Work and Fluid Force: Calculus II

[think4432]

I worked out these two problems and got an answer but was not sure about the answer...

If you could just check it and if it is not write tell me what I did wrong or give me a step by step, it would be greatly appreciated.

1. A tank is formed by revolving y = 3x^2, x = [0,2] around the y-axis is filled to the 4 feet point with water (w = 62.4 lb/ft^3). Find the work necessary to pump the water out of the tank over the top.

I got the integral being from a = 0 and b = 4 and integrating [12y^(1/2) - y^(3/2)]dy with w and pi as constants outside of the integral.

Would this be correct?

2. A plate shaped as in the figure [question 2.jpg is attached] is submerged vertically in a fluid as indicted. Find the fluid force on the plate if the fluid has weight density 62.4 lb/ft^3

The integral I set up was the limits being -5 to -1 and the integral being (5-y)(-7/4y - 7/4) dy.

Which is integrated out with the limits it comes out to be 665/6, and I was wondering if it was correct or not [the way I did it]

Thank you.

Can someone please step by step show me how to solve this if I am not correct.

Please help! I greatly appreciate it!

 

[LCKurtz]

I worked out these two problems and got an answer but was not sure about the answer...

If you could just check it and if it is not write tell me what I did wrong or give me a step by step, it would be greatly appreciated.

1. A tank is formed by revolving y = 3x^2, x = [0,2] around the y-axis is filled to the 4 feet point with water (w = 62.4 lb/ft^3). Find the work necessary to pump the water out of the tank over the top.

I got the integral being from a = 0 and b = 4 and integrating [12y^(1/2) - y^(3/2)]dy with w and pi as constants outside of the integral.

Would this be correct?

No, it wouldn't be correct. Figure out the mass of a circular slab of water of thickness dy and how far it must be lifted. Then integrate that.

 

[LCKurtz]

2. A plate shaped as in the figure [question 2.jpg is attached] is submerged vertically in a fluid as indicted. Find the fluid force on the plate if the fluid has weight density 62.4 lb/ft^3

The integral I set up was the limits being -5 to -1 and the integral being (5-y)(-7/4y - 7/4) dy.

Which is integrated out with the limits it comes out to be 665/6, and I was wondering if it was correct or not.

Why are you using 5 - y as the depth? When y = -5 that would be 10.

 

[think4432]

No, it wouldn't be correct. Figure out the mass of a circular slab of water of thickness dy and how far it must be lifted. Then integrate that.

Mass of the circular slab? Like a slice? How do we find the mass of that? Is that not just the 'w' or the 'weight density'

?

 

[think4432]

Why are you using 5 - y as the depth? When y = -5 that would be 10.

I am using 5-y for the depth because the top of the fluid to the bottom of the plate is y-5?

Is that not correct? I don't understand how it can be 10-y?

Is the other part of the integral correct for that part?

 

[LCKurtz]

Mass of the circular slab? Like a slice? How do we find the mass of that? Is that not just the 'w' or the 'weight density'

?

I should have said "weight". But anyway, you must multiply the weight density by the volume to get the weight. Your tank has circular cross sections being formed by revolving the curve. Your element of volume would be a cylindrical shaped disk of thickness dy. What is its volume dV?

 

[LCKurtz]

I am using 5-y for the depth because the top of the fluid to the bottom of the plate is y-5?

Is that not correct? I don't understand how it can be 10-y?

Is the other part of the integral correct for that part?

Did I say it should be 10 - y??

Apparently you have taken your origin to be at the water level above the point, judging by your equation of the slanted line. That's fine. What is the depth when y = -1? -2? -3? -4?

Use a formula that gives correct answers.

Yes, the other part looks OK.

 

[think4432]

I should have said "weight". But anyway, you must multiply the weight density by the volume to get the weight. Your tank has circular cross sections being formed by revolving the curve. Your element of volume would be a cylindrical shaped disk of thickness dy. What is its volume dV?

We set up the integral in the dy, right?

So we would use pi(r)^2 for the base of the circle/cylindrical disk
Meaning pi[ 1/[2(sqrt)3] y^(-1/2)]

Because I solved y=3x^2 for y and squared it, right?

