Checking the Orientation of an Integral on a Surface Bounded by a Sphere

[mathmari]

Hey!

I want to calculate $$\iint_{\Sigma}\left (ydy\land dz+zdz\land dx+zdx\land dy\right )$$ where $\Sigma$ is the surface that is described by $x^2+y^2+z^2=1$ and $y\geq 0$ and has such an orientation that the perpendicular vectors that implies have a direction away from the point $(0,0,0)$.

Do we use here spherical coordinates?
$$x=\cos\theta\sin\phi , \ y=\sin\theta\sin\phi, \ z=\cos \phi$$ Since $y\geq 0$ we get that $\theta, \phi\in [0,\pi]$, right?

We have that $$\iint_{\Sigma}(P dy\land dz+Q dz\land dx + Rdx\land dy)=\iint_D\left (P(x(\theta, \phi), y(\theta, \phi), z(\theta, \phi))\frac{\partial{(y,z)}}{\partial{(\theta, \phi)}}+Q(x(\theta, \phi), y(\theta, \phi), z(\theta, \phi))\frac{\partial{(z,x)}}{\partial{(\theta, \phi)}}+R(x(\theta, \phi), y(\theta, \phi), z(\theta, \phi))\frac{\partial{(x,y)}}{\partial{(\theta, \phi)}}\right )d\theta d\phi$$

where $$\frac{\partial{(y,z)}}{\partial{(\theta, \phi)}}=-\cos\theta\sin^2\phi, \ \frac{\partial{(z,x)}}{\partial{(\theta, \phi)}}=-\sin\theta\sin^2\phi, \ \frac{\partial{(x,y)}}{\partial{(\theta, \phi)}}=-\sin\phi\cos\phi$$

Therefore, we get \begin{align*}\iint_{\Sigma}\left (ydy\land dz+zdz\land dx+zdx\land dy\right )&=\iint_D\left (-\sin\theta\sin\phi\cos\theta\sin^2\phi-\cos \phi\sin\theta\sin^2\phi-\cos \phi\sin\phi\cos\phi\right )d\theta d\phi \\ & =\iint_D\left (-\sin\theta\sin\phi\cos\theta\sin^2\phi-\cos \phi\sin\theta\sin^2\phi-\sin\phi\cos^2\phi\right )d\theta d\phi\end{align*}

Is everything correct so far? How do we use the given information about the orientation? (Wondering)

 

[DrWahoo]

Re: Integral - Oriantation

Just a suggestion, you can always check using a gradient vector field to see what happens at that point. Check for the normal vector. What signs and which direction do they point?

Hey!

I want to calculate $$\iint_{\Sigma}\left (ydy\land dz+zdz\land dx+zdx\land dy\right )$$ where $\Sigma$ is the surface that is described by $x^2+y^2+z^2=1$ and $y\geq 0$ and has such an orientation that the perpendicular vectors that implies have a direction away from the point $(0,0,0)$.

Do we use here spherical coordinates?
$$x=\cos\theta\sin\phi , \ y=\sin\theta\sin\phi, \ z=\cos \phi$$ Since $y\geq 0$ we get that $\theta, \phi\in [0,\pi]$, right?

We have that $$\iint_{\Sigma}(P dy\land dz+Q dz\land dx + Rdx\land dy)=\iint_D\left (P(x(\theta, \phi), y(\theta, \phi), z(\theta, \phi))\frac{\partial{(y,z)}}{\partial{(\theta, \phi)}}+Q(x(\theta, \phi), y(\theta, \phi), z(\theta, \phi))\frac{\partial{(z,x)}}{\partial{(\theta, \phi)}}+R(x(\theta, \phi), y(\theta, \phi), z(\theta, \phi))\frac{\partial{(x,y)}}{\partial{(\theta, \phi)}}\right )d\theta d\phi$$

where $$\frac{\partial{(y,z)}}{\partial{(\theta, \phi)}}=-\cos\theta\sin^2\phi, \ \frac{\partial{(z,x)}}{\partial{(\theta, \phi)}}=-\sin\theta\sin^2\phi, \ \frac{\partial{(x,y)}}{\partial{(\theta, \phi)}}=-\sin\phi\cos\phi$$

Therefore, we get \begin{align*}\iint_{\Sigma}\left (ydy\land dz+zdz\land dx+zdx\land dy\right )&=\iint_D\left (-\sin\theta\sin\phi\cos\theta\sin^2\phi-\cos \phi\sin\theta\sin^2\phi-\cos \phi\sin\phi\cos\phi\right )d\theta d\phi \\ & =\iint_D\left (-\sin\theta\sin\phi\cos\theta\sin^2\phi-\cos \phi\sin\theta\sin^2\phi-\sin\phi\cos^2\phi\right )d\theta d\phi\end{align*}

