Mixing
Help find work done to pump fluid
$begingroup$
A tank is full of oil weighing 20 lb/ft${^3}$. The tank is an inverted right rectangular pyramid (with the base at the top) with a width of 2 feet, a depth of 2 feet, and a height of 5 feet. Find the work required to pump the water to a height of 3 feet above the top of the tank.
I was thinking that the answer would be
Work = $int_0^{8} (20)($'some equation I need help finding'$)(8-y)dy $
How would I find the missing equation and the work?
calculus integration area
asked Jan 15 at 20:53
SomeGuy2312SomeGuy2312
224
$endgroup$
add a comment |
$begingroup$
A tank is full of oil weighing 20 lb/ft${^3}$. The tank is an inverted right rectangular pyramid (with the base at the top) with a width of 2 feet, a depth of 2 feet, and a height of 5 feet. Find the work required to pump the water to a height of 3 feet above the top of the tank.
I was thinking that the answer would be
Work = $int_0^{8} (20)($'some equation I need help finding'$)(8-y)dy $
How would I find the missing equation and the work?
calculus integration area
asked Jan 15 at 20:53
SomeGuy2312SomeGuy2312
224
$endgroup$
$begingroup$
It is very difficult to tell what you are asking. You have a tank full of oil and a problem that asks for the work required to pump "the water". You, however, don't seem to be interested in that problem. Instead, you ask for the area of the tank.
$endgroup$
– John Douma
Jan 15 at 21:01
$begingroup$
What I am asking is if the equation I have posted to find the work is correct, and if so, how would I find what "Area" is equal to? Does this make sense?
$endgroup$
– SomeGuy2312
Jan 15 at 21:07
$begingroup$
@SomeGuy2312 You might be able to eliminate the confusion for readers by changing the subject line to something like "Help find work done to pump fluid". Also, what exactly do you mean by "area" of a 3D tank?
$endgroup$
– Aditya Dua
Jan 15 at 21:34
$begingroup$
Thank you for the help, and what I mean by "area" is that I want to find the area of the cross-section in the shape.
$endgroup$
– SomeGuy2312
Jan 15 at 21:44
$begingroup$
@AdityaDua I tried editing to hopefully make more sense!
$endgroup$
– SomeGuy2312
Jan 15 at 22:01
add a comment |
$begingroup$
A tank is full of oil weighing 20 lb/ft${^3}$. The tank is an inverted right rectangular pyramid (with the base at the top) with a width of 2 feet, a depth of 2 feet, and a height of 5 feet. Find the work required to pump the water to a height of 3 feet above the top of the tank.
I was thinking that the answer would be
Work = $int_0^{8} (20)($'some equation I need help finding'$)(8-y)dy $
How would I find the missing equation and the work?
calculus integration area
asked Jan 15 at 20:53
SomeGuy2312SomeGuy2312
224
$endgroup$
A tank is full of oil weighing 20 lb/ft${^3}$. The tank is an inverted right rectangular pyramid (with the base at the top) with a width of 2 feet, a depth of 2 feet, and a height of 5 feet. Find the work required to pump the water to a height of 3 feet above the top of the tank.
I was thinking that the answer would be
Work = $int_0^{8} (20)($'some equation I need help finding'$)(8-y)dy $
How would I find the missing equation and the work?
calculus integration area
calculus integration area
asked Jan 15 at 20:53
SomeGuy2312SomeGuy2312
224
asked Jan 15 at 20:53
SomeGuy2312SomeGuy2312
224
edited Jan 15 at 21:54
SomeGuy2312
asked Jan 15 at 20:53
SomeGuy2312SomeGuy2312
224
asked Jan 15 at 20:53
SomeGuy2312SomeGuy2312
224
asked Jan 15 at 20:53
SomeGuy2312SomeGuy2312
224
224
$begingroup$
It is very difficult to tell what you are asking. You have a tank full of oil and a problem that asks for the work required to pump "the water". You, however, don't seem to be interested in that problem. Instead, you ask for the area of the tank.
$endgroup$
– John Douma
Jan 15 at 21:01
$begingroup$
What I am asking is if the equation I have posted to find the work is correct, and if so, how would I find what "Area" is equal to? Does this make sense?
$endgroup$
– SomeGuy2312
Jan 15 at 21:07
$begingroup$
@SomeGuy2312 You might be able to eliminate the confusion for readers by changing the subject line to something like "Help find work done to pump fluid". Also, what exactly do you mean by "area" of a 3D tank?
$endgroup$
– Aditya Dua
Jan 15 at 21:34
$begingroup$
Thank you for the help, and what I mean by "area" is that I want to find the area of the cross-section in the shape.
