- #1
SPhy
- 25
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Even problem in textbook.
"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".
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, I am 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!
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, I am 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!