Work required to pump water out of a tank in the shape of a right circular cone

A tank in the shape of a right circular cone is full of water. The tank is 6ft. across the top and 8 ft. high. How much work is done in pumping water over the top edge.

(a) Set up the integral

(b) Solve using the graphing calculator.

I only need help with part (a). My professor gave the answer in class but I can't seem to get my answers to match. Her answer is: 4929

$$Water: 62.4 lb/ft^3$$ $$Radius: 3ft$$ $$Height: 8ft$$ $$\frac38 = \frac xy => x=\frac38y$$ $$w = \int_0^8 \pi x^2 dy (62.4)(8-y)$$ $$w = 62.4 \pi \int_0^8 \left({\frac38y} \right)^2(8-y)dy$$

Then I plug that into calculator. My answer is 1323.2 Which is wrong. I know my integral isn't set up correctly. So if you can help me out.

• As a start, define x and also y clearly first. – Mick Sep 29 '14 at 2:51
• x is $\frac38y$. What do you mean define x? I substituted x into the integral. – GiBiT 09 Sep 29 '14 at 3:07
• I mean what does x represent. Is it radius? height? amount of water pumped out? What about y? – Mick Sep 29 '14 at 3:38
• Your integral appears to be set up correctly. When you evaluate it, I think you get $(62.4\pi)(48)$. – user84413 Sep 29 '14 at 17:47
• @Mick x represents the radius and Y is the height. – GiBiT 09 Sep 29 '14 at 19:42

(1) Set up a representative disc as shown. It should be clear that its volume is $πx^2(dy)$.
(2) At that instant, $x = (3/8)y$.
(3) To empty all the water out, integrate $\int_0^8 πx^2(dy)$.
Note:- If work done is equal to rate times amount of water, then it is equal to $62.4 [(1/3) π 3^2 (8)]$ directly.