To compute the divergence in spherical coordinates, we need to understand the formula for divergence in this coordinate system. Spherical coordinates are defined by three parameters: the radial distance $r$, the polar angle $\theta$ (measured from the positive z-axis), and the azimuthal angle $\phi$ (measured in the x-y plane from the positive x-axis).
To compute the divergence in spherical coordinates, we need to understand the formula for divergence in this coordinate system. Spherical coordinates are defined by three parameters: the radial distance $r$, the polar angle $\theta$ (measured from the positive z-axis), and the azimuthal angle $\phi$ (measured in the x-y plane from the positive x-axis).
The divergence of a vector field $\vec{F} = F_r \hat{r} + F_\theta \hat{\theta} + F_\phi \hat{\phi}$ in spherical coordinates is given by the following formula:
Here, $F_r$, $F_\theta$, and $F_\phi$ are the components of the vector field $\vec{F}$ in the radial, polar, and azimuthal directions, respectively.
Let's go through the steps to compute the divergence of a vector field in spherical coordinates:
Step 1: Identify the components of the vector field First, you need to have the vector field $\vec{F}$ expressed in terms of its spherical components $F_r$, $F_\theta$, and $F_\phi$.
Step 2: Compute the partial derivative with respect to $r$ Calculate the partial derivative of the radial component $F_r$ with respect to $r$, and then multiply by $r^2$. This gives you the first term in the divergence formula.
Step 3: Compute the partial derivative with respect to $\theta$ Calculate the partial derivative of the polar component $F_\theta$ with respect to $\theta$, multiply by $\sin \theta$, and then divide by $r \sin \theta$. This gives you the second term in the divergence formula.
Step 4: Compute the partial derivative with respect to $\phi$ Calculate the partial derivative of the azimuthal component $F_\phi$ with respect to $\phi$, and then divide by $r \sin \theta$. This gives you the third term in the divergence formula.
Step 5: Sum the terms Add the three terms obtained from steps 2, 3, and 4 to get the divergence of the vector field in spherical coordinates.
Step 6: Simplify the expression Finally, simplify the expression obtained in step 5 to get the divergence in its simplest form.
By following these steps, you can compute the divergence of any vector field in spherical coordinates. Remember that the divergence is a scalar field that represents the magnitude of a vector field's source or sink at a given point.
Exercise 6-Computing the Surface Integral of a Vector Field on a Body
Exercise 7-Convert (5,0,𝝅) from spherical to cartesian coordinates.
Exercise 5-Calculating the Volume of Overlapping Spheres using Triple Integrals in Spherical Coordinates
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Exercise 6-Computing the Surface Integral of a Vector Field on a Body
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In this exercise, we have to compute the following type to surface integral.
N is the outward unit normal and the vector field F is given here.
We have to compute it over the surface S,
which is the outer wrapper, if you like,
of the body bounded by,
well here the equations.
Let me call the body B.
B is like the 3D region and S is just the outer.
It's a hemisphere and a disk at the bottom.
Why did I say it's a hemisphere?
Because this is equivalent to,
if you square both sides and move stuff over,
you get x squared plus y squared,
plus z squared equals a squared,
or rather you get this from this by extracting z but only taking
the positive square root leaves us just with the top part.
Anyway, this is the body,
and because we're in the chapter on divergence theorem,
you can guess what we're not going to compute the surface integral,
we're going to call in the theorem and compute the triple integral over B.
In fact, the divergence theorem here says that the integral above the double integral
of F.ndS is equal to the triple integral.
S is the border,
the boundary of body B of
the divergence dv for 3D integrals.
Usually we label each of the functions.
This one, this one and this one.
Some like to use [inaudible] ,
I like to use p, q, and r,
and in the event the divergence of F would be this thing with respect to x,
partial derivative plus second one with respect to y,
plus the third one,
in our case r with respect to z.
In our case, what do we get?
P with respect to x is just z squared,
q with respect to y. I'd like to apologize,
I miss copied the question.
