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[Furtuka]

How come when integrating over the volume of a sphere you have to multiply the function by p^2 * sin(phi)?

[Furtuka]

How do I integrate a function over the surface of a cube?

[i2amroy]

How do I integrate a function over the surface of a cube?

Lots of double integrals put together (or a single triple integral with very creative bounds). Integrate over each face of the cube separately with an appropriate double integral, then add the results all together. If the cube surface has a thickness then you would need to use triple integrals over each face, and ensure that your bounds didn't overlap at the corners.

An alternative way (and probably easier way) to do it if the cube has a thickness would be to triple integrate over a solid cube of the outer face, then subtract the triple integral of a cube the size of the inner face to "remove" that part of the total cube.

[Helgoland]

Don't we have Stokes for these things?

[3man75]

Hi everyone!

I'm taking Algebra again in college again next semester and wanted to ask about radical solutions.

I have an example in my book where it feels that the numbers simply come out of the blue for some reason and I don't know why they exist. "What do you mean?" you might be asking, well its this.

Imagine a radical stretching over (X-1) and it continues to "equal X-7". From here I understand to get rid of the radical you do the opposite which is exponents.

Which nets me X-1=X(to the second) - 14x +49.  (Note: I don't know how to do exponents on my keyboard, I apologize in advance).

At this point I have to ask why 14x and 49 are existing simultaneously. I guess it has to do with the X being multiplied by the 2 but if thats true then there should only be 2 X's and not 3.

[Helgoland]

I'm taking Algