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So in the previous two examples, we have rotated a quasi one dimensional stick, either around its

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end or around the middle of the stick.

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And in both cases, the numerical solution was really easy.

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It was basically just creating this list of our points and then just calculating into with a single

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line of code until this one.

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And still, you may say it's not really worth doing it because the analytical solution is so simple.

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It's just a one dimensional integral where you just have to be careful and what you plug in, but then

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you just have to integrate over our square, which is really easy to do.

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It's just one-third.

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Times are to the power of three.

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So and I would agree to this statement, it wasn't really that useful, to be honest, because I would

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also say this one is equally as easy.

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However, now we're going to more difficult examples where the analytical result will become more and

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more difficult and numerical results will basically stay just as easy.

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And the first example of we're going to solve here is a sphere which is three dimensional, but it's

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still very symmetric, and therefore we can use some tricks and make it a bit more easy.

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So the first thing that I want to show you not in detail, but just briefly, is the analytical solution

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to calculate the moment of inertia, of rotating a sphere around an axis which I just call the axis,

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which goes through the center of the sphere.

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And the result will be two over five m times the radius of the sphere squared.

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So once again, we start from the same equation as we did before integral over R squared and R is not

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the radius.

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This is really important to remember.

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This is the distance of the point that we are considering with respect to the z axis.

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So always the shortest distance.

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So it's not the radius, not the distance to the center of the sphere necessarily.

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And now, since this time we have a really a three dimensional object, we can take the actual density

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instead of the line density.

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So I wrote to volume density, which is, of course, mass divided by volume.

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And then the is real times.

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So we write here that this is a row which is over three times the integral Devi Square.

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So I don't know your background.

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And if you have done some integration in cylindrical coordinates, but here it's actually pretty easy

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to do.

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Maybe you ask me now, why are we not using spherical coordinates?

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And the reason is that this hour, as I just mentioned, is not the radius of this, you know, of the

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sphere, but it's the distance to the axis.

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So it's basically a distance in cylindrical coordinates.

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And therefore we use cylindrical coordinates and we express this volume element DV in terms of cylindrical

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coordinates.

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This is something you can look up in any text book.

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This is just our times.

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DRDO z define where it is.

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It's the polar angle and the and then of course, we have to consider also this all square here.

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So in total, we get our to the power of three times d.r and easy defy.

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And no, basically the only difficult part will be to get the boundaries here, correct.

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So for five, this is the angle in the X y plane, basically, which yeah, we in Germany call fi,

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but I heard that in other parts of the world, you call this theta.

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So please don't be confused here.

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So fi goes from zero to two pi to describe a full rotation.

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So this is really easy to do.

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This is what you see here, basically.

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And then once we calculated it's just two pi, and for the other two integrals, it becomes a bit more

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difficult because we integrate over our end of a z.

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And so first, we have to integrate over our because we have to r here as well and our goes from zero.

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So basically, from the Z axes to a certain point, which I called R2, the which is basically the yeah,

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the distance from the axis to the shell of the sphere.

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And this depends on the Z coordinates.

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So if we are at Z equals zero, so this is just in the X Y plane, then this our tilde will be equal

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to our.

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But if we are at the top of the sphere just at the tip, then it will be zero because they're at a sphere

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cuts the z axis.

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So when you do some trigonometry or some, yeah, Protagoras.

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Then you get that this hour, tilde is square root of our square minus C square, because the radius

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of the sphere, the actual radius capital R is equal to z squared plus R to the square.

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And now it's basically just mathematics.

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You calculate, first of all, the integral of R2 to power three, which is one of four times, are

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to the power of four.

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Then you use these two boundaries, which are these two here?

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The other one is just zero, and the upper one becomes all square minus C square square root to the

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poorer, for which you can also write like this.

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Then you have to use some binomial formula to get these three terms, and these are also really easy

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to integrate.

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You basically have.

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Here are two to four times Z.

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Here you get a term that is that gives one over three times z to the power of three, which in total

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gives this one.

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With these two factors, and this one just gives one of a five z to the power of five.

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Then you use the boundaries Z goes from minus capital R to plus capital R.

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And so all of these terms simplify and gives you a power of R2, the power of five times some prefecture.

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Now you have to just be careful with the fractions.

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And here this one gives you basically 15 over 15.

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This one gives you minus 10 over 15.

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And this one gives you three over 15.

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So if you add all of this up, it gives you eight over 15 together with these twos, which cancel,

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actually.

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So this will be the results, though we can further simplify this by using the volume of the sphere,

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which is, of course, four four three times PI times R two to three.

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So you get this fraction here.

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And so basically, you just have to divide these two fractions.

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So you have eight over 15 times the reciprocal, which is three over four.

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So the three and the 15 council, which gives them one over five and the eight and the four cancel,

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which gives two.

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So maybe this was a bit fast, but as I said, this is not here theoretical physics or mathematics.

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So we actually don't want to go through all of this.

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I just wanted you to show.

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Show you this ones and to explain to you or to make clear to you that when you do this, especially

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for the first time, it's pretty difficult that it takes a long time to write this down and to figure

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this out.

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And maybe you even do mistakes on the way.

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And yeah, so all in all, it's just pretty demanding and it would be a good idea to solve this numerically

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instead.