Is everything correct so far? How do we use the given information about the orientation? (Wondering)

 

[I like Serena]

$$\frac{\partial{(z,x)}}{\partial{(\theta, \phi)}}=-\sin\theta\sin^2\phi$$

Shouldn't that be $\frac{\partial{(z,x)}}{\partial{(\theta, \phi)}}=+\sin\theta\sin^2\phi$? (Wondering)

How do we use the given information about the orientation? (Wondering)

We have the coordinates $(\theta,\phi)$, with $\theta$ first.
The normal vector at some point $\mathbf r$ on the sphere is $\pd {\mathbf r}\theta \times \pd {\mathbf r}\phi$.
Is it away from the origin? (Wondering)

 

[mathmari]

Shouldn't that be $\frac{\partial{(z,x)}}{\partial{(\theta, \phi)}}=+\sin\theta\sin^2\phi$? (Wondering)

We have the following: \begin{align*}\frac{\partial{(z,x)}}{\partial{(\theta,\phi)}}&=\begin{vmatrix}
\frac{\partial z}{\partial \theta} & \frac{\partial z}{\partial \phi}\\
\frac{\partial x}{\partial \theta} & \frac{\partial x}{\partial \phi}
\end{vmatrix} =\begin{vmatrix}
0 & -\sin \phi \\
-\sin\theta\sin\phi & \cos\theta\cos\phi
\end{vmatrix} \\ & =0\cdot \cos\theta\cos\phi-(-\sin \phi)\cdot (-\sin\theta\sin\phi) \\ & =-\sin \phi\cdot \sin\theta\sin\phi \\ & =-\sin\theta\sin\phi^2\end{align*}
or not? (Wondering)

We have the coordinates $(\theta,\phi)$, with $\theta$ first.
The normal vector at some point $\mathbf r$ on the sphere is $\pd {\mathbf r}\theta \times \pd {\mathbf r}\phi$.
Is it away from the origin? (Wondering)

Do you mean that $\mathbf r=(\cos\theta\sin\phi, \sin\theta\sin\phi, \cos\phi )$ ?

Do we check this with the sign of $\pd {\mathbf r}\theta \times \pd {\mathbf r}\phi$?

We have that $\pd {\mathbf r}\theta \times \pd {\mathbf r}\phi=\sin\phi (x,y,z)$, which has the same sign as $(x,y,z)$, since $\sin\phi>0$ for $\phi\in[0,\pi]$, right?

(Wondering)

 

[I like Serena]

We have the following: \begin{align*}\frac{\partial{(z,x)}}{\partial{(\theta,\phi)}}&=\begin{vmatrix}
\frac{\partial z}{\partial \theta} & \frac{\partial z}{\partial \phi}\\
\frac{\partial x}{\partial \theta} & \frac{\partial x}{\partial \phi}
\end{vmatrix} =\begin{vmatrix}
0 & -\sin \phi \\
-\sin\theta\sin\phi & \cos\theta\cos\phi
\end{vmatrix} \\ & =0\cdot \cos\theta\cos\phi-(-\sin \phi)\cdot (-\sin\theta\sin\phi) \\ & =-\sin \phi\cdot \sin\theta\sin\phi \\ & =-\sin\theta\sin\phi^2\end{align*}
or not? (Wondering)

Oh yes. (Blush)

Do you mean that $\mathbf r=(\cos\theta\sin\phi, \sin\theta\sin\phi, \cos\phi )$ ?

Do we check this with the sign of $\pd {\mathbf r}\theta \times \pd {\mathbf r}\phi$?

We have that $\pd {\mathbf r}\theta \times \pd {\mathbf r}\phi=\sin\phi (x,y,z)$, which has the same sign as $(x,y,z)$, since $\sin\phi>0$ for $\phi\in[0,\pi]$, right? (Wondering)

Yep. (Nod)

 

[mathmari]

Yep. (Nod)

Since $\pd {\mathbf r}\theta \times \pd {\mathbf r}\phi=\sin\phi (x,y,z)$ has the same sign as $(x,y,z)$, since $\sin\phi>0$ for $\phi\in[0,\pi]$, this means that we have an orientation such that the perpendicular vectors that implies have a direction away from the point $(0,0,0)$,right?

