# Hydrostatic Force set-up

Even problem in text book.

## Homework Statement

"Set up the integral to find the hydrostatic force on the face of the aquarium water tank, whose cross sectional area can be described by, y = e^-x^2 on 0.5≤x≤4.5 meters, resting at the bottom of the water 4 meters deep. Assume the bound y=0".

## The Attempt at a Solution

Using the formula

F = bounds[0,a] ∫ρg(a-y)(w(y))dy , essentially, depth x width x gravity x water density

To find width, write y= e^-x^2 in terms of x, yields, x=√ln(1/y) consider only positive values.

integral so far,

F= bounds[0,a] ∫pg(a-y)√ln(1/y) dy

My issue is finding a. Since the book gives the condition y=0, I assume the height of the water = height of tank, so can I say a=4?

Hopefully that made sense. In many problems done so far that are similar to this one, im finding the force on a window, so my A value on the integral bound is different from my A value in the depth expression. I might be thinking about this problem completely backwards, so any help or suggestions would be welcomed!

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haruspex
Homework Helper
Gold Member
It is not clear to me whether y is a vertical or horizontal coordinate. "Resting at the bottom" suggests x and y are both horizontal, but then the problem reduces to finding an area.

The textbook gives no diagram. Although on a problem where the Y axis was horizontal, a diagram was given. So I'm assuming the X axis is horizontal here. My guess is the tank is resting at the bottom of the body of water, 4 meters deep, so height of water=height of tank?

haruspex