Mixing
Triple integral with tetrahedron [duplicate]
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This question already has an answer here:
Compute volume of tetrahedron using a triple integral
1 answer
$$iiint xyz ,dV$$ T is the solid tetrahedron with vertices (0,0,0) (1,0,0) (1,1,0) (1,0,1).
How do I figure out what z=f(x,y) is with these points? The equation of a tetrahedron is x/a+y/b+z/c=1
But with the 4 points given I don't know how to use the formula.
calculus
asked Feb 1 at 23:48
Alexandra M Alexandra M
274
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marked as duplicate by Lord Shark the Unknown, Adrian Keister, mrtaurho, YiFan, Gibbs Feb 3 at 0:10
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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show 1 more comment
$begingroup$
This question already has an answer here:
Compute volume of tetrahedron using a triple integral
1 answer
$$iiint xyz ,dV$$ T is the solid tetrahedron with vertices (0,0,0) (1,0,0) (1,1,0) (1,0,1).
How do I figure out what z=f(x,y) is with these points? The equation of a tetrahedron is x/a+y/b+z/c=1
But with the 4 points given I don't know how to use the formula.
calculus
asked Feb 1 at 23:48
Alexandra M Alexandra M
274
$endgroup$
marked as duplicate by Lord Shark the Unknown, Adrian Keister, mrtaurho, YiFan, Gibbs Feb 3 at 0:10
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
$begingroup$
The equation you mention is for a plane, not a tetrahedron. You must obtain the equation of the plane passing through $((0,0,0), (1,1,0)$ and $(1,0,1)$ (sketch the tetrahedron and you'll see why). The integral becomes $int_0^1int_0^x int_0^{ax+by+c} xyz dz , dy, dx$. You just need to compute a,b,c.
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– PierreCarre
Feb 2 at 0:05
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@PierreCarre Why those points ? Why not (0,0,0) (1,0,0) (1,1,0)? Or any of the other combinations you can make with the 4 points?
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– Alexandra M
Feb 2 at 0:09
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@AkashPatel No this is not a duplicate. The points are not even the same.
$endgroup$
– Alexandra M
Feb 2 at 0:09
1
$begingroup$
You have to draw the tetrahedron to see which of the four points lie on the surface.
$endgroup$
– John Wayland Bales
Feb 2 at 0:11
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@ John Wayland Bales , @PierreCarre Thank you !
$endgroup$
– Alexandra M
Feb 2 at 0:13
|
show 1 more comment
$begingroup$
This question already has an answer here:
Compute volume of tetrahedron using a triple integral
1 answer
$$iiint xyz ,dV$$ T is the solid tetrahedron with vertices (0,0,0) (1,0,0) (1,1,0) (1,0,1).
How do I figure out what z=f(x,y) is with these points? The equation of a tetrahedron is x/a+y/b+z/c=1
But with the 4 points given I don't know how to use the formula.
calculus
asked Feb 1 at 23:48
Alexandra M Alexandra M
274
$endgroup$
This question already has an answer here:
Compute volume of tetrahedron using a triple integral
1 answer
$$iiint xyz ,dV$$ T is the solid tetrahedron with vertices (0,0,0) (1,0,0) (1,1,0) (1,0,1).
How do I figure out what z=f(x,y) is with these points? The equation of a tetrahedron is x/a+y/b+z/c=1
But with the 4 points given I don't know how to use the formula.
This question already has an answer here:
Compute volume of tetrahedron using a triple integral
1 answer
calculus
calculus
asked Feb 1 at 23:48
Alexandra M Alexandra M
274
asked Feb 1 at 23:48
Alexandra M Alexandra M
274
asked Feb 1 at 23:48
Alexandra M Alexandra M
274
asked Feb 1 at 23:48
Alexandra M Alexandra M
274
asked Feb 1 at 23:48
Alexandra M Alexandra M
274
274
marked as duplicate by Lord Shark the Unknown, Adrian Keister, mrtaurho, YiFan, Gibbs Feb 3 at 0:10
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Lord Shark the Unknown, Adrian Keister, mrtaurho, YiFan, Gibbs Feb 3 at 0:10
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
$begingroup$
The equation you mention is for a plane, not a tetrahedron. You must obtain the equation of the plane passing through $((0,0,0), (1,1,0)$ and $(1,0,1)$ (sketch the tetrahedron and you'll see why). The integral becomes $int_0^1int_0^x int_0^{ax+by+c} xyz dz , dy, dx$. You just need to compute a,b,c.
$endgroup$
– PierreCarre
Feb 2 at 0:05
$begingroup$
@PierreCarre Why those points ? Why not (0,0,0) (1,0,0) (1,1,0)? Or any of the other combinations you can make with the 4 points?
$endgroup$
– Alexandra M
Feb 2 at 0:09
$begingroup$
@AkashPatel No this is not a duplicate. The points are not even the same.
$endgroup$
– Alexandra M
Feb 2 at 0:09
1
$begingroup$
You have to draw the tetrahedron to see which of the four points lie on the surface.
$endgroup$
– John Wayland Bales
Feb 2 at 0:11
$begingroup$
@ John Wayland Bales , @PierreCarre Thank you !
$endgroup$
– Alexandra M
Feb 2 at 0:13
|
show 1 more comment
1
$begingroup$
The equation you mention is for a plane, not a tetrahedron. You must obtain the equation of the plane passing through $((0,0,0), (1,1,0)$ and $(1,0,1)$ (sketch the tetrahedron and you'll see why). The integral becomes $int_0^1int_0^x int_0^{ax+by+c} xyz dz , dy, dx$. You just need to compute a,b,c.
$endgroup$
– PierreCarre
Feb 2 at 0:05
$begingroup$
@PierreCarre Why those points ? Why not (0,0,0) (1,0,0) (1,1,0)? Or any of the other combinations you can make with the 4 points?
$endgroup$
– Alexandra M
Feb 2 at 0:09
$begingroup$
@AkashPatel No this is not a duplicate. The points are not even the same.
$endgroup$
– Alexandra M
Feb 2 at 0:09
1
$begingroup$
You have to draw the tetrahedron to see which of the four points lie on the surface.
$endgroup$
– John Wayland Bales
Feb 2 at 0:11
$begingroup$
@ John Wayland Bales , @PierreCarre Thank you !
$endgroup$
– Alexandra M
Feb 2 at 0:13
1
1
$begingroup$
The equation you mention is for a plane, not a tetrahedron. You must obtain the equation of the plane passing through $((0,0,0), (1,1,0)$ and $(1,0,1)$ (sketch the tetrahedron and you'll see why). The integral becomes $int_0^1int_0^x int_0^{ax+by+c} xyz dz , dy, dx$. You just need to compute a,b,c.
$endgroup$
– PierreCarre
Feb 2 at 0:05
$begingroup$
The equation you mention is for a plane, not a tetrahedron. You must obtain the equation of the plane passing through $((0,0,0), (1,1,0)$ and $(1,0,1)$ (sketch the tetrahedron and you'll see why). The integral becomes $int_0^1int_0^x int_0^{ax+by+c} xyz dz , dy, dx$. You just need to compute a,b,c.
$endgroup$
– PierreCarre
Feb 2 at 0:05
$begingroup$
@PierreCarre Why those points ? Why not (0,0,0) (1,0,0) (1,1,0)? Or any of the other combinations you can make with the 4 points?
$endgroup$
– Alexandra M
Feb 2 at 0:09
$begingroup$
@PierreCarre Why those points ? Why not (0,0,0) (1,0,0) (1,1,0)? Or any of the other combinations you can make with the 4 points?
$endgroup$
– Alexandra M
Feb 2 at 0:09
$begingroup$
@AkashPatel No this is not a duplicate. The points are not even the same.
$endgroup$
– Alexandra M
Feb 2 at 0:09
$begingroup$
@AkashPatel No this is not a duplicate. The points are not even the same.
$endgroup$
– Alexandra M
Feb 2 at 0:09
1
1
$begingroup$
You have to draw the tetrahedron to see which of the four points lie on the surface.
$endgroup$
– John Wayland Bales
Feb 2 at 0:11
$begingroup$
You have to draw the tetrahedron to see which of the four points lie on the surface.
$endgroup$
– John Wayland Bales
Feb 2 at 0:11
$begingroup$
@ John Wayland Bales , @PierreCarre Thank you !
$endgroup$
– Alexandra M
Feb 2 at 0:13
$begingroup$
@ John Wayland Bales , @PierreCarre Thank you !
$endgroup$
– Alexandra M
Feb 2 at 0:13
|
show 1 more comment
1 Answer
1
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oldest
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You have three points $A(0,0,0), C(1,1,0)$ and $B(1,0,1)$ lying on the plane $z=f(x,y)$, so you need to find a direction vector for the plane.
Two vectors lying in the plane are $(1,0,1)$ and $(1,1,0)$ and their cross-product gives you a normal vector $(-1,1,1)$. Using the plane equation
$$ a(x-x_0)+b(y-y_0)+c(z-z_0)=0 $$
where $(x_0,y_0,z_0)$ is a point of the plane and $(a,b,c)$ is a normal vector we get, using point $(1,1,0)$
$$ -(x-1)+(y-1)+z=0 $$
Therefore
$$z = x-y$$
answered Feb 2 at 0:03
John Wayland BalesJohn Wayland Bales
15.1k21238
$endgroup$
$begingroup$
Note that if you know that the volume of a cone is one-third the product of the area of the base times the height, then you know that you should obtain a volume of one-third of one-half, that is, one-sixth.
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– John Wayland Bales
Feb 2 at 0:30
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You have three points $A(0,0,0), C(1,1,0)$ and $B(1,0,1)$ lying on the plane $z=f(x,y)$, so you need to find a direction vector for the plane.
Two vectors lying in the plane are $(1,0,1)$ and $(1,1,0)$ and their cross-product gives you a normal vector $(-1,1,1)$. Using the plane equation
$$ a(x-x_0)+b(y-y_0)+c(z-z_0)=0 $$
where $(x_0,y_0,z_0)$ is a point of the plane and $(a,b,c)$ is a normal vector we get, using point $(1,1,0)$
$$ -(x-1)+(y-1)+z=0 $$
Therefore
$$z = x-y$$
answered Feb 2 at 0:03
John Wayland BalesJohn Wayland Bales
15.1k21238
$endgroup$
$begingroup$
Note that if you know that the volume of a cone is one-third the product of the area of the base times the height, then you know that you should obtain a volume of one-third of one-half, that is, one-sixth.
$endgroup$
– John Wayland Bales
Feb 2 at 0:30
add a comment |
$begingroup$
You have three points $A(0,0,0), C(1,1,0)$ and $B(1,0,1)$ lying on the plane $z=f(x,y)$, so you need to find a direction vector for the plane.
Two vectors lying in the plane are $(1,0,1)$ and $(1,1,0)$ and their cross-product gives you a normal vector $(-1,1,1)$. Using the plane equation
$$ a(x-x_0)+b(y-y_0)+c(z-z_0)=0 $$
where $(x_0,y_0,z_0)$ is a point of the plane and $(a,b,c)$ is a normal vector we get, using point $(1,1,0)$
$$ -(x-1)+(y-1)+z=0 $$
Therefore
$$z = x-y$$
answered Feb 2 at 0:03
John Wayland BalesJohn Wayland Bales
15.1k21238
$endgroup$
$begingroup$
Note that if you know that the volume of a cone is one-third the product of the area of the base times the height, then you know that you should obtain a volume of one-third of one-half, that is, one-sixth.
$endgroup$
– John Wayland Bales
Feb 2 at 0:30
add a comment |
$begingroup$
You have three points $A(0,0,0), C(1,1,0)$ and $B(1,0,1)$ lying on the plane $z=f(x,y)$, so you need to find a direction vector for the plane.
Two vectors lying in the plane are $(1,0,1)$ and $(1,1,0)$ and their cross-product gives you a normal vector $(-1,1,1)$. Using the plane equation
$$ a(x-x_0)+b(y-y_0)+c(z-z_0)=0 $$
where $(x_0,y_0,z_0)$ is a point of the plane and $(a,b,c)$ is a normal vector we get, using point $(1,1,0)$
$$ -(x-1)+(y-1)+z=0 $$
Therefore
$$z = x-y$$
answered Feb 2 at 0:03
John Wayland BalesJohn Wayland Bales
15.1k21238
$endgroup$
You have three points $A(0,0,0), C(1,1,0)$ and $B(1,0,1)$ lying on the plane $z=f(x,y)$, so you need to find a direction vector for the plane.
Two vectors lying in the plane are $(1,0,1)$ and $(1,1,0)$ and their cross-product gives you a normal vector $(-1,1,1)$. Using the plane equation
$$ a(x-x_0)+b(y-y_0)+c(z-z_0)=0 $$
where $(x_0,y_0,z_0)$ is a point of the plane and $(a,b,c)$ is a normal vector we get, using point $(1,1,0)$
$$ -(x-1)+(y-1)+z=0 $$
Therefore
$$z = x-y$$
answered Feb 2 at 0:03
John Wayland BalesJohn Wayland Bales
15.1k21238
edited Feb 2 at 0:18
answered Feb 2 at 0:03
John Wayland BalesJohn Wayland Bales
15.1k21238
answered Feb 2 at 0:03
John Wayland BalesJohn Wayland Bales
15.1k21238
answered Feb 2 at 0:03
John Wayland BalesJohn Wayland Bales
15.1k21238
15.1k21238
$begingroup$
Note that if you know that the volume of a cone is one-third the product of the area of the base times the height, then you know that you should obtain a volume of one-third of one-half, that is, one-sixth.
$endgroup$
– John Wayland Bales
Feb 2 at 0:30
add a comment |
$begingroup$
Note that if you know that the volume of a cone is one-third the product of the area of the base times the height, then you know that you should obtain a volume of one-third of one-half, that is, one-sixth.
$endgroup$
– John Wayland Bales
Feb 2 at 0:30
$begingroup$
Note that if you know that the volume of a cone is one-third the product of the area of the base times the height, then you know that you should obtain a volume of one-third of one-half, that is, one-sixth.
$endgroup$
– John Wayland Bales
Feb 2 at 0:30
$begingroup$
Note that if you know that the volume of a cone is one-third the product of the area of the base times the height, then you know that you should obtain a volume of one-third of one-half, that is, one-sixth.
$endgroup$
– John Wayland Bales
Feb 2 at 0:30
add a comment |
1
$begingroup$
The equation you mention is for a plane, not a tetrahedron. You must obtain the equation of the plane passing through $((0,0,0), (1,1,0)$ and $(1,0,1)$ (sketch the tetrahedron and you'll see why). The integral becomes $int_0^1int_0^x int_0^{ax+by+c} xyz dz , dy, dx$. You just need to compute a,b,c.
$endgroup$
– PierreCarre
Feb 2 at 0:05
$begingroup$
@PierreCarre Why those points ? Why not (0,0,0) (1,0,0) (1,1,0)? Or any of the other combinations you can make with the 4 points?
$endgroup$
– Alexandra M
Feb 2 at 0:09
$begingroup$
@AkashPatel No this is not a duplicate. The points are not even the same.
$endgroup$
– Alexandra M
Feb 2 at 0:09
1
$begingroup$
You have to draw the tetrahedron to see which of the four points lie on the surface.
$endgroup$
– John Wayland Bales
Feb 2 at 0:11
$begingroup$
@ John Wayland Bales , @PierreCarre Thank you !
$endgroup$
– Alexandra M
Feb 2 at 0:13