Therefore, we get the following:
\begin{align*}\iint_{\Sigma}\left (ydy\land dz+zdz\land dx+zdx\land dy\right ) & =\iint_D\left (-\sin\phi\cos\theta\sin^3\phi-\cos \phi\sin\theta\sin^2\phi-\sin\phi\cos^2\phi\right )d\theta d\phi \\ & = \int_0^{\pi}\int_0^{\pi}\left (-\sin\phi\cos\theta\sin^3\phi-\cos \phi\sin\theta\sin^2\phi-\sin\phi\cos^2\phi\right )d\theta d\phi\end{align*}
At an other exercise we have:
$\Sigma$ is the surface that is described by $\frac{x^2}{4}+\frac{y^2}{9}=1$ and $0\leq z\leq 1$ and has such an orientation that the perpendicular vectors that implies have a direction away from the $z$-axis.

Using cylindrical coordinates we have $\Sigma (\theta , z)=(2\cos\theta , \ 3\sin\theta, \ z)$ and so $\Sigma_{\theta}\times\Sigma_z=(3\cos\theta, 2\sin\theta, 0)$.
What should hold here so that we have an orientation so that the perpendicular vectors that implies have a direction away from the $z$-axis?

 

[I like Serena]

The perpendicular vector needs to be in the same general direction as the vector to the point, or rather, it needs to be in the same half space.
We can check by taking their dot product, which must be positive then. (Thinking)

 

[mathmari]

At an other exercise we have:
$\Sigma$ is the surface that is described by $\frac{x^2}{4}+\frac{y^2}{9}=1$ and $0\leq z\leq 1$ and has such an orientation that the perpendicular vectors that implies have a direction away from the $z$-axis.
Using cylindrical coordinates we have $\Sigma (\theta , z)=(2\cos\theta , \ 3\sin\theta, \ z)$ and so $\Sigma_{\theta}\times\Sigma_z=(3\cos\theta, 2\sin\theta, 0)$.
What should hold here so that we have an orientation so that the perpendicular vectors that implies have a direction away from the $z$-axis?

The perpendicular vector needs to be in the same general direction as the vector to the point, or rather, it needs to be in the same half space.
We can check by taking their dot product, which must be positive then. (Thinking)

So, we have to take the interval for $\theta$ in such a way so that $\Sigma_{\theta}\times\Sigma_z=(3\cos\theta, 2\sin\theta, 0)>0$ ? So, we take $\theta\in\left [0, \frac{\pi}{2}\right ]$, or not? (Wondering)

 

[I like Serena]

So, we have to take the interval for $\theta$ in such a way so that $\Sigma_{\theta}\times\Sigma_z=(3\cos\theta, 2\sin\theta, 0)>0$ ? So, we take $\theta\in\left [0, \frac{\pi}{2}\right ]$, or not? (Wondering)

We should take $\theta$ such that $(3\cos\theta, 2\sin\theta, 0)\cdot (\cos\theta,\sin\theta,z)>0$. (Thinking)

 

[mathmari]

We should take $\theta$ such that $(3\cos\theta, 2\sin\theta, 0)\cdot (\cos\theta,\sin\theta,z)>0$. (Thinking)

Do you maybe mean $(3\cos\theta, 2\sin\theta, 0)\cdot (2\cos\theta,3\sin\theta,z)>0$ ?

If yes, we have the following:
$$(3\cos\theta, 2\sin\theta, 0)\cdot (2\cos\theta,3\sin\theta,z)>0 \Rightarrow 6\cos^2\theta+3\sin^2\theta>0 \Rightarrow 6>0$$ Which holds for every vlue of $\theta$, so $\theta\in [0, 2\pi]$, right? (Wondering)

 

[I like Serena]

Nope. I meant the dot product of the perpendicular vector with the vector to the point (x, y, z) where we have the perpendicular vector.

 

[mathmari]

Nope. I meant the dot product of the perpendicular vector with the vector to the point (x, y, z) where we have the perpendicular vector.

Although we have defined $x=2\cos\theta, \ y=3\sin\theta, \ z=z$ ?

We have the following:
$$(3\cos\theta, 2\sin\theta, 0)\cdot (\cos\theta,\sin\theta,z)>0\Rightarrow 3\cos^2\theta+2\sin^2\theta>0\Rightarrow \cos^2\theta+2>0$$ This holds for every $\theta\in [0, 2\pi]$, right?

 

[I like Serena]

Although we have defined $x=2\cos\theta, \ y=3\sin\theta, \ z=z$ ?

