Tutor me

>>8827322
back to >>>/pol/ stormfrontfag

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>>8827322
This is all you need to know. Everything else you can easily prove yourself.

>>8827325
That's just the definition of an integral, that's not helpful.

>>8827322
>>>/pol/

>>8827331
If you can't derive everything else from that, you are a brainlet.

>>8827338
If I wasn't a brainlet I wouldn't be asking for help.

>>8827322
>>>/pol/

>>8827341
>>8827336
>>8827323

I'm not talking about racial integration, I'm talking about antiderivatives.

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>>8827345
you need to go back
>>>/pol/

>>8827345
what's the point of solving an indefinite multiple integral?

>>8827355
You can calculate volumes of shapes with varying sizes. Joking aside, I do need help.

>>8827338
>derive
it's differentiate you stupid fucking brainlet

>>8827357
but computing the anti derivative doesn't give you volume, it just gives you the anti derivative. you have to evaluate multiple anti derivatives at some points and use FTC to get a volume
is this what's confusing you?

>>8827367
I know you have to compute the antiderivative for each integral, I just don't know how to properly set up the limits of integration.

>>8827370
just figure out the bounds of each variable by analyzing the equations you're given, convert to whatever coordinate system makes the integration easiest (if need be), then integrate multiple times dudde

>>8827375
I'm stuck on the analyze part. Converting coordinate systems isn't too hard.

>>8827379
draw pictures
eg for calculating volume under a surface f(x,y), draw a picture of the domain on the xy plane, which should clearly show the limits of x and y

>>8827382
Do you always project along the axis you're integrating?

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>>8827382
For example say I was integrating the volume of a vertical cucumber in rectangular coordinates; what would the limits of integration be for the outer z-axis integral?

>>8827390
you're drawing a picture of the projection of the part of the surface that's bounded by some other function(s)

you want to project onto the plane that contains the domain of the function

>>8827427
>>8827424

Would I integrate this over the (...)-Z plane?

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>>8827424
>>8827424

well, if the cucumber is like in pic related,
your limits for z would be the functions of x and y (h(x,y and u(x,y)) that bound the top and bottom of the cucumber

you would use the projection of the cucumber onto the xy plane to determine your bounds for x and y

>>8827438
So for example if the top and the bottom are paraboloids, then the limits of integration are the two paraboloids?

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>>8827462
yes

now, the cucumber could also be like pic related, along the y axis, in which case you have functions in terms of different variables

your two paraboloids are now functions of x and z and make up the limits for y
the cucumber is projected onto the xz plane, you use this projection to find your limits for x and z

of course this extends to case where you cucumber is along the x axis

>>8827322
>Hi do you guys want to tutor me on Calculus III while I'm online?

i make 30$/hr doing that on Wyzant, why would i do it for free?

>>8827483
So for the xz projection, the limits of integration are +/- root(z^2 - r^2)? Whence cometh antiderivatives? We just integrate dzdydx in this scenario, right?

>>8827485
Because I'm such a great guy.

One time in Calculus 3, I spent 15 minutes working out a triple integral, only to get an answer of 0. That was the right answer. But the experience made me seriously question why I was in this class (besides that it was required).

>>8827492
I spent multiple hours on single problems involving Green's theorem.

>>8827490
if it's projected onto the xy plane, your limits of integration with respect to y will be the functions of the two paraboloids along the y axis

in my previous picture, h(x,z) = y = x^2 + z^2 and u(x,z) = y = -x^2 - z^2

>>8827507
I mean if it's projected onto xz plane, like you said, and like in >>8827483

>>8827507
Right, the limits of integration with respect to y are the paraboloids in terms of x and z. You calculate that integral and then integrate along either of the remaining axes, with the next limits of integration now being the top and bottom of the projected circle? Then you're done?

>>8827518

Yeah, but you're going to want to probably use cylindrical coordinates to make the integration easier

>>8827528
I just used the cucumber as a simple example. If you're using rectangular coordinates though, what does the stuff in the place of dydx do? Is it like volumetric density?

>>8827531
what do you mean the stuff in place of dydx?

dydx = dA

in a triple integral you have dydxdz = dAdz = dV

>>8827543
Say for example I replaced dydx with (y^2)(e^x)dydx. What would that represent?

>>8827553
(Not necessarily in terms of its particular graph, just what it does in general.)

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>>8827553

if it's in a double integral like so (forgive me for these paint drawings, I don't know latex), it would represent the volume of the solid under f(x,y) = (y^2)(e^x) and above the region R

>>8827565
Now say the limits of integration were the same as for the horizontal 'cucumber' you drew. Would it be the volume of the cucumber intersection lying below the surface (y^2)(e^x)?

