How To Parametrize Background In Behave

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How tin we parametrize the triangle?

• MHB
• Thread starter mathmari
• Start date Nov xx, 2017

Hey!!

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

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

Simply how tin can we define in this example the function $\Sigma$, how can we parametrize the triangle? (Wondering)

Answers and Replies

Hey!!

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

We take to ascertain the surface $\Sigma(u,5) = (x(u,5), y(u,5), z(u,v))$ then we become $$\iint_{\Sigma}fdA=\iint_Df(\Sigma(u,v))\|\frac{\partial{\Sigma}}{\fractional{u}}(u,five)\times \frac{\partial{\Sigma}}{\partial{5}}(u,v)\|dudv$$

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

How-do-you-do mathmari!! (Smile)

How about $ten(u,v)=u, y(u,v)=v, z(u,5)=1-u-v$? (Wondering)

How most $x(u,v)=u, y(u,v)=5, z(u,five)=1-u-v$? (Wondering)

How practise nosotros get that $z$ ? (Wondering)

The edges of the triangle are the post-obit:
$(1,0,0)+t[(0,1,0)-(i,0,0)]=(one,0,0)+t(-1,1,0)=(1-t, t, 0)$
$(0,ane,0)+t[(0,0,1)-(0,1,0)]=(0,1,0)+t(0,-1,i)=(0, 1-t, t)$
$(0,0,1)+t[(ane,0,0)-(0,0,one)]=(0,0,1)+t(i,0,-i)=(t, 0, 1-t)$
correct? Practice we employ these equations to become $z(u,v)=ane-u-v$ ? (Wondering)

How practise nosotros go that $z$ ? (Wondering)

The edges of the triangle are the post-obit:
$(1,0,0)+t[(0,1,0)-(1,0,0)]=(1,0,0)+t(-1,1,0)=(one-t, t, 0)$
$(0,1,0)+t[(0,0,1)-(0,one,0)]=(0,1,0)+t(0,-ane,ane)=(0, 1-t, t)$
$(0,0,1)+t[(one,0,0)-(0,0,1)]=(0,0,1)+t(1,0,-1)=(t, 0, 1-t)$
right? Do we employ these equations to get $z(u,v)=1-u-5$ ?

The triangle is in a plane.
Therefore z must be a linear combination of ten 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 10=y=0, shows that it must be z=1-x-y. (Thinking)

Alternatively, we tin indeed kickoff with the parametrized edges that you've listed.
We can pick 1 and replace $t$ by $u$, choice another one and replace $t$ by $5$ or $one-v$.
And so, when we add together them together, we get all linear combinations that will encompass the plane of the triangle.
If we start with the third, nosotros become $10=u$, and if we so option the second and substitute $t=1-five$, we get $y=v$ and leave $x$ unchanged, and so adding them together gives united states the same $z$ I plant previously. (Thinking)

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

The triangle is in a plane.
Therefore z must exist a linear combination of x and y (assuming both 10 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=ane-x-y. (Thinking)

The triangle is in a airplane, considering there is a plne that passes through the three vertices?
The equation of the airplane in which the triangle is is $ax+past+cz=d$ where $(a,b,c)$ is the normal vector to the aeroplane.
We tin can get 2 vectors in the plane by subtracting pairs of points in the airplane: $(i,0,0)-(0,1,0)=(1,-ane,0)$ and $(0,1,0)-(0,0,1)=(0,1,-ane)$.
The cross production of these two vectors will exist orthogonal to both, and hence information technology will be a normal vector to the airplane:
$(1,-i,0)\times (0,ane,-one)=(ane,i,ane)$.
So, for $a=b=c=1$ we get the equation of the plane is $10+y+z=d$.
Since $(1,0,0)$ is a signal on the plane nosotros go $1+0+0=d \Rightarrow d=1$.
Therefore, the equation of the airplane is $ten+y+z=1$. If we solve for $z$ we get $z=1-ten-y$.

Is everything correct? (Wondering)

Alternatively, we tin indeed first with the parametrized edges that you've listed.
Nosotros can selection 1 and replace $t$ by $u$, selection some other 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 and then pick the second and substitute $t=1-v$, we go $y=v$ and go out $x$ unchanged, and then adding them together gives us the same $z$ I found previously. (Thinking)

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

So, the points on the triangle are of course $(ane-u, u, 0)$, $(0, i-v, v)$, $(w, 0, i-due west)$ with $u,5,due west\in [0,ane]$, correct?

So, do we accept to add these three points to get all linear combinations that will encompass the aeroplane of the triangle? (Wondering)

The triangle is in a aeroplane, considering 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 not-collinear, determine a unique triangle and simultaneously, a unique airplane (i.east. a two-dimensional Euclidean space). In other words, at that place 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 $ax+past+cz=d$ where $(a,b,c)$ is the normal vector to the plane.
We can get 2 vectors in the plane by subtracting pairs of points in the plane: $(1,0,0)-(0,1,0)=(ane,-1,0)$ and $(0,1,0)-(0,0,one)=(0,1,-1)$.
The cross production of these two vectors volition be orthogonal to both, and hence information technology will be a normal vector to the aeroplane:
$(1,-one,0)\times (0,1,-one)=(1,one,1)$.
So, for $a=b=c=1$ we get the equation of the aeroplane is $x+y+z=d$.
Since $(1,0,0)$ is a point on the airplane we get $1+0+0=d \Rightarrow d=1$.
Therefore, the equation of the plane is $x+y+z=1$. If we solve for $z$ nosotros get $z=1-x-y$.

