Calculus II- Pressure and Force Question

[1LastTry]

Homework Statement

Find the force acting on one end of a cylindrical drum with 3m. If the rim is submerged horizontally into water so that the bottom is 10m deep.

Homework Equations

A(x) - Area
p(x) - Pressure = density * Depth

g= 9.8

A(x) =2$\sqrt{9-x^{2}}$

P(x) = 1000(7+x) * g

dF = 2000*g*$\sqrt{9-x^{2}}$(7+x) dx

F = 2000g $\int^{3}_{-3}$$\sqrt{9-x^{2}}$(7+x)dx

Finally, my question is why is the limit goes from -3 to 3? Where does it come from? A picture is in the attachment.

The Attempt at a Solution

I have no clue about this. I am ok with Integrals and finding the equations for the problem. However, I am not sure on how to find the limits of the definite integral, so hints and information on how to find the limits (such as -3 to 3) is very much appreciated. Thank you for your time.

 

[SammyS]

Homework Statement

Find the force acting on one end of a cylindrical drum with 3m. If the rim is submerged horizontally into water so that the bottom is 10m deep.

Homework Equations

A(x) - Area
p(x) - Pressure = density * Depth

g= 9.8

A(x) =2$\sqrt{9-x^{2}}$

P(x) = 1000(7+x) * g

dF = 2000*g*$\sqrt{9-x^{2}}$(7+x) dx

F = 2000g $\int^{3}_{-3}$$\sqrt{9-x^{2}}$(7+x)dx

Finally, my question is why is the limit goes from -3 to 3? Where does it come from? A picture is in the attachment.

The Attempt at a Solution

I have no clue about this. I am ok with Integrals and finding the equations for the problem. However, I am not sure on how to find the limits of the definite integral, so hints and information on how to find the limits (such as -3 to 3) is very much appreciated. Thank you for your time.

How is x defined in this problem?

Where is x = -3 ?

Where is x = 3 ?


By the way, I think you intended to say that the cylindrical drum has a radius of 3m .

 

[1LastTry]

How is x defined in this problem?

Where is x = -3 ?

Where is x = 3 ?By the way, I think you intended to say that the cylindrical drum has a radius of 3m .

Yes that is what I meant the radius is 3. And x is marked in the attachment.

And that is what I am asking: where did the prof get the -3 to 3 for the definite integral.

 

[SteamKing]

Quote: And that is what I am asking: where did the prof get the -3 to 3 for the definite integral.

Hint: the cylinder has a radius of 3 m. Doesn't that suggest why the limits are -3 to 3?

Do you understand how the integral is set up?

 

[1LastTry]

Quote: And that is what I am asking: where did the prof get the -3 to 3 for the definite integral.

Hint: the cylinder has a radius of 3 m. Doesn't that suggest why the limits are -3 to 3?

Do you understand how the integral is set up?

I know it suggests that it has something to do with the radius. I am unclear on how the integral is set up for this question, but I know what each number or variable is representing. The main thing I am not clear about is why -3 to 3? Why -3 to 3? Why not 0-6 or 4-10? Does it have to do with the area where we represented with x?

 

[SteamKing]

The reference point for x is the center of the circle. The circle has a radius of 3 m. The depth of water at the center of the circle is 4m + 3m = 7m. The pressure of the water is constant along a horizontal line, like the one shown in the picture. The pressure acting at x from the center of the circle is (x + 7)*1000*g N/m^2.
The equation of a circle with radius 3 is x^2+y^2 = 9. This implies that the half-width y of the circle at x is
sqrt (9 - x^2). The force at x is P*dA and dA = 2y*dx. In order to find the total force on the end of the cylinder, you must integrate from 3 m above the center of the circle to 3 m below. Given the definition of water depth relative to the center of the circle, -3 m is at the top closest to the water surface and 3 m is at the bottom.