Calc question: Find the force against the vertical end of a tank containing water weighing 62.4 lb/ft^3 as an integral with respect to y
The tank end is shaped like an isosceles triangle. The base is 6ft long, and it is 8ft tall. The water only comes up to the 5ft mark

Question

Calc question: Find the force against the vertical end of a tank containing water weighing 62.4 lb/ft^3 as an integral with respect to y
The tank end is shaped like an isosceles triangle. The base is 6ft long, and it is 8ft tall. The water only comes up to the 5ft mark

anonymous · Answer

|dw:1425227014606:dw|

anonymous · Answer

@phi

anonymous · Answer

We were given the equation F=whA where F is the fluid force, w is the weight density of the water, h is the depth, and A is the area

phi · Answer

I don't understand what the shape of the tank is.  I see a triangle, but tanks are 3-d ?

anonymous · Answer

The tanks are 3D, but they only want us to find the fluid force against one side of the tank

phi · Answer

if we use x for width and y for height, then something like
$$ w \int \int y \ dx \ dy $$

anonymous · Answer

Why the second integral?

phi · Answer

I'm thinking a tiny area is dx * dy
we might be able to get by with one integral, but we would have to think about it.

anonymous · Answer

I tried the problem and got $$\int\limits_{0}^{5}62.4(8-y)(2)(-3/8 y+3)dy$$ but apparently the depth is supposed to be 5-y instead of 8-y, and I didn't understand why that is

phi · Answer

|dw:1425227979952:dw|
are you doing something like this ?