All right, so now I want to show you this really useful to consider alternative coordinate systems.
So we will do here is we will consider integration using spherical coordinates so you can see a problem
of a Cartesian coordinate system.
So I have discussed this in the previous lecture.
So the problem could, for example, be you want to calculate the volume of a sphere and you realize
that is very difficult to even determine the boundaries of your three dimensional integral, because
what you would have to do is you would have to parametrized this sphere in terms of X, Y and Z, which
is really difficult.
And the whole problem arises because your volume element is a cube, there is your shape that you want
to integrate over is a sphere.
So it's really difficult to fit this cube in a sphere.
And what would be much better is if this volume element would have already the shape of a sphere or
at least some portion of the sphere.
So what we want to do next is we want to derive the volume element in spherical coordinates.
So I can already tell you it's not that simple.
It's not just the extra Dizzie.
It's what we need in this case is we need the partial derivatives of the position vector with respect
to the individual coordinates and you will soon see why this is the case.
So here it's very easy, actually.
So if you if you calculate the derivative with respect to the length, you just get rid of these ares
here.
So this is what you get.
If you calculate the derivative with respect to theta, you have to differentiate sign of theta, which
is cosine of Seeta here the same.
And here you calculate the derivative of cosine of Sleater, which is minus sign of Theta.
And then the last one where you differentiate with respect to the polar angle, you have to differentiate
cosine of PHI, which is minus sign of Phi.
Here it is the derivative of sign-off which is plus cosine of five.
And here you do not have any phi dependent's.
So the partial derivative is zero.
The next thing we need is the length of these partial derivatives.
So what do you see here is of course still a vector.
You still have these three components.
So if we calculate the length of this vector, we have to take the X component squared plus the Y component
squared, plus the Z component square and then calculate the square root.
And while this one here may look difficult, it really turns out to be one that's actually a great exercise
for you if you want to practice this.
And the only thing that you need is that the cosine square plus the square of any angle is always one.
That's the only thing you need for this.
All right.
So we have the length of this derivative to be one.
This one here is R and this one is R times the sign of data here.
You can see it once again.
And now we have to consider that our position vector is not uniform in space, so in our gret it's not
uniform.
And this is expressed by these lengths of the partial derivatives.
We need to account for these when we determine the volume integral, but also the line integral and
other things.
This is because, for example, if you express a line integral for a line integration, you write down
that this is just D.R.
But now of course, the question is what does the R look like in our coordinate system?
Well, in any cordin system, what you can do is you can use such a chain rule, so to say.
So in very rough terms, this is not correct.
So please be careful about in very rough terms, you see that you could cancel this one in this one
out, so you would still have D-R.
But this is just like to memorize and to verify that this is true here.
So what do you have here?
Is the the chain rule basically of differentiation?
What you have to do is here you have to differentiate your position vector with respect to every single
coordinate and then have this differential here and use some of all the three components.
So in our case, this would be here, the R d to define.
And now, since this is a vector, you can also express it in terms of the length of times the unit
vector.
So we now need to use this one does and designer of sinus Tito.
And these unit vectors are E.R. Zitter and IFI.
So what we get in our case is that the ah is the scale of the ah times factor here plus ah there's one
here times Dr. Elita plus our science theater times DFI is fine.
This is our line element.
And similarly you can consider the surface elements.
So here, if you want to consider a surface element along the radial direction, then you have to of
course multiply the.
Yeah.
The other two perpendicular components.
So this would be the DFI.
But you also have to take into account that this base is not uniform anymore.
So you have to take into account these two absolute values of these derivatives.
So this gives us, in this case, our times, our times sign of detail.
And for the volume element, it's also very similar.
So here we have to multiply the three individual yeah, these are the three to divide elements, defi
elements.
Sorry, but we have to account for these lengths.
So we have to multiply all of these three terms here, which gives us an additional factor of our square
sign of theta.
So most important for our purposes is probably this the development here, then sometimes also this
D-R element and sometimes just DSR element, which is the surface for surface integration.
And so what this means is we now have found a way to express our volume elements.
We do not have a cube anymore, but we have something that looks like a part of our sphere and we know
how to express this.
So here is what our volume element looks like.
So you see here, it's really not surrounded by straight lines, it's really taking a part of the surface
of the sphere and then extruding it towards the radio direction so you can really build your whole sphere
by these volume elements that are no cubes anymore.
And the same applies to this, to the surface.
You really have a surface element that is not a square anymore.
And you can take multiple of these elements and fill up the whole sphere.
But please also consider that your element and also this element will look different for different positions
in space, which is due to the non uniformity of our coordinate system.
