The ends of a "parabolic" water tank are the shape of the region inside the graph of $y = x^2$ for $0 \leq y \leq 9$; the cross sections parallel to the top of the tank (and the ground) are rectangles.
At its center the tank is $9$ feet deep and $6$ feet across. The tank is $9$ feet long.
Rain has filled the tank and water is removed by pumping it up to a spout that is $2$ feet above the top of the tank.
Set up a definite integral to find the work $W$ that is done to lower the water to a depth of 4 feet and then find the work.
My integral:
$$W = \int_4^9 62.5\times9\times2\times\sqrt{9-y^2}\times9\times(11 – y) \ dy $$
I did a cross-sectional slice of rectangles so $A(x) = L\times W\times H$
Length and height were givens, I found width using triangles and using the Pythagorean theorem.
Why is this wrong. I've seen that it's suppose to be:
[ 62.5] [ 18 ] [ 11 √y – y^(3/2) ] dy over y in [ 4 , 9 ]
But not sure why.
Please help
Thank you
Best Answer
The height of the water, $y$ (not the height of the tank) varies as water is pumped out of the tank.
The width of the tank for any $y$ is $2\sqrt y$
This comes from $y=x^2$ and considering the cross section of that parabola.
The length of tank is 9 feet.
The area of each cross section is $18\sqrt y$
The spout is 2 feet above the top of the tank (or 11 feet above the bottom).
$\rho$ is the density of water.
And we only need to pump the water down to the 4 foot mark.
$\rho\int_4^9 (11-y)(18\sqrt y)\ dy$