>>8827576
Would it represent*

>>8827576

>volume of the cucumber intersection lying below the surface (y^2)(e^x)?

I don't follow. Is the entire cucumber (i.e., the cucumber shaped region in space bounded by 2 paraboloids) lying under the surface?
What do you mean by intersection? Intersection with what?

>>8827592
The intersection of the volumes of each; the intersection of the volume under e^x * y^2 with the internal volume of the cucumber, regardless of where the cucumber is positioned.

>>8827322
Work from the inside out based on which comes first dx,dy,dz etc.

Treat any variable other than the one you are integrating over as a constant.

Pretty straight forward mate.

>>8827610
That doesn't explain how to set up the integral in the first place. Thanks anyway.

>>8827602
Well if I understand correctly, this wouldn't work in a double integral, since the limits of integration for y in the case of the horizontal cucumber are functions of 2 variables (the two paraboloids).

If we wanted the volume of the cucumber intersection itself, we would just need to calculate the volume of the cucumber with a triple integral like before
the fact that it's under a surface is irrelevant

>>8827622
unless of course, the cucumber is oriented in such a way that it's "sticking out" of the solid under the surface, in which case we would need to solve for the volume of the space enclosed within the cucumber and the solid under the surface.

>>8827635
I'm sorry, I really don't feel I'm qualified enough to tutor for money, I'd be able to explain in much more clarity if I were. I hope I was able to help a little.

Do you have a text book for this stuff? Honestly some of the best advice I can give is read sections on multiple integration in the text book. If not do you know about Paul's Online Math notes? It is also pretty helpful.

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>>8827635
>that email THAT YOU DELETED
I FUCKING KNEW YOU WERE FROM /POL/ FAGGOT
GET OUT OUT OUT

>>8827637
I've read the book front to back and I honestly don't understand it. I could re-read it but it takes so long to read it. It would be much easier to get tutoring but all I have to my name is a really awesome app that I'm going to put on the market soon. I'll pay someone in market share if they can tutor me in more advanced calculus.

>>8827642
What is the deal between /sci/ and /pol/?

>>8827646
Are you in a university? If so you should have resources there you can take advantage of (office hours, math help centers, etc)

>>8827654
There's so much I'm unclear about on the subject that I'd rather just power through it with a tutor. If I went to the math help center I'd basically be going through the majority of the course with them. If anyone else is interested in app stock I'll take you up on it and give you a demo.. I would definitely be on the losing end of the deal big time.

>>8827661
What's there to be confused about?

Basically you can integrate independent variables freely.

So for example, if I have something like (x^2 + y^2)^1/2, so long as x and y are not functions of each other, I can integrate for dx, then for dy, or vice/versa.

The limits for one may contain the other variable. For example, I may integrate y between x and -x, setting up my next integral for x

It really is quite simple. This can be used to calculate areas, volumes, Fourier functions....basically anything.

>>8827661
>>8827661
Here is a fun little challenge problem for you. I want you to do this without looking it up.

Try to find the gravitational force from a sphere with uniform mass distribution. I'll give you a hint, use a cylindrical coordinate system

>>8827667
It's pretty clear that the limits of integration are always the "top and bottom" of the function along the axis of integration, then for each integral you just use FTC and do the next integral. That describes the volume in terms of "tops and bottoms", but it's unclear what the equation inside the integral (... dzdydx) do combined with the limits of integration.

I'm also confused about vector integrals; I never comprehended what the integral of for example V (dot) dx represents.

>>8827673
I've already done that. I've calculated escape velocity too (but I forget how I did it).

>>8827685
does*

>>8827685
ok ok my friend, let me explain this to you

An integral is a sum. Always go back to the riemann definition. Let's take the integral of f(x,y) dy from y=-x to y=x

What you are doing is taking f(x,-x) + f(x, -x +h) + f(x, -x + 2h) + ........ f(x, x).
This is how you should visualize it. You are summing up little differential elements

Now I must say I'm very impressed you've already done that problem, as it's pretty difficult and involves trig substitution. You can do stuff with electric fields to practice as well (what does the electric field look like around a wire with a uniform charge distribution?)

As far as vector integrals go, there are many different ones to look at. The one you are probably thinking of is the work integral. Which is the integral of the force vector dot dx

When dx is represented as a vector, it is the tangent vector to a path. So imagine a line curving through three space, like a looping wire. Dx would be the tangent vector to any place on that path. Usually these sorts of things are represented parametrically, so you can represent dx that way too (although you don’t have to).