The two should go here like this, sorry about that.
The derivative with respect to y is x squared and the
derivative with respect to z will be just y squared.
What we have to compute now is the triple integral over the upper half ball
B. I can change the order to suite me to write it as x
squared plus y squared plus z squared dv.
The way I'm going to do this is to use spherical coordinates, remember those.
I'll just write the word spherical.
We have 3 coordinates that we use in 3D.
Cartesian cylindrical, spherical.
We have a sphere shape or part of a sphere centered at the origin.
Also, x squared plus y squared plus z squared is very good for spherical.
I'll just write some of the equations.
We need an iterative equation for x,
for y, for z, and for dv.
Now actually, I'm not even going to use this equation.
I'll leave this if I need it,
I'll go back and tell you what it is.
I do need dv,
which is r squared sine of
Pi d r d Pi d Theta.
Usually there's an extra equation that goes with these that it turns out that x squared
plus y squared plus z squared equals r squared.
What we get here, it is as follows.
We have to convert the region b,
the upper half bull,
and see what happens to each of the 3 new variables,
Theta, where does it go from?
Pi? Where does it go?
R, where does it go?
Well, Theta is like the longitude.
Actually we start from the x-axis and go around,
and we get one complete circle around.
We go from 0 to 2Pi.
Right here, 0 to 2Pi.
Now, Pi is like latitude only.
We don't start from the equator,
we start from the North Pole.
Here it's Pi is 0,
and then it goes all the way
down to the negative z-axis, the South Pole.
Here it's a 180 degrees or Pi,
but we're only getting half the way around 90 degrees around the world.
This is Pi over 2 and Pi goes from 0 to Pi over 2.
R is just the distance from the origin.
It goes from the origin to the sphere.
It goes from 0 to a.
This is the r, 0 to a.
Now, x squared plus y squared plus z squared from this formula is r squared and dv
from here is
r squared sine Pi
d r d Pi d Theta.
Well, you weren't looking at, I filled an x, y, and z,
I didn't feel right about not leaving them out, but we didn't use them.
Now back here, I'd like to make some minor changes.
What I'd like to do is, well, two things.
I'd like to combine,
let me just write these and wanted to leave a bit more space here. There we go.
I want to put sine Pi in front.
I like to pull things to the front because this doesn't depend on r,
and then r squared with r squared is r to the fourth,
and then I just have d r d Pi d Theta.
I need to still write the limits 0 to 2Pi,
0 to Pi over 2, 0 to a.
We start from the innermost, here.
I'll do this one on the side.
What I get is r to the fifth, just doing this one,
r to the fifth over 5 from 0 to a,
and that is equal to a to the fifth over 5 minus 0,
and this is a constant.
I can actually bring this right in front a to the fifth over 5.
There's nothing left here.
I have integral 0 to 2Pi,
integral from 0 to Pi over 2 of
just sine Pi d Pi d Theta.
Now, the innermost one is this one.
I'll do that at the side also.
Over here what we get is the integral of sine Pi is minus cosine Pi,
and this we want to take from 0 to Pi over 2.
This equals cosine of Pi over 2 is 0.
I have minus 0,
minus and cosine of 0 is 1 minus minus 1,
so this is just equal to 1.
If this is 1, then all I'm left with now is a to the fifth
over 5 times the
integral from 0 to 2 Pi of just 1d Theta.
Now the integral of one is always the upper limit minus the lower limit is 2 Pi.
What I get is 2 Pi a to the fifth over 5,
because like I said, this bit here came out to be just 2Pi,
I multiplied by this,
and this is the answer to the question. We're done.
This video explains how to compute a surface integral over a hemisphere and a disk at the bottom. The body is bounded by the equation x² + y² + z² = a². The integral is computed using the divergence theorem, which states that the integral of F.ndS is equal to the triple integral of the divergence of F with respect to x, y, and z. The integral is computed using spherical coordinates, and the result is 2πa⁵/5.
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