Hydrostatic Force and Center of a Mass - Calc 2

Keldroc
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Hey guys, I have a couple of problems about hydrostatic force and center of a mass that I was hoping someone could help me out with.

Center of mass

Homework Statement


Sketch the region bounded by the curves y=2x^3, x+y=3 and y=0. Find the coordinates of the centroid.

Homework Equations


Sorry I don't know how to make equations appear on here so I'll just type them out:
M=int f(x)-g(x)dx from a to b
x bar= 1/M * int x[f(x)-g(x)]dx from a to b
y bar= 1/M * int (1/2)[f(x)^2-g(x)^2]dx from a to b

The Attempt at a Solution


I used y=3-x for f(x) and y=2x^3 for g(x) and plugged them into the equations, used a=0 and b=3 and I got the wrong answer so I'm not sure what to do now.
Hydrostatic Force

Homework Statement


Find the following questions at the depth of 25m. The gravitational acceleration is g=9.8m/s^2 and the density of water is =1000kg/m^3:

Find the hydrostatic force on a square steel plate with sides 3m placed vertically.

Homework Equations


Pressure=\rhogd
F=Pressure * Area

The Attempt at a Solution


For pressure I did P=(1000kg/m^3)(9.8m/s^2)(25m)= 245000

Then, did Force=(245000 Pa)(3m * 3m)=2205000 N and it was the wrong answer.
 
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Keldroc said:
Hey guys, I have a couple of problems about hydrostatic force and center of a mass that I was hoping someone could help me out with.

Center of mass

Homework Statement


Sketch the region bounded by the curves y=2x3 x+y=3 and y=0. Find the coordinates of the centroid.


Homework Equations


Sorry I don't know how to make equations appear on here so I'll just type them out:
M=int f(x)-g(x)dx from a to b
x bar= 1/M * int x[f(x)-g(x)]dx from a to b
y bar= 1/M * int (1/2)[f(x)^2-g(x)^2]dx from a to b


The Attempt at a Solution


I used y=3-x for f(x) and y=2x^3 for g(x) and plugged them into the equations, used a=0 and b=3 and I got the wrong answer so I'm not sure what to do now.
I believe you are working with the wrong region. Did you sketch a graph of it? If you use vertical strips to calculate the area/mass, you will need two integrals, because the upper boundary changes at (1, 2). The region you should be working with has a sort of triangular shape (but with one curved side) and is bounded on its left side by y = 2x2 and on its right side by y = -x + 3.
Keldroc said:
Hydrostatic Force

Homework Statement


Find the following questions at the depth of 25m. The gravitational acceleration is g=9.8m/s^2 and the density of water is =1000kg/m^3:

Find the hydrostatic force on a square steel plate with sides 3m placed vertically.


Homework Equations


Pressure=\rhogd
F=Pressure * Area


The Attempt at a Solution


For pressure I did P=(1000kg/m^3)(9.8m/s^2)(25m)= 245000

Then, did Force=(245000 Pa)(3m * 3m)=2205000 N and it was the wrong answer.
 
Mark44 said:
I believe you are working with the wrong region. Did you sketch a graph of it? If you use vertical strips to calculate the area/mass, you will need two integrals, because the upper boundary changes at (1, 2). The region you should be working with has a sort of triangular shape (but with one curved side) and is bounded on its left side by y = 2x2 and on its right side by y = -x + 3.

Thanks for the response but I don't quite understand what you mean by setting up two integrals? Could you explain it to me?

Also I could still use help for the second problem. Thanks alot!
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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