Is everything correct? (Wondering)

Yes. (Nod)

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

So, the points on the triangle are of grade $(1-u, u, 0)$, $(0, 1-v, five)$, $(due west, 0, one-westward)$ with $u,5,w\in [0,1]$, right?

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

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

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

I see!! (Nerd)

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

So, nosotros have the surface $\Sigma (10,y)=(ten, y, 1-x-y)$. From the vertices of the triangle nosotros run across that $x$ and $y$ go from $0$ to $i$. And so, we have that $\Sigma: D\rightarrow \mathbb{R}^3$, where $D=[0,ane]\times [0,1]$. Is this correct? (Wondering)

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

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

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

From the vertices we run across that $0\leq x\leq 1$. It likewise holds that $0\leq z\leq 1\Rightarrow 0\leq ane-x-y\leq i \Rightarrow -1+x\leq -y\leq ten\Rightarrow -10\leq y\leq 1-x$. Since from the vertices $y$ is greater that $0$ we get $0\leq y\leq 1-ten$.

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

From the vertices we see that $0\leq x\leq 1$. It also holds that $0\leq z\leq 1\Rightarrow 0\leq i-x-y\leq 1 \Rightarrow -1+x\leq -y\leq x\Rightarrow -x\leq y\leq 1-10$. Since from the vertices $y$ is greater that $0$ we go $0\leq y\leq i-10$.

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

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

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

So nosotros have to begin with $x$ and $z$ that they are between $0$ and $one$ considering $z$ contains also $y$ and if we would begin with $x$ and $y$ that they are between $0$ and $1$ we would go likewise the result $z=1-1-1=-ane$ ? (Wondering)

To calculate the integral, I have done the following:

We accept the surface $\Sigma (10,y) = (x,y, 1-10-y)$. Then we have that $\Sigma_x=(ane,0,-1)$ and $\Sigma_y=(0,1,-1)$. So nosotros become that $\Sigma_x\times\Sigma_y=(one,ane,1)$ and so $\|\Sigma_x\times\Sigma_y\|=\sqrt{i^2+1^2+one^two}=\sqrt{3}$.

And so we get the following:
$$\iint_{\Sigma}xdA=\iint_Dx\cdot \sqrt{3}dxdy=\sqrt{3}\int_0^ane\int_0^{i-x}xdydx=\sqrt{three}\int_0^ane 10\cdot (1-x)=\sqrt{3}\int_0^1(10-x^two)dx=\frac{\sqrt{3}}{6}$$

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

So we take to begin with $x$ and $z$ that they are between $0$ and $1$ because $z$ contains likewise $y$ and if nosotros would begin with $10$ and $y$ that they are betwixt $0$ and $1$ we would go also the result $z=1-one-1=-ane$ ? (Wondering)

I guess and then...
... every bit long as we conclude that x=y=ane has no corresponding point in the triangle right? (Thinking)

To summate the integral, I have done the following:

We have the surface $\Sigma (x,y) = (x,y, 1-10-y)$. So we have that $\Sigma_x=(ane,0,-1)$ and $\Sigma_y=(0,one,-1)$. Then we go that $\Sigma_x\times\Sigma_y=(i,1,ane)$ and so $\|\Sigma_x\times\Sigma_y\|=\sqrt{i^2+ane^two+one^2}=\sqrt{iii}$.

So we get the following:
$$\iint_{\Sigma}xdA=\iint_Dx\cdot \sqrt{3}dxdy=\sqrt{3}\int_0^ane\int_0^{1-10}xdydx=\sqrt{3}\int_0^1 x\cdot (1-x)=\sqrt{3}\int_0^1(ten-x^two)dx=\frac{\sqrt{three}}{6}$$

The given result is $\frac{\sqrt{2}}{vi}$. What take I done incorrect? (Wondering)

Your adding looks right to me.
Perhaps the given result is wrong? (Wondering)

Your calculation looks right to me.
Perchance the given outcome is incorrect? (Wondering)

Ah ok!! Thank you very much!! (Mmm)

I am thinking about it again (Thinking)

A plane is a surface with nix volume, correct? Does it then concur also that a triangle is a surface with zippo volume?

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

To parametrize the triangle, also using the equation of the airplane and the style 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 infinite $D$ would be a square and non a triangle, right?

(Wondering)

A aeroplane is a surface with zero volume, right? Does it so hold besides that a triangle is a surface with zero book?

Yep.

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

3 points in infinite that are not co-linear ascertain 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 airplane 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 airplane equation is $ax+past+cz=d$.
Substitute each of the vertices to notice $a=b=c=d$.
Since (a,b,c) cannot be the cipher vector we tin can divide past $a$ to notice the equation $10+y+z=1$.
It follows that $z=1-x-y$ giving us the parametrization (x,y,ane-ten-y). (Thinking)

We don't accept the boundaries $0\leq ten\leq 1$ and $0\leq y \leq 1$ considering and then the infinite $D$ would exist a foursquare and non a triangle, right?

Yep. (Nod)

3 points in infinite that are non co-linear define a plane exercise they not?
They likewise define a triangle that is defined to be inside that plane. (Thinking)

From iii points in infinite we get two vectors in space. The cantankerous production of these vectos is normal to both of them. So we go the airplane $ax+by+cz=d$, where $(a,b,c)$ is the upshot of the cantankerous product and we become $d$ by substituting a indicate.

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

Is everything correct? (Wondering)

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

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

Since the aeroplane 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)

And so, the parametrization of the triangle has to satisfy the equation of the airplane.

Yep. (Nod)

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How To Parametrize Background In Behave,

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