MHB Divergence Theorem: Applying to Sphere $\hat{i}x+\hat{j}y+\hat{k}z$

[evinda]

Hello again! (Wave)

I am looking at an exercise of the divergence theorem..

We want to apply the divergence theorem for the sphere $x^2+y^2+z^2=a^2$ in the case when the vector field is $\overrightarrow{F}=\hat{i}x+\hat{j}y+\hat{k}z$.$\displaystyle{\nabla \cdot \overrightarrow{F}=\frac{\partial{F}}{\partial{x}} \hat{i}+\frac{\partial{F}}{\partial{y}} \hat{j}+\frac{\partial{F}}{\partial{z}} \hat{k}=3}$

So $\displaystyle{\iiint_D \nabla \cdot \overrightarrow{F} dV=\iiint_D 3 dV=4 \pi a^3}$, which is $\displaystyle{3 \cdot \text{Volume of a sphere}}$.$\displaystyle{G=x^2+y^2+z^2=a^2}$
$\displaystyle{\hat{n}=\frac{\nabla G}{|\nabla G|}=\frac{2x \hat{i}+2y \hat{j}+2z \hat{k}}{2 \sqrt{x^2+y^2+z^2}}=\frac{x \hat{i}+y \hat{j}+z \hat{k}}{a}}$

$\displaystyle{\overrightarrow{F} \cdot \hat{n}=\frac{x^2+y^2+z^2}{a}=\frac{a^2}{a}=a}$

$\displaystyle{\iint_S \overrightarrow{F} \cdot \hat{n} d \sigma=a \iint_S d \sigma=a(4 \pi a^2)=4 \pi a^3}$, which is $\displaystyle{a \cdot \text{Area of a sphere}}$
But... since $D$ is a sphere, why is $S$ also a sphere and not a circle?

 

[I like Serena]

Hi again! (Happy)

Hello again! (Wave)

I am looking at an exercise of the divergence theorem..

We want to apply the divergence theorem for the sphere $x^2+y^2+z^2=a^2$ in the case when the vector field is $\overrightarrow{F}=\hat{i}x+\hat{j}y+\hat{k}z$.$\displaystyle{\nabla \cdot \overrightarrow{F}=\frac{\partial{F}}{\partial{x}} \hat{i}+\frac{\partial{F}}{\partial{y}} \hat{j}+\frac{\partial{F}}{\partial{z}} \hat{k}=3}$

That should be $\displaystyle{\nabla \cdot \overrightarrow{F}=\frac{\partial{F_x}}{\partial{x}} +\frac{\partial{F_y}}{\partial{y}} +\frac{\partial{F_z}}{\partial{z}} =3}$.

So $\displaystyle{\iiint_D \nabla \cdot \overrightarrow{F} dV=\iiint_D 3 dV=4 \pi a^3}$, which is $\displaystyle{3 \cdot \text{Volume of a sphere}}$.$\displaystyle{G=x^2+y^2+z^2=a^2}$
$\displaystyle{\hat{n}=\frac{\nabla G}{|\nabla G|}=\frac{2x \hat{i}+2y \hat{j}+2z \hat{k}}{2 \sqrt{x^2+y^2+z^2}}=\frac{x \hat{i}+y \hat{j}+z \hat{k}}{a}}$

$\displaystyle{\overrightarrow{F} \cdot \hat{n}=\frac{x^2+y^2+z^2}{a}=\frac{a^2}{a}=a}$

$\displaystyle{\iint_S \overrightarrow{F} \cdot \hat{n} d \sigma=a \iint_S d \sigma=a(4 \pi a^2)=4 \pi a^3}$, which is $\displaystyle{a \cdot \text{Area of a sphere}}$

Good! (Smile)

But... since $D$ is a sphere, why is $S$ also a sphere and not a circle?

The volume $D$ is a solid sphere (aka ball), whereas $S$ is its boundary which is the surface of the sphere. The word "ball" is usually used to indicate the volume whereas the word "sphere" usually indicates only its surface.

Similarly we have a circle disk and its boundary which is a circle. (Nerd)

 

[evinda]

That should be $\displaystyle{\nabla \cdot \overrightarrow{F}=\frac{\partial{F_x}}{\partial{x}} +\frac{\partial{F_y}}{\partial{y}} +\frac{\partial{F_z}}{\partial{z}} =3}$.

A ok! I understand! :)

The volume $D$ is a solid sphere (aka ball), whereas $S$ is its boundary which is the surface of the sphere. The word "ball" is usually used to indicate the volume whereas the word "sphere" usually indicates only its surface.

Similarly we have a circle disk and its boundary which is a circle. (Nerd)

I thought that it would be a circle,because I had seen an example about the surface integral,where we take the projection of the surface at the $xy-$ plane..

So,we have to use the formula: $\displaystyle{ \iint_R g(x,y,z) d \sigma}=\iint_S g(x,y,z) \cdot \frac{|\nabla{F}|}{|\nabla{F} \cdot \hat{n}|}dA$.

The exercise I mean is this:

$$F=x^2+y^2+z^2=a^2, g=z , z \geq 0$$
$$\nabla{F}=2(x,y,z)$$
$$|\nabla{F}|=2r=2a$$
$$|\nabla{F} \cdot \hat{k}|=2z$$
$$\iint_R g d \sigma=\iint_S z \frac{2a}{2z} dA=\iint_S a dA=a \iint_S dA= \pi a^3$$So,in this case $R$ is a hemisphere and $S$ a circle.. At the proof of the divergence theorem, we also take the projection of the surface at the $xy-$ plane..

So,isn't it similar to the exercise I have asked..??

 

[I like Serena]

I thought that it would be a circle,because I had seen an example about the surface integral,where we take the projection of the surface at the $xy-$ plane..

So,we have to use the formula: $\displaystyle{ \iint_R g(x,y,z) d \sigma}=\iint_S g(x,y,z) \cdot \frac{|\nabla{F}|}{|\nabla{F} \cdot \hat{n}|}dA$.

The exercise I mean is this:

$$F=x^2+y^2+z^2=a^2, g=z , z \geq 0$$
$$\nabla{F}=2(x,y,z)$$
$$|\nabla{F}|=2r=2a$$
$$|\nabla{F} \cdot \hat{k}|=2z$$
$$\iint_R g d \sigma=\iint_S z \frac{2a}{2z} dA=\iint_S a dA=a \iint_S dA= \pi a^3$$So,in this case $R$ is a hemisphere and $S$ a circle.. At the proof of the divergence theorem, we also take the projection of the surface at the $xy-$ plane..

Ahaa.. I see what you mean. (Mmm)

So,isn't it similar to the exercise I have asked..??

Nope. I'm afraid not.

Your problem statement is about the integration of a 3-dimensional volume, which you can recognize by the triple integral $\iiint$ and also by $dV$, which represents an infinitesimal volume element.
The example you just gave is about the integration of a 2-dimensional surface in 3 dimensions, which you can recognize by the double integral $\iint$ and also by $d\sigma$ respectively $dA$ that both represent an infinitesimal surface element. (Wink)

 

[evinda]

Ahaa.. I see what you mean. (Mmm)

Nope. I'm afraid not.

Your problem statement is about the integration of a 3-dimensional volume, which you can recognize by the triple integral $\iiint$ and also by $dV$, which represents an infinitesimal volume element.
The example you just gave is about the integration of a 2-dimensional surface in 3 dimensions, which you can recognize by the double integral $\iint$ and also by $d\sigma$ respectively $dA$ that both represent an infinitesimal surface element. (Wink)

A ok... Thank you very much!