And we multiply that by by (12-y)?

Would I have the weight (density X pi) / 18 root(3) outside the integral and y^(1/2)(12-y) dy inside to integrate?

I think I am more confused then I started. :[

 

[think4432]

Did I say it should be 10 - y??

Apparently you have taken your origin to be at the water level above the point, judging by your equation of the slanted line. That's fine. What is the depth when y = -1? -2? -3? -4?

Use a formula that gives correct answers.

Yes, the other part looks OK.

Wait...so, the answer I have is correct?

I don't understand this post by you.

Im sorry!

 

[LCKurtz]

We set up the integral in the dy, right?

So we would use pi(r)^2 for the base of the circle/cylindrical disk
Meaning pi[ 1/[2(sqrt)3] y^(-1/2)]

Because I solved y=3x^2 for y and squared it, right?

When you revolve it around the y-axis the radius is measured in the x direction.

 

[think4432]

When you revolve it around the y-axis the radius is measured in the x direction.

I do have it in the x direction...right?

Im not sure what you're trying to say...

 

[LCKurtz]

Did I say it should be 10 - y??

Apparently you have taken your origin to be at the water level above the point, judging by your equation of the slanted line. That's fine. What is the depth when y = -1? -2? -3? -4?

Use a formula that gives correct answers.

Yes, the other part looks OK.

Wait...so, the answer I have is correct?

I don't understand this post by you.

Im sorry!

No, your formula of 5 - y for the depth is not correct. What does it give when you try y = -1 or -2 or -3 or -4? Does it give the correct depth? Can you fix it?

 

[LCKurtz]

I do have it in the x direction...right?

Im not sure what you're trying to say...

I'm saying that the radius of your little disk is the x value not the y value of the point on your curve. Have you drawn a picture?

 

[think4432]

No, your formula of 5 - y for the depth is not correct. What does it give when you try y = -1 or -2 or -3 or -4? Does it give the correct depth? Can you fix it?

Why would I try y = -1 or -2 or -3 or -4?


I really appreciate your help!

 

[LCKurtz]

Why would I try y = -1 or -2 or -3 or -4?


I really appreciate your help!

Because in the coordinate system you have chosen, those are example values of y where the triangle is located.

 

[think4432]

Because in the coordinate system you have chosen, those are example values of y where the triangle is located.

So you want me to plug in those values in where to find the exact location of the depth?

 

[think4432]

I'm saying that the radius of your little disk is the x value not the y value of the point on your curve. Have you drawn a picture?

So do I solve the equation for y? And then do the integral?

I think setting up the integral is the hardest part! And I am getting so confused by looking in the book and trying to figure it out and trying to do what you are telling me to!

:[

 

[LCKurtz]

So you want me to plug in those values in where to find the exact location of the depth?

You are using the formula 5 - y for the depth. I am telling you that it doesn't work and you can check it for yourself by trying those example numbers and comparing what you get with the actual depth. You need to use a formula that gives the correct depth.

 

[think4432]

You are using the formula 5 - y for the depth. I am telling you that it doesn't work and you can check it for yourself by trying those example numbers and comparing what you get with the actual depth. You need to use a formula that gives the correct depth.

Oh. Ok, So the actual depth is 5 ft, correct?

So if I put 0 into y it will give me the actual depth? Is that what you're saying?

But how does that help at all?

 

[LCKurtz]

Oh. Ok, So the actual depth is 5 ft, correct?

So if I put 0 into y it will give me the actual depth? Is that what you're saying?

But how does that help at all?

In your choice of coordinates, the triangle sits at a level where the y values are between -5 at the bottom and and -1 at the top of the triangle, right? And the actual depth of points in the triangle range from 5 at the bottom to 1 at the top, measured in how far it is under water.

You need a formula that for any y between -5 and -1 gives the correct depth. 5-y doesn't do it. You are making it much harder than it is.

 

[LCKurtz]

I'm saying that the radius of your little disk is the x value not the y value of the point on your curve. Have you drawn a picture?

So do I solve the equation for y? And then do the integral?

I think setting up the integral is the hardest part! And I am getting so confused by looking in the book and trying to figure it out and trying to do what you are telling me to!