Didn't we have $\Sigma_\theta\times \Sigma_z=(2\cos\theta, 3\sin\theta, z)$ instead? (Wondering)

The idea is that we calculate $(\Sigma_\theta\times \Sigma_z)\cdot \mathbf r$, where $\mathbf r$ is the vector to the point where we calculate the perpendicular vector $\Sigma_\theta\times \Sigma_z$.

We have the following:
$$(3\cos\theta, 2\sin\theta, 0)\cdot (\cos\theta,\sin\theta,z)>0\Rightarrow 3\cos^2\theta+2\sin^2\theta>0\Rightarrow \cos^2\theta+2>0$$ This holds for every $\theta\in [0, 2\pi]$, right?

Yep. (Nod)

 

[mathmari]

Didn't we have $\Sigma_\theta\times \Sigma_z=(2\cos\theta, 3\sin\theta, z)$ instead? (Wondering)

We have the following:

$\Sigma$ is the surface that is described by $\frac{x^2}{4}+\frac{y^2}{9}=1$ and $0\leq z\leq 1$ and has such an orientation that the perpendicular vectors that implies have a direction away from the $z$-axis.

Using cylindrical coordinates we have $\Sigma (\theta , z)=(2\cos\theta , \ 3\sin\theta, \ z)$ and so $\Sigma_{\theta}\times\Sigma_z=(3\cos\theta, 2\sin\theta, 0)$.

 

[I like Serena]

We have the following:

Oh yes. You were right all along.

 

[mathmari]

Oh yes. You were right all along.

Great!

So, always the dot product must be positive, or not? To check if I understood it well:
We consider $\Sigma$ to be the boundary of the bounded space $D$ that is defined by $0\leq z\leq 1-x^2-y^2$ and has such an orientation that the perpendicular vectors have direction to the outside of the space $D$.

We use here cylindrical coordinates, or not?
So, $\Sigma (\theta , z)=(R\cos\theta , \ R\sin\theta, \ z)$. Then we have that $0\leq z\leq 1-R^2$.
The perpendicular vector is $\Sigma_{\theta}\times\Sigma_z=(R\cos\theta, \ R\sin\theta, \ 0)$. Then the dot product is equal to $$(\Sigma_{\theta}\times\Sigma_z)\cdot \Sigma=(R\cos\theta, \ R\sin\theta, \ 0)\cdot (R\cos\theta , \ R\sin\theta, \ z)=R^2>0, \ \forall \theta\in [0,2\pi]$$ So, $\theta\in [0,2\pi]$.

Is everything correct? (Wondering)

 

[I like Serena]

So, always the dot product must be positive, or not?

That depends on what we want to check.
We are currently assuming a convex region around the origin respectively the z-axis.
This is the case for a sphere around the origin respectively for an elliptic cylinder around the z-axis.
In such cases the dot product must be positive everywhere on the surface.
Moreover, we should not limit the range of e.g. $\theta$ because of it. Instead we should verify that it is satisfied for every $\theta$. (Nerd)

If these assumptions are not satisfied (not convex or not around the origin or an axis), we need a different way to check. (Worried)

To check if I understood it well:
We consider $\Sigma$ to be the boundary of the bounded space $D$ that is defined by $0\leq z\leq 1-x^2-y^2$ and has such an orientation that the perpendicular vectors have direction to the outside of the space $D$.

We use here cylindrical coordinates, or not?

I would recommend spherical since z is bounded by a sphere.

So, $\Sigma (\theta , z)=(R\cos\theta , \ R\sin\theta, \ z)$. Then we have that $0\leq z\leq 1-R^2$.
The perpendicular vector is $\Sigma_{\theta}\times\Sigma_z=(R\cos\theta, \ R\sin\theta, \ 0)$. Then the dot product is equal to $$(\Sigma_{\theta}\times\Sigma_z)\cdot \Sigma=(R\cos\theta, \ R\sin\theta, \ 0)\cdot (R\cos\theta , \ R\sin\theta, \ z)=R^2>0, \ \forall \theta\in [0,2\pi]$$ So, $\theta\in [0,2\pi]$.

Is everything correct? (Wondering)

Shouldn't that be $D=\{(R\cos\theta , \ R\sin\theta, \ z) : 0\le R\le 1, 0\leq z\leq 1-R^2\}$ and $\Sigma (\theta , z)=(\sqrt{1-z^2}\cos\theta , \ \sqrt{1-z^2}\sin\theta, \ z)$? (Wondering)

Then we'll get a different perpendicular vector and dot product.