So basically, take this path through space and cut it up into tiny little segments. Like a bunch of little vectors put end-to-end. Then make the tiny vectors differentially small. That is the dx vector. So remember work = force * distance. Represented in vector form, this is (F-vector) dot (displacement-vector). But if our displacement direction is continually changing, we dot F with the differential element, and then add up all the little elements

>>8827715
not him, but the force is due to a force vector field, correct?
so the integral of F dot dx along the curve you've described would be a line integral

>>8827729
Yeah they call it a line integral.

Which, if you know something about vector fields, will simplify nicely if the vector field is conservative

>>8827715
That vector explanation makes sense (I think that's a line integral) but there are other things to understand too like surface integrals. I think electric fields are represented as vector fields and those are hard to conceptualize too. Your explanation of an integral over f(x,y) is hard to picture; I can picture rectangles from z=0 to z=f with width h, but are they summed by holding x constant for each riemann sum along the y-axis?

Vectors are the hardest are for me; I don't really understand conservative vector fields or integrals with the circle around them, let alone a real understanding of Green's theorem and such.

>>8827752
Or I should say, by holding each dx constant (since it seems to be a rectangular box instead of a rectangle).

>>8827752
The circle simply means you're integrating over a closed curve. It doesn't change the integral, it's just there to signify what the curve is like.

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>>8827752
Ok, let me explain

A "vector field" is just a catalog. It tells you two things
>The magnitude
>The direction
of a vector at any given point in space.

So let's take an vector F(x,y,z). Plug in any coordinates x, y, and z, and this function will spit out a vector with a magnitude and direction. You can represent this graphically like pic related

Instead of little rectangles, picture little cubes. The length of one side is dx, the length of another is dy, and the length of the final side is dz.

When you integrate over one dimension (let's say dx), you add up all the little dx's, like stacking a bunch of boxes on top of each other. Now you are left with a series of columns. When you integrate across another dimension (let's say dy), it's like you're stacking all those columns together to form a sheet. And then when you integrate over the final dimension, you are stacking the sheets to create an object.

Now, it turns out through the miracle of integration that when you do a line integral of a gradient, the result is independent of the path you took. This is sort of like how when you take a standard integral, the final result is simply a function of the endpoints.

So a “conservative vector field” is a vector field which can be represented as a gradient. Therefore any associated line integral will be path independent (and therefore very very easy).

Since the curl of a gradient is 0, if you take the curl of your vector field and get 0, you know it can be represented as a gradient (and is therefore a conservative vector field)

>>8827779
As far as Green's theorem and stuff like this, I would just suggest you work through the proof a couple times. It makes perfect sense.

Basically it's saying that an integral through a region can be represented by an integral around the border of that region. It's basically like the Fundamental Theorem of Calculus

>>8827779
I don't have any intuition of what curl is; if I recall it's the amount that the vector field goes in a complete circle. My book made an analogy to a spinning windmill of a sort. But what are the implications if a gradient does this?

I can see the analogy of taking the line integral of a gradient; if you think of the gradient as a mountain it's just the difference in height of the endpoints, so I guess if you could go inside the mountain on your path it would somewhat make sense.. I guess the thing that's 'conserved' is just the outcome.

The last thing that really escapes me is curvature. I can interpret the normal vector to a path as centripetal acceleration but is that all there is to it and is this vector just the orthogonal component of the second derivative of the path?

I'll take your word on Green's theorem; maybe there's a video that will take me through it line by line. It's one of the topics that I was completely lost about by the time I got to it.

>>8827358
>differentiate an integral over an arbitrary measure space

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(1/2)
Ok, what when you get into mathematics you need to realize that trying to visualize and intuit every little thing starts to become impossible. It's difficult to "see" the curl in your head, but it makes mathematical sense on paper.

The book that does the best job I've ever seen of explaining all this is Griffiths "Introduction to Electrodynamics". In my opinion, this is one of the most amazing, beautifully written textbooks ever. If you can get your hands on it, I would highly recommend it (I have the 4th edition).

So let's say you have a function of F, which depends nominally on three variables. So it looks like F(x, y, z). Now let's say you're going to take the derivative of this function with respect to some arbitrary third parameter....let's go with t. F does not directly depend on t, but x, y, and z do. So by chain rule dF/dt = (dF/dx)(dx/dt) + (dF/dy)(dy/dt) + (dF/dz)(dz/dt). Now you have this parametrized function.