:[

You have to use x for the radius. If you are doing a dy integral you will have to get it in terms of y to integrate it, won't you.

 

[think4432]

In your choice of coordinates, the triangle sits at a level where the y values are between -5 at the bottom and and -1 at the top of the triangle, right? And the actual depth of points in the triangle range from 5 at the bottom to 1 at the top, measured in how far it is under water.

You need a formula that for any y between -5 and -1 gives the correct depth. 5-y doesn't do it. You are making it much harder than it is.

Would -5-y work as the formula? Because if I plug in -1 into y, it would be -4 which is still in the triangle? Is that what you're trying to say?

I don't know why, I just can't understand these 2 problems! I completed everything on the homework sheet, except these two!

 

[LCKurtz]

In your choice of coordinates, the triangle sits at a level where the y values are between -5 at the bottom and and -1 at the top of the triangle, right? And the actual depth of points in the triangle range from 5 at the bottom to 1 at the top, measured in how far it is under water.

You need a formula that for any y between -5 and -1 gives the correct depth. 5-y doesn't do it. You are making it much harder than it is.

Would -5-y work as the formula? Because if I plug in -1 into y, it would be -4 which is still in the triangle? Is that what you're trying to say?

I don't know why, I just can't understand these 2 problems! I completed everything on the homework sheet, except these two!

Here's what I want you to do. Fill in this table (actually do it):

\begin{array}{ccc}<br /> y-value &amp;|&amp; actual depth\\<br /> -1&amp;|&amp;?\\<br /> -2&amp;|&amp;?\\<br /> -3&amp;|&amp;?\\<br /> -4&amp;|&amp;?\\<br /> -5&amp;|&amp;?<br /> \end{array}

Then see if you can figure out a formula in terms of y that works.

 

[think4432]

You have to use x for the radius. If you are doing a dy integral you will have to get it in terms of y to integrate it, won't you.

Ok. So if I solve y = 3x^2 for x

I get x = y^(1/2) / root 3

And the formula is wpi(r)^2 dy

So I get w [pi (y^(1/2) / root 3))^2 dy]

So w [pi (y / (root 3)^2 dy]

?

[Probably totally wrong...but I gave it a shot with the equation solved for x, so it can be dy according to the picture I drew.

 

[think4432]

Here's what I want you to do. Fill in this table (actually do it):

\begin{array}{ccc}<br /> y-value &amp;|&amp; actual depth\\<br /> -1&amp;|&amp;?\\<br /> -2&amp;|&amp;?\\<br /> -3&amp;|&amp;?\\<br /> -4&amp;|&amp;?\\<br /> -5&amp;|&amp;?<br /> \end{array}

Then see if you can figure out a formula in terms of y that works.

6
7
8
9
10

y-10?

 

[LCKurtz]

Ok. So if I solve y = 3x^2 for x

I get x = y^(1/2) / root 3

And the formula is wpi(r)^2 dy

So I get w [pi (y^(1/2) / root 3))^2 dy]

So w [pi (y / (root 3)^2 dy]

?

[Probably totally wrong...but I gave it a shot with the equation solved for x, so it can be dy according to the picture I drew.

Of course, (root(3))² is just 3, no? So now you have the weight of the circular slab correct. (*Whew!*).

Now, given that, show me the integral you need to do to calculate the work lifting the water out the top.

 

[LCKurtz]

Here's what I want you to do. Fill in this table (actually do it):

\begin{array}{ccc}<br /> y-value &amp;|&amp; actual depth\\<br /> -1&amp;|&amp;?\\<br /> -2&amp;|&amp;?\\<br /> -3&amp;|&amp;?\\<br /> -4&amp;|&amp;?\\<br /> -5&amp;|&amp;?<br /> \end{array}

Then see if you can figure out a formula in terms of y that works.

6
7
8
9
10

y-10?

Why are you giving me these numbers for the depths? No part of your triangle is that deep.

 

[think4432]

Of course, (root(3))² is just 3, no? So now you have the weight of the circular slab correct. (*Whew!*).

Now, given that, show me the integral you need to do to calculate the work lifting the water out the top.

