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### 1 Answer

You need to evaluate the area of plate in y direction, such that:

`A = 4sqrt(1 - y^2)`

Evaluating the depth yields:

depth = 0 - y = -y

`F = pressure*A`

`pressure = (mass)/rho`

Considering the density of freshwater is `rho = 1000 Kg/m^3` and the mass = 1Kg yields:

`Force = 1/1000*4 sqrt(1 - y^2)*(-y)`

`F = int_(-1)^0 0.004 sqrt(1 - y^2)*(-y) dy`

`F = 0.004int_(-1)^0 -y*sqrt(1 - y^2)dy`

You should use integration by substitution, such that:

`1 - y^2 = t=> -2ydy = dt => -ydy = (dt)/2`

Changing the variable, yields:

`int (sqrt t)*(dt)/2 = (1/2) int t^(1/2)*dt `

`(1/2) int t^(1/2)*dt = (1/2)*(2/3)(t^(1/2+1))`

`(1/2) int t^(1/2)*dt = (1/3)t*sqrtt`

Replacing back `1 - y^2` for t yields:

`F = 0.004*(1/3)(1 - y^2)sqrt(1 - y^2)|_(-1)^0`

Using the fundamental formula of calculus, yields:

`F = 0.0013*(1 - (1 - 1)sqrt(1 - 1))`

`F = 0.0013 KN`

**Hence, evaluating the force exerted on one face of the plate yields, under the given conditions, consideringÂ the density of fresh water `rho = 1000 Kg/m^3` , **`F = 0.0013 KN.`