6 since this strange cone-shaped tank is 6 m high. g = 9.8 m/s^2, times the change in y which is (y-6) since the cone is 6m high

My problem is that I have no idea how to compute the cross-sectional area of a cone. The cuts are circles which have an area of pi*r^2. The tricky thing with this problem though is the radius does not remain constant from top to bottom.

My solutions manual gives the area as pi * (y^2)/16. I have no idea how they got to this. Any help would be very much appreciated.
 
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Draw the cone on your paper as a triangle facing down. Pick some point along the center axis. The radius is the distance from the center to the edge, and the height is the distance from the bottom to the current location.

Now you can use trigonometry to solve for the radius in terms of the height.
 
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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