Yes. Whew. One part down...the other part to go.

Integrating a = 0, and b = 2, so integrating wpi [y/3 - (y-4)]

 

[think4432]

Why are you giving me these numbers for the depths? No part of your triangle is that deep.

y-1, if you plug in all the values...it would be on the triangle, except for -5.

For -5 to be on it...the formula just has to be 'y'

Which is probably not right...?

 

[LCKurtz]

Yes. Whew. One part down...the other part to go.

Integrating a = 0, and b = 2, so integrating wpi [y/3 - (y-4)]

No, not even close. For one thing, a dy integral requires y limits.

You have this disk slab of water whose weight is wπ/3. This disk is at location y vertically. How far does it have to be lifted to get out the top? Because, remember, work is force times distance.

 

[LCKurtz]

y-1, if you plug in all the values...it would be on the triangle, except for -5.

For -5 to be on it...the formula just has to be 'y'

Which is probably not right...?

You are just guessing. Fill in the table with the correct numbers and show it to me.

 

[think4432]

You are just guessing. Fill in the table with the correct numbers and show it to me.

The thing is I don't know where to plug in the numbers to fill in the table, which is why I am trying to think of a formula to plug in those numbers that would be on the triangle.

 

[think4432]

No, not even close. For one thing, a dy integral requires y limits.

You have this disk slab of water whose weight is wπ/3. This disk is at location y vertically. How far does it have to be lifted to get out the top? Because, remember, work is force times distance.

It has to be lifted the the height of the function? Which is just 3x^2?

But in terms of y,

x = y^(1/2)/root 3

So integration of a=0 to b=2 and the integration of wpi [ y/3 - y^(1/2)/root 3 ]

?

 

[LCKurtz]

The thing is I don't know where to plug in the numbers to fill in the table, which is why I am trying to think of a formula to plug in those numbers that would be on the triangle.

Draw the line y = -2 across your triangle. How deep is that line under the waterline? Put that number in the table for the actual depth. Do the other values of y. Once you have filled in the table, then see if you can figure out the formula. (It's obvious, you're going to kick yourself).

 

[LCKurtz]

No, not even close. For one thing, a dy integral requires y limits.

You have this disk slab of water whose weight is wπ/3. This disk is at location y vertically. How far does it have to be lifted to get out the top? Because, remember, work is force times distance.

It has to be lifted the the height of the function? Which is just 3x^2?

You were given a fixed parabolic container. It doesn't go on forever. How tall is it (a specific number)? How far does a slab of water at height y have to be lifted to get to the top?

So integration of a=0 to b=2 ..

Did you even read what I said about y limits? Those are still x limits.

 

[think4432]

Draw the line y = -2 across your triangle. How deep is that line under the waterline? Put that number in the table for the actual depth. Do the other values of y. Once you have filled in the table, then see if you can figure out the formula. (It's obvious, you're going to kick yourself).

If I draw a y = -2 line, it would be 7 ft under the top of the liquid.

So 7 ft is the actual depth.

And I get numbers like 6, 5, 4, 3, and 2. Which are all about the y = -2 line..and under the liquid, correct?

So would my height just by 7 + y?

Thats probably not it..because its not obvious to me, that it would be 7 + y

 

[think4432]

You were given a fixed parabolic container. It doesn't go on forever. How tall is it (a specific number)? How far does a slab of water at height y have to be lifted to get to the top?



Did you even read what I said about y limits? Those are still x limits.

Y limits...so I would just plug in the x limits in the y = 3x^2 and get [0, 12] for the limits?

So if the y limits are 0, 12 ...the distance would be 12-y?

 

[LCKurtz]

If I draw a y = -2 line, it would be 7 ft under the top of the liquid.

What picture are you looking at? Isn't it true in the picture you posted that no part of the triangle is more than 5 ft under the waterline??

 

[LCKurtz]

Y limits...so I would just plug in the x limits in the y = 3x^2 and get [0, 12] for the limits?

So if the y limits are 0, 12 ...the distance would be 12-y?

Yes for the distance of 12 - y. But the container isn't full of water to begin with and you don't have to lift water from where there isn't any in the first place. So the upper limit is ? And your final integral is what?