But let's remove the parametrization, and just get this thing in terms of arbitrary differential elements.
dF = (dF/dx)dx + (dF/dy)dy + (dF/dz)dz.
Now when we think about integrating thing, we have to consider the path this integral will take through 3-space. Whatever path we choose will define our parametrization (in other words, we will choose a 't' such that x(t), y(t), and z(t) give us a path we want). In the meantime, we just leave it like this as an arbitrary differential element.
It's sort of like representing
dF = (dF/dx)dx
(Maybe you've seen this before)

Anyway, it turns out that a convenient way to represent this function is as two vectors dotted together.
dF = [dF/dx, dF/dy, dF/dz] dot [dx, dy, dz]

We will call G{} = [dF/dx, dF/dy, dF/dz] the gradient of F. A vector quantity which (if you think about it) points in the direction of maximum change of F.
(by the way, I'm using {} to denote the thing as a vector. Since I can't draw an arrow over it)

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>>8827936
>>8827837
Now, because people in Electrodynamics like to abstract things into comfortable symbols and operators, let's pull out the F like this
G{} = [d/dx, d/dy, d/dz]F
We now will call [d/dx, d/dy, d/dz] the "del operator" which operates on F, transforming it into a vector field function. The del operator looks like an upside down triangle.

It turns out that this provides for some very neat and comfortable notation, as you'll soon see.

So this del operator a vector. What if we took the dot product with another vector?
Del{} dot V{}
= [d/dx, d/dy, d/dz] dot [Vx, Vy, Vz]
= dVx/dx + dVy/dy + dVz/dz
We call this the “divergence” of V

Seems useful right? What else can we do with this del operator?

Well let’s try a cross product.
del{} cross V{}
= [d/dx, d/dy, d/dz] cross [Vx, Vy, Vz]
= [dVz/dy – dVy/dz, dVx/dz – dVz/dx, dVy/dx – dVx/dy]
This is known as the “curl” of V

Interesting huh? Now using this operator we can prove some identities to help us use it in an algebraic context. For example, we can prove:
del{} dot (del{} cross V{}) = 0
>the divergence of the curl of V is 0
del{} cross (del{}F) = 0
>the cross product of the gradient is 0

And there’s lots of others. Pic related.

Now it turns out that by expressing the laws of electromagnetism in terms of the del operator, we can simplify our notation greatly. And also take advantage of all the identities in pic related.

Some other useful things with the del operator. Look these up yourself
>Gradient theorem
>Divergence theorem
>Curl theorem

These theorems massively simplify many of the laws of physics

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>>8827940
Here you go, Maxwell's laws written in differential form using the del

Apparently Maxwell himself did not have the benefit of this notation, and it hampered him greatly

>>8827936

That was a good writeup of that topic, but I read most of that in my book. I'm at a loss for what most of it means other than the interpretation of cross product, dot product and gradient. I can run with the idea that del is just an operator but I don't understand how it interacts with some things. Like >>8827338 said, I'm too much of a brainlet to memorize equations; I need to use visual memory to store the ideas in my head. Khan Academy supplements well for areas I'm weak in but they don't have everything there.

>>8827995
Del is an operator that provides a convenient way for simplifying notation (even though a lot of the equations it is involved in get really long and complex). When you're working with del, you can't treat it like a normal variable or something with the regular rules of algebra. It has special algebraic properties that must be applied when it is involved, (that image I posted for example)

But anyway, that's why the curl is a thing. It's just another thing to do with the del operator.
I don't memorize equations either. I go through and prove them once, then keep them written down for easy reference.

Anyway the reason I went into the electrodynamics stuff is because that's what this math is geared towards. It was basically developed for the purpose of simplifying the laws of electromagnetism. You can use it in other topics, but that's sort of it's "home base".

>>8828007
Thanks for the explanations anyway, they helped a lot. Maybe I'll be able to polish my Calculus III and finally do some advanced calculus.

The reason you aren't getting help is everyone on this forum is a "cult of science" pseudo intellectual.

In all honesty you're going to need to go to the calc III section of kahn academy or go to a tutoring center. This part of calc III really sucks and everyone has a hard time with it. I passed but am not good enough to try and explain it. Don't worry about all these assholes who google "hard math problems" and post results acting like they're something special.

>>8828011
Ok. Just some parting words

People think that math is some sort of super mystical, complicated thing that they'll never be able to understand as long as they live. When in reality it's pretty straightforward. If you're confused about something, work through the proof and you'll find out that actually, it makes a lot of sense.

Math is easy. People make it out to be way harder than it is. Don't buy this hollywood imagery of the tortured math genius with swirling complex images in his head. That isn't what it's like at all. You don't have to be a genius to understand it, you don't need some sort of special power. You just need to follow the reasoning.

It's like people who think they will never ever be able to play the piano. Then they sit down and practice, and a month later they can play fluently. Same story with math, anyone can do it

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>>8828015

Agreed in total. See you around, sir.

No. I do not