How can we parametrize the triangle?

[mathmari]

Hey!

I want to calculate $\iint_{\Sigma}xdA$ on the triangle with vertices $(1,0,0)$, $(0,1,0)$ and $(0,0,1)$.

We have to define the surface $\Sigma(u,v) = (x(u,v), y(u,v), z(u,v))$ then we get $$\iint_{\Sigma}fdA=\iint_Df(\Sigma(u,v))\|\frac{\partial{\Sigma}}{\partial{u}}(u,v)\times \frac{\partial{\Sigma}}{\partial{v}}(u,v)\|dudv$$

But how can we define in this case the function $\Sigma$, how can we parametrize the triangle? (Wondering)

 

[I like Serena]

Hey!

I want to calculate $\iint_{\Sigma}xdA$ on the triangle with vertices $(1,0,0)$, $(0,1,0)$ and $(0,0,1)$.

We have to define the surface $\Sigma(u,v) = (x(u,v), y(u,v), z(u,v))$ then we get $$\iint_{\Sigma}fdA=\iint_Df(\Sigma(u,v))\|\frac{\partial{\Sigma}}{\partial{u}}(u,v)\times \frac{\partial{\Sigma}}{\partial{v}}(u,v)\|dudv$$

But how can we define in this case the function $\Sigma$, how can we parametrize the triangle? (Wondering)

Hi mathmari! (Smile)

How about $x(u,v)=u, y(u,v)=v, z(u,v)=1-u-v$? (Wondering)

 

[mathmari]

How about $x(u,v)=u, y(u,v)=v, z(u,v)=1-u-v$? (Wondering)

How do we get that $z$ ? (Wondering)

The edges of the triangle are the following:
$(1,0,0)+t[(0,1,0)-(1,0,0)]=(1,0,0)+t(-1,1,0)=(1-t, t, 0)$
$(0,1,0)+t[(0,0,1)-(0,1,0)]=(0,1,0)+t(0,-1,1)=(0, 1-t, t)$
$(0,0,1)+t[(1,0,0)-(0,0,1)]=(0,0,1)+t(1,0,-1)=(t, 0, 1-t)$
right? Do we use these equations to get $z(u,v)=1-u-v$ ? (Wondering)

 

[I like Serena]

How do we get that $z$ ? (Wondering)

The edges of the triangle are the following:
$(1,0,0)+t[(0,1,0)-(1,0,0)]=(1,0,0)+t(-1,1,0)=(1-t, t, 0)$
$(0,1,0)+t[(0,0,1)-(0,1,0)]=(0,1,0)+t(0,-1,1)=(0, 1-t, t)$
$(0,0,1)+t[(1,0,0)-(0,0,1)]=(0,0,1)+t(1,0,-1)=(t, 0, 1-t)$
right? Do we use these equations to get $z(u,v)=1-u-v$ ?

The triangle is in a plane.
Therefore z must be a linear combination of x and y (assuming both x and y have multiple values in the plane, which they do). Something like $z=a+bx+cy$.
So the only question is: which linear equation?
Checking/substituting each of the vertices, starting with x=y=0, shows that it must be z=1-x-y. (Thinking)

Alternatively, we can indeed start with the parametrized edges that you've listed.
We can pick one and replace $t$ by $u$, pick another one and replace $t$ by $v$ or $1-v$.
Then, when we add them together, we get all linear combinations that will cover the plane of the triangle.
If we start with the 3rd, we get $x=u$, and if we then pick the second and substitute $t=1-v$, we get $y=v$ and leave $x$ unchanged, then adding them together gives us the same $z$ I found previously. (Thinking)Btw, to integrate, we still need to set up a double integral with appropriate boundaries. (Thinking)

 

[mathmari]

The triangle is in a plane.
Therefore z must be a linear combination of x and y (assuming both x and y have multiple values in the plane, which they do). Something like $z=a+bx+cy$.
So the only question is: which linear equation?
Checking/substituting each of the vertices, starting with x=y=0, shows that it must be z=1-x-y. (Thinking)

The triangle is in a plane, because there is a plne that passes through the three vertices?
The equation of the plane in which the triangle is is $ax+by+cz=d$ where $(a,b,c)$ is the normal vector to the plane.
We can get two vectors in the plane by subtracting pairs of points in the plane: $(1,0,0)-(0,1,0)=(1,-1,0)$ and $(0,1,0)-(0,0,1)=(0,1,-1)$.
The cross product of these two vectors will be orthogonal to both, and hence it will be a normal vector to the plane:
$(1,-1,0)\times (0,1,-1)=(1,1,1)$.
So, for $a=b=c=1$ we get the equation of the plane is $x+y+z=d$.
Since $(1,0,0)$ is a point on the plane we get $1+0+0=d \Rightarrow d=1$.
Therefore, the equation of the plane is $x+y+z=1$. If we solve for $z$ we get $z=1-x-y$.

Is everything correct? (Wondering)

Alternatively, we can indeed start with the parametrized edges that you've listed.
We can pick one and replace $t$ by $u$, pick another one and replace $t$ by $v$ or $1-v$.
Then, when we add them together, we get all linear combinations that will cover the plane of the triangle.
If we start with the 3rd, we get $x=u$, and if we then pick the second and substitute $t=1-v$, we get $y=v$ and leave $x$ unchanged, then adding them together gives us the same $z$ I found previously. (Thinking)

We have that
$(1,0,0)+u[(0,1,0)-(1,0,0)]=(1,0,0)+u(-1,1,0)=(1-u, u, 0)$
$(0,1,0)+v[(0,0,1)-(0,1,0)]=(0,1,0)+v(0,-1,1)=(0, 1-v, v)$
$(0,0,1)+w[(1,0,0)-(0,0,1)]=(0,0,1)+w(1,0,-1)=(w, 0, 1-w)$

So, the points on the triangle are of form $(1-u, u, 0)$, $(0, 1-v, v)$, $(w, 0, 1-w)$ with $u,v,w\in [0,1]$, right?

So, do we have to add these three points to get all linear combinations that will cover the plane of the triangle? (Wondering)

 

[I like Serena]

The triangle is in a plane, because there is a plane that passes through the three vertices?

A triangle in euclidean space is by definition in a plane. (Nerd)
See wiki:

In Euclidean geometry any three points, when non-collinear, determine a unique triangle and simultaneously, a unique plane (i.e. a two-dimensional Euclidean space). In other words, there is only one plane that contains that triangle, and every triangle is contained in some plane.

The equation of the plane in which the triangle is is $ax+by+cz=d$ where $(a,b,c)$ is the normal vector to the plane.
We can get two vectors in the plane by subtracting pairs of points in the plane: $(1,0,0)-(0,1,0)=(1,-1,0)$ and $(0,1,0)-(0,0,1)=(0,1,-1)$.
The cross product of these two vectors will be orthogonal to both, and hence it will be a normal vector to the plane:
$(1,-1,0)\times (0,1,-1)=(1,1,1)$.
So, for $a=b=c=1$ we get the equation of the plane is $x+y+z=d$.
Since $(1,0,0)$ is a point on the plane we get $1+0+0=d \Rightarrow d=1$.
Therefore, the equation of the plane is $x+y+z=1$. If we solve for $z$ we get $z=1-x-y$.

Is everything correct? (Wondering)

Yep. (Nod)

We have that
$(1,0,0)+u[(0,1,0)-(1,0,0)]=(1,0,0)+u(-1,1,0)=(1-u, u, 0)$
$(0,1,0)+v[(0,0,1)-(0,1,0)]=(0,1,0)+v(0,-1,1)=(0, 1-v, v)$
$(0,0,1)+w[(1,0,0)-(0,0,1)]=(0,0,1)+w(1,0,-1)=(w, 0, 1-w)$

So, the points on the triangle are of form $(1-u, u, 0)$, $(0, 1-v, v)$, $(w, 0, 1-w)$ with $u,v,w\in [0,1]$, right?

So, do we have to add these three points to get all linear combinations that will cover the plane of the triangle? (Wondering)

We can.
Note that one of them is redundant, being a linear combination of the other two. (Thinking)

 

[mathmari]

A triangle in euclidean space is by definition in a plane. (Nerd)
See wiki:

I see! (Nerd)

Yep. (Nod)

Btw, to integrate, we still need to set up a double integral with appropriate boundaries. (Thinking)

So, we have the surface $\Sigma (x,y)=(x, y, 1-x-y)$. From the vertices of the triangle we see that $x$ and $y$ go from $0$ to $1$. So, we have that $\Sigma: D\rightarrow \mathbb{R}^3$, where $D=[0,1]\times [0,1]$. Is this correct? (Wondering)

 

[I like Serena]

So, we have the surface $\Sigma (x,y)=(x, y, 1-x-y)$. From the vertices of the triangle we see that $x$ and $y$ go from $0$ to $1$. So, we have that $\Sigma: D\rightarrow \mathbb{R}^3$, where $D=[0,1]\times [0,1]$. Is this correct? (Wondering)

Nope.
For a given $x$ coordinate between $0$ and $1$, the $y$ coordinate is limited to $[0, 1-x]$. (Sweating)

 

[mathmari]

Nope.
For a given $x$ coordinate between $0$ and $1$, the $y$ coordinate is limited to $[0, 1-x]$. (Sweating)

From the vertices we see that $0\leq x\leq 1$. It also holds that $0\leq z\leq 1\Rightarrow 0\leq 1-x-y\leq 1 \Rightarrow -1+x\leq -y\leq x\Rightarrow -x\leq y\leq 1-x$. Since from the vertices $y$ is greater that $0$ we get $0\leq y\leq 1-x$.

Do we get the boundaries in that way? Or do we get them otherwise? (Wondering)

 

[I like Serena]

From the vertices we see that $0\leq x\leq 1$. It also holds that $0\leq z\leq 1\Rightarrow 0\leq 1-x-y\leq 1 \Rightarrow -1+x\leq -y\leq x\Rightarrow -x\leq y\leq 1-x$. Since from the vertices $y$ is greater that $0$ we get $0\leq y\leq 1-x$.

Do we get the boundaries in that way? Or do we get them otherwise? (Wondering)

Yes. We can get them that way. (Nod)

 

[mathmari]

Yes. We can get them that way. (Nod)

So we have to begin with $x$ and $z$ that they are between $0$ and $1$ because $z$ contains also $y$ and if we would begin with $x$ and $y$ that they are between $0$ and $1$ we would get also the result $z=1-1-1=-1$ ? (Wondering)
To calculate the integral, I have done the following:

We have the surface $\Sigma (x,y) = (x,y, 1-x-y)$. Then we have that $\Sigma_x=(1,0,-1)$ and $\Sigma_y=(0,1,-1)$. Then we get that $\Sigma_x\times\Sigma_y=(1,1,1)$ and so $\|\Sigma_x\times\Sigma_y\|=\sqrt{1^2+1^2+1^2}=\sqrt{3}$.

Then we get the following:
$$\iint_{\Sigma}xdA=\iint_Dx\cdot \sqrt{3}dxdy=\sqrt{3}\int_0^1\int_0^{1-x}xdydx=\sqrt{3}\int_0^1 x\cdot (1-x)=\sqrt{3}\int_0^1(x-x^2)dx=\frac{\sqrt{3}}{6}$$

The given result is $\frac{\sqrt{2}}{6}$. What have I done wrong? (Wondering)

 

[I like Serena]

So we have to begin with $x$ and $z$ that they are between $0$ and $1$ because $z$ contains also $y$ and if we would begin with $x$ and $y$ that they are between $0$ and $1$ we would get also the result $z=1-1-1=-1$ ? (Wondering)

I guess so...
... as long as we conclude that x=y=1 has no corresponding point in the triangle right? (Thinking)

To calculate the integral, I have done the following:

We have the surface $\Sigma (x,y) = (x,y, 1-x-y)$. Then we have that $\Sigma_x=(1,0,-1)$ and $\Sigma_y=(0,1,-1)$. Then we get that $\Sigma_x\times\Sigma_y=(1,1,1)$ and so $\|\Sigma_x\times\Sigma_y\|=\sqrt{1^2+1^2+1^2}=\sqrt{3}$.

Then we get the following:
$$\iint_{\Sigma}xdA=\iint_Dx\cdot \sqrt{3}dxdy=\sqrt{3}\int_0^1\int_0^{1-x}xdydx=\sqrt{3}\int_0^1 x\cdot (1-x)=\sqrt{3}\int_0^1(x-x^2)dx=\frac{\sqrt{3}}{6}$$

The given result is $\frac{\sqrt{2}}{6}$. What have I done wrong? (Wondering)

Your calculation looks correct to me.
Perhaps the given result is wrong? (Wondering)

 

[mathmari]

Your calculation looks correct to me.
Perhaps the given result is wrong? (Wondering)

Ah ok! Thank you very much! (Mmm)

 

[mathmari]

I am thinking about it again (Thinking)

A plane is a surface with zero volume, right? Does it then hold also that a triangle is a surface with zero volume?

I haven't really understood why every triangle is contained in some plane. Could you explain it further to me?

To parametrize the triangle, besides using the equation of the plane and the way I started using the parametric equations of the edges, is there also an other way?

We don't take the boundaries $0\leq x\leq 1$ and $0\leq y \leq 1$ because then the space $D$ would be a square and not a triangle, right?

(Wondering)

 

[I like Serena]

A plane is a surface with zero volume, right? Does it then hold also that a triangle is a surface with zero volume?

Yes.

I haven't really understood why every triangle is contained in some plane. Could you explain it further to me?

3 points in space that are not co-linear define a plane do they not?
They also define a triangle that is defined to be inside that plane. (Thinking)

To parametrize the triangle, besides using the equation of the plane and the way I started using the parametric equations of the edges, is there also an other way?

We can simplify the approach with the plane equation.
The plane equation is $ax+by+cz=d$.
Substitute each of the vertices to find $a=b=c=d$.
Since (a,b,c) cannot be the null vector we can divide by $a$ to find the equation $x+y+z=1$.
It follows that $z=1-x-y$ giving us the parametrization (x,y,1-x-y). (Thinking)

We don't take the boundaries $0\leq x\leq 1$ and $0\leq y \leq 1$ because then the space $D$ would be a square and not a triangle, right?

Yep. (Nod)

 

[mathmari]

3 points in space that are not co-linear define a plane do they not?
They also define a triangle that is defined to be inside that plane. (Thinking)

From 3 points in space we get two vectors in space. The cross product of these vectos is normal to both of them. So we get the plane $ax+by+cz=d$, where $(a,b,c)$ is the result of the cross product and we get $d$ by substituting a point.

Since the plane is infnite, we know that the triangle is inside the plane.
So, the parametrization of the triangle has to satisfy the equation of the plane.

Is everything correct? (Wondering)

 

[I like Serena]

From 3 points in space we get two vectors in space. The cross product of these vectos is normal to both of them. So we get the plane $ax+by+cz=d$, where $(a,b,c)$ is the result of the cross product and we get $d$ by substituting a point.

Yep.
Moreover, if (a,b,c) has unit length, then d is the distance of the plane to the origin. (Nerd)

Since the plane is infnite, we know that the triangle is inside the plane.

Not exactly sure what that means. It sounds okay'ish, but I would suggest to stick to the definition of triangle on wiki. (Nerd)

So, the parametrization of the triangle has to satisfy the equation of the plane.

Yep. (Nod)