$endgroup$
– SomeGuy2312
Jan 15 at 21:44
$begingroup$
@AdityaDua I tried editing to hopefully make more sense!
$endgroup$
– SomeGuy2312
Jan 15 at 22:01
add a comment |
$begingroup$
It is very difficult to tell what you are asking. You have a tank full of oil and a problem that asks for the work required to pump "the water". You, however, don't seem to be interested in that problem. Instead, you ask for the area of the tank.
$endgroup$
– John Douma
Jan 15 at 21:01
$begingroup$
What I am asking is if the equation I have posted to find the work is correct, and if so, how would I find what "Area" is equal to? Does this make sense?
$endgroup$
– SomeGuy2312
Jan 15 at 21:07
$begingroup$
@SomeGuy2312 You might be able to eliminate the confusion for readers by changing the subject line to something like "Help find work done to pump fluid". Also, what exactly do you mean by "area" of a 3D tank?
$endgroup$
– Aditya Dua
Jan 15 at 21:34
$begingroup$
Thank you for the help, and what I mean by "area" is that I want to find the area of the cross-section in the shape.
$endgroup$
– SomeGuy2312
Jan 15 at 21:44
$begingroup$
@AdityaDua I tried editing to hopefully make more sense!
$endgroup$
– SomeGuy2312
Jan 15 at 22:01
$begingroup$
It is very difficult to tell what you are asking. You have a tank full of oil and a problem that asks for the work required to pump "the water". You, however, don't seem to be interested in that problem. Instead, you ask for the area of the tank.
$endgroup$
– John Douma
Jan 15 at 21:01
$begingroup$
It is very difficult to tell what you are asking. You have a tank full of oil and a problem that asks for the work required to pump "the water". You, however, don't seem to be interested in that problem. Instead, you ask for the area of the tank.
$endgroup$
– John Douma
Jan 15 at 21:01
$begingroup$
What I am asking is if the equation I have posted to find the work is correct, and if so, how would I find what "Area" is equal to? Does this make sense?
$endgroup$
– SomeGuy2312
Jan 15 at 21:07
$begingroup$
What I am asking is if the equation I have posted to find the work is correct, and if so, how would I find what "Area" is equal to? Does this make sense?
$endgroup$
– SomeGuy2312
Jan 15 at 21:07
$begingroup$
@SomeGuy2312 You might be able to eliminate the confusion for readers by changing the subject line to something like "Help find work done to pump fluid". Also, what exactly do you mean by "area" of a 3D tank?
$endgroup$
– Aditya Dua
Jan 15 at 21:34
$begingroup$
@SomeGuy2312 You might be able to eliminate the confusion for readers by changing the subject line to something like "Help find work done to pump fluid". Also, what exactly do you mean by "area" of a 3D tank?
$endgroup$
– Aditya Dua
Jan 15 at 21:34
$begingroup$
Thank you for the help, and what I mean by "area" is that I want to find the area of the cross-section in the shape.
$endgroup$
– SomeGuy2312
Jan 15 at 21:44
$begingroup$
Thank you for the help, and what I mean by "area" is that I want to find the area of the cross-section in the shape.
$endgroup$
– SomeGuy2312
Jan 15 at 21:44
$begingroup$
@AdityaDua I tried editing to hopefully make more sense!
$endgroup$
– SomeGuy2312
Jan 15 at 22:01
$begingroup$
@AdityaDua I tried editing to hopefully make more sense!
$endgroup$
– SomeGuy2312
Jan 15 at 22:01
add a comment |
1 Answer
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$begingroup$
The work to move something with mass $m$ up a height $h$ against gravity $g$ is $mgh$.
Mass is density $rho$ times volume, so the differential mass is $rhotext{ }dV = rhotext{ }dxtext{ }dytext{ }dz$.
You're given the density, so to get the total work you calculate
$$W=iiint_V rho gh(z) text{ }dV,$$
where $h(z)$ is the height moved up as a function of $z$. You already have $h(z)$ in your question if the tip of the pyramid is at the origin.
You can simplify things a bit by noting that the side of the cross section varies linearly from the tip to the base. It's zero at the origin, and $2$ feet at five feet above the origin. So the $x$ and $y$ integrations end up giving you the area of the cross section:
$$W=int_0^5 rho g (8-z) (2z/5)^2 text{ } dz.$$
answered Jan 15 at 22:36
JohnJohn
22.7k32550
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add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The work to move something with mass $m$ up a height $h$ against gravity $g$ is $mgh$.
Mass is density $rho$ times volume, so the differential mass is $rhotext{ }dV = rhotext{ }dxtext{ }dytext{ }dz$.
