calculus continued

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calculus continued

Mathematics
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Alright, good. Here goes the explanation, and I apologize in advance for the probably atrocious drawing.
When we evaluate integrals, we break up domains into tiny chunks, as we talked about previously. When it's a line integral, we break it up into little tiny lines. When it's a surface integral, we break it up into little tiny bits of surface area, and for a volume integral, we break it up into little tiny volumes. Therefore, fundamentally, we want to find this: \[\iiint F\space dV\] Now, when we use cartesian coordinates, we need to express both the function F and the volume element dV in terms of the rectangular coordinates x,y, and z. We then get \[\iiint F(x,y,z) \space dxdydz \]
right!

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But who says I want to use cartesian coordinates? Spherical coordinates are also lovely. Transforming the function is one thing, which you can do already, but what about transforming the volume element? When we use cartesian coordinates, the volume element is a cube, with side lengths dx,dy, and dz. What about in spherical coordinates?
This is a bit more complicated, so lets look at a drawing: |dw:1328599322865:dw|
|dw:1328599356326:dw|
and you said your drawings would be autrocious..
atrocious*
Imagine a little tiny volume at the end of that. It has basically three components, just like the cartesian cube, but the lengths are a bit more complicated. First, when we fiddle with r, then we change this element by an amount dr:|dw:1328599510907:dw|
When we fiddle with phi, it's a little more complicated. If you recall the idea of arc length of a circle, the length of the arc we trace out when we have a circle of radius r is equal to \[r\Delta \theta\] That means that the dimension we get when we fiddle with phi is:|dw:1328599725183:dw|
I have never looked so deep into this theory before, wow.
Lastly, we apply the same reasoning to the last dimension. However, this is the most complicated of all, because when we mess with theta, we're rotating around the z axis rather than the origin. Therefore, our circle doesn't have a radius r, it has a radius equal to r's shadow in the x-y plane -- i.e., a radius of \[r\sin(\phi)\] So all together, we have \[ dr \text{ , } r d\phi \text{ , and } r\sin(\phi) d\theta\] All multiplied together, that yields \[ dV = r^2\sin(\phi) dr d\phi d\theta\]
Jesus.. I'm going to be referring back to these notes you gave me for the rest of the term, and I'm 100% serious. if you want I'll create like 20 threads to give you medals ahha.. you deserve ... i don't even know. Something amazing
That won't be necessary :) It's enough to know that I was able to give you some more insight. I'm glad it was useful. If you feel motivated, try the same thing with cylindrical coordinates. It isn't too different from this exercise, and you'll find that \[ dV = r\space dr d\theta dz \] Calculus really isn't as obtuse as people make it, but sometimes they just refuse to explain it properly.... oh well. Such is life. :)
Yeah, consider teaching! You make so much more sense than my calculus professor, and he's considered a pretty smart guy! heck you're already some kind of professor I bet though.
Just know your time was well spent and greatly appreciated by THIS GUY. I think that'll be enough for me tonight, hopefully this insight will stick with me as I'm writing my nice little quiz tomorrow. (as well as finishing my assignment). You take care, as well I. Calculus is fun when it makes sense :)
Not a professor, will be entering Graduate school in Physics this fall ;) Thank you though. And good luck!
Jemurray - awesome awesome work my friend. Great answers, and great use of the drawing tool. Amazing.
And mattt9 - way to stick in there and give out a medal - well played sir.
Thanks cshalvey, I appreciate it!
Goodnight guys, hopefully see you wandering around on here some other late math filled evening
Hey, it's people (like both of you) that make OS an awesome place. Can't thank you both enough =)

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