 

[think4432]

What picture are you looking at? Isn't it true in the picture you posted that no part of the triangle is more than 5 ft under the waterline??

Im looking at the same picture.

Maybe I am just really really slow at this or something, but I thought a line a y= -2 would be below the triangle...but that has nothing to do with it.

:[

Im glad you're very patient.

OH. If I do y + 5...and plug all those values in..then they would all be in that region! [above the triangle...not under like y + 7]

Right?!?

 

[think4432]

Yes for the distance of 12 - y. But the container isn't full of water to begin with and you don't have to lift water from where there isn't any in the first place. So the upper limit is ? And your final integral is what?

Upper limit is 4! Because 4 ft is where the water is filled too!

(62.4 ft/lb^3)pi from a = 0, b = 4 integrating [y/3 - (12-y)] << final integral. I hope.

 

[LCKurtz]

Im looking at the same picture.

Maybe I am just really really slow at this or something, but I thought a line a y= -2 would be below the triangle...but that has nothing to do with it.

When you started this problem, one thing you did was figure out the equation of the slanted side. When you did that you established your coordinate system. Judging from the equation you got I think your origin is at the water level 1 unit above the point of the triangle, with y positive upward. You have to use that coordinate system for all of your measurements. And use it to fill in that table.

Im glad you're very patient.

No problem, I had a little time to kill this afternoon. So show me what you get when you fill in the table now.

 

[LCKurtz]

Upper limit is 4! Because 4 ft is where the water is filled too!

(62.4 ft/lb^3)pi from a = 0, b = 4 integrating [y/3 - (12-y)] << final integral. I hope.

Close. Work is force times distance, not force minus distance.

 

[think4432]

When you started this problem, one thing you did was figure out the equation of the slanted side. When you did that you established your coordinate system. Judging from the equation you got I think your origin is at the water level 1 unit above the point of the triangle, with y positive upward. You have to use that coordinate system for all of your measurements. And use it to fill in that table.


No problem, I had a little time to kill this afternoon. So show me what you get when you fill in the table now.

I thought I had it with y + 5, Now I am back at square 1.

I plugged in the values and got 4,3,2,1

Confused once again.

 

[think4432]

Close. Work is force times distance, not force minus distance.

Oh! So the integral is:

(62.4 ft/lb^3)pi from a = 0, b = 4 integrating [y/3 X (12-y)] << final integral. I hope.

or integrating [(12y-y^2)/3] from the limits 0,4 with 62.4pi out in front of the integral?

 

[LCKurtz]

I thought I had it with y + 5, Now I am back at square 1.

I plugged in the values and got 4,3,2,1

Confused once again.

Show me the whole table using your coordinate system and looking at your picture:

y = -2 depth = ?
y = -3 depth = ?
...

 

[LCKurtz]

Oh! So the integral is:

(62.4 ft/lb^3)pi from a = 0, b = 4 integrating [y/3 X (12-y)] << final integral. I hope.

or integrating [(12y-y^2)/3] from the limits 0,4 with 62.4pi out in front of the integral?

Yes. Ta Daaah! I hope you understand it well enough now that you could explain it to others.
(Do you have the right units on the 62.4?)

 

[think4432]

Yes. Ta Daaah! I hope you understand it well enough now that you could explain it to others.
(Do you have the right units on the 62.4?)

Once I read over this forum a couple of times, I will be able to explain it to others.

It says in the problem that w = 62.4 lb/ft^3

So that's what I use for the w, correct?

THANK YOU SO VERY MUCH FOR YOUR HELP ON THIS PROBLEM!

I don't know how many times I would have been wrong with your help!

Thank you so much...one problem down, one more to go!

 

[LCKurtz]

You're welcome. I'm off to dinner now. Will check back in later to see if you have the depth formula figured out.

 

[think4432]

Show me the whole table using your coordinate system and looking at your picture:

y = -2 depth = ?
y = -3 depth = ?
...

Would the depth be to the top of the liquid?

If that's the case, then when y = -2 then the depth would be 7...and that's not in the picture.

y = -3 then 8

But that's obviously not right...right?