You're given the density, so to get the total work you calculate
$$W=iiint_V rho gh(z) text{ }dV,$$
where $h(z)$ is the height moved up as a function of $z$. You already have $h(z)$ in your question if the tip of the pyramid is at the origin.
You can simplify things a bit by noting that the side of the cross section varies linearly from the tip to the base. It's zero at the origin, and $2$ feet at five feet above the origin. So the $x$ and $y$ integrations end up giving you the area of the cross section:
$$W=int_0^5 rho g (8-z) (2z/5)^2 text{ } dz.$$
answered Jan 15 at 22:36
JohnJohn
22.7k32550
$endgroup$
add a comment |
$begingroup$
The work to move something with mass $m$ up a height $h$ against gravity $g$ is $mgh$.
Mass is density $rho$ times volume, so the differential mass is $rhotext{ }dV = rhotext{ }dxtext{ }dytext{ }dz$.
You're given the density, so to get the total work you calculate
$$W=iiint_V rho gh(z) text{ }dV,$$
where $h(z)$ is the height moved up as a function of $z$. You already have $h(z)$ in your question if the tip of the pyramid is at the origin.
You can simplify things a bit by noting that the side of the cross section varies linearly from the tip to the base. It's zero at the origin, and $2$ feet at five feet above the origin. So the $x$ and $y$ integrations end up giving you the area of the cross section:
$$W=int_0^5 rho g (8-z) (2z/5)^2 text{ } dz.$$
answered Jan 15 at 22:36
JohnJohn
22.7k32550
$endgroup$
add a comment |
$begingroup$
The work to move something with mass $m$ up a height $h$ against gravity $g$ is $mgh$.
Mass is density $rho$ times volume, so the differential mass is $rhotext{ }dV = rhotext{ }dxtext{ }dytext{ }dz$.
You're given the density, so to get the total work you calculate
$$W=iiint_V rho gh(z) text{ }dV,$$
where $h(z)$ is the height moved up as a function of $z$. You already have $h(z)$ in your question if the tip of the pyramid is at the origin.
You can simplify things a bit by noting that the side of the cross section varies linearly from the tip to the base. It's zero at the origin, and $2$ feet at five feet above the origin. So the $x$ and $y$ integrations end up giving you the area of the cross section:
$$W=int_0^5 rho g (8-z) (2z/5)^2 text{ } dz.$$
answered Jan 15 at 22:36
JohnJohn
22.7k32550
$endgroup$
The work to move something with mass $m$ up a height $h$ against gravity $g$ is $mgh$.
Mass is density $rho$ times volume, so the differential mass is $rhotext{ }dV = rhotext{ }dxtext{ }dytext{ }dz$.
You're given the density, so to get the total work you calculate
$$W=iiint_V rho gh(z) text{ }dV,$$
where $h(z)$ is the height moved up as a function of $z$. You already have $h(z)$ in your question if the tip of the pyramid is at the origin.
You can simplify things a bit by noting that the side of the cross section varies linearly from the tip to the base. It's zero at the origin, and $2$ feet at five feet above the origin. So the $x$ and $y$ integrations end up giving you the area of the cross section:
$$W=int_0^5 rho g (8-z) (2z/5)^2 text{ } dz.$$
answered Jan 15 at 22:36
JohnJohn
22.7k32550
answered Jan 15 at 22:36
JohnJohn
22.7k32550
answered Jan 15 at 22:36
JohnJohn
22.7k32550
answered Jan 15 at 22:36
JohnJohn
22.7k32550
22.7k32550
add a comment |
add a comment |
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$begingroup$
It is very difficult to tell what you are asking. You have a tank full of oil and a problem that asks for the work required to pump "the water". You, however, don't seem to be interested in that problem. Instead, you ask for the area of the tank.
$endgroup$
– John Douma
Jan 15 at 21:01
$begingroup$
What I am asking is if the equation I have posted to find the work is correct, and if so, how would I find what "Area" is equal to? Does this make sense?
$endgroup$
– SomeGuy2312
Jan 15 at 21:07
$begingroup$
@SomeGuy2312 You might be able to eliminate the confusion for readers by changing the subject line to something like "Help find work done to pump fluid". Also, what exactly do you mean by "area" of a 3D tank?
$endgroup$
– Aditya Dua
Jan 15 at 21:34
$begingroup$
Thank you for the help, and what I mean by "area" is that I want to find the area of the cross-section in the shape.
$endgroup$
– SomeGuy2312
Jan 15 at 21:44
$begingroup$
@AdityaDua I tried editing to hopefully make more sense!
$endgroup$
– SomeGuy2312
Jan 15 at 22:01