OK, but that was enough theory.
Let's use this.
Let's see what we can do with it finally.
So, yes, yes.
It's also, again, the equation for the surface element from the previous slide.
So now we take these two elements and we integrate over them.
So we want to calculate the volume of this of the sphere, which was our problem from the beginning.
So what we do is we integrate over the sphere with a radius capital R and we integrate over our volume
elements.
So we take this one here.
So, yeah, that's very simple.
Now, what we have to consider is that this is actually a three dimensional integral so we can split
it up into three individual integrals and we have to make sure that we get the correct boundaries.
So for the angle of five, we have to boundaries zero to two PI because this is the full circle and
the pull up.
Right then we have for the star angle from zero to PI, which is from the positive direction to the
negative C direction.
And for the radius we go from zero to the Capitol to our radius, which is the radius of us of our sphere.
And here you can see already what is so nice about spherical coordinates and what is so useful.
You can see that our boundaries do not depend on any of these quantities here.
For example, if we would calculate the same thing in Cartesian coordinates and our boundaries would
be X, Y and Z dependent so we could never write it down in terms of a product as I did here.
So what this means is we have now three independent, independent, one dimensional problems.
We can first calculate this integral than this one than this one.
And that is something that we can do that's very easy to do, actually.
So let's go ahead and calculate the in the find an integral integral of these things here.
So here we would write one time as defined.
And the integral of one is.
Fight, if you will, right here, the X then would be X.
So here it's five.
And now you have to take into account is the integration boundaries, which goes from zero to two PI.
Here we have to integrate a sign of theta, which gives us a minus cosine of Seeta, because the opposite
of the opposite way would mean which function do you have to take and differentiate to get the sign
of theta?
Well, of course you have to take minus cosine of data and then here the boundaries.
And for the last integral we have to think about which function do we have to differentiate to get on
square.
And this is of course, one of the three are to the power of three because then we would get three times
this whole thing.
But here with the power of two, which is exactly how square.
So now let's take into account our duration boundaries, so here we get to PI, minus zero is two PI.
Here we get the cosine of Pi is minus one, but we have another minus sign here, which is plus one.
And then we have to calculate minus minus the cost of zero, which is one.
So we have one minus minus one, which is two.
And he is very simple.
We just have one third times capital R to the power of three minus zero.
So these are our three individual integrals.
And so our result is that the volume of a sphere is four over three times pi times are to the power
of three.
So maybe you see now why this is really, really great, but maybe you may also say now that is trivial.
Everybody knows what the volume of a sphere is.
But the truth is, without these three dimensional integration in spherical coordinates, no one would
be able to tell about the volume of a sphere.
I mean, you get this taught in school that you have to know what the volume of the sphere is, but
you never learn where this comes from.
And now you know where this comes from.
It comes from the spherical coordinates and from the integration of for the surface elements.
The whole procedure is very similar.
So this time we calculate the surface of the boundary of atmosphere.
So we have a closed integral over this boundary, which I just indicate indicated here by this partial
symbol.
And then here we have again the radius of our sphere, which is the capital R, and we integrate over
the, um, the the surface area element along the radio direction.
So we have to take this one here and use Capital R instead of this normal R here and write this down
here under the integral.
And now what you can see is we have an integral over to two angles, Theta and Phi.
We have no integration over the radios because we know already that we only want to consider a certain
radius of our sphere.
So this means there's our square can be pulled out of the integration.
So we have our square times and then we have these two integrations here that we can factories because
the boundaries do not depend on any of these variables here.
So this means we have a product of our squared times, this integral times, this integral here.
And this is, again, very simple because we already know what this one is here is the same as that
one.
So we get to pie in this case.
And here we have to do it in a very similar way as here.
In fact, it's, again, the same thing.
So we get minus cosine theta as the undefined integral.
And this gives us a two if we consider the boundaries.
So what we get is our squared times two pi times two.
So the area of a sphere is four PI Times Square, which is also something that you probably have learned
already in school in the very early year.
But you were never really told where this comes from.
And in fact, it's not possible for a student in school to know where this comes from because you need
to consider integration and it's very much easier to consider spherical coordinates.
So I hope that now you see that sometimes it can really be useful to consider alternative coordinate
systems and you will find the use of these coordinate systems, especially in physics, in many problems.
I've already introduced two of them to you.
We have a wire where we calculate the magnetic field that we typically use cylindrical coordinates,
or we could also calculate the electric field or four point charge where we would use the spherical
coordinate system because our whole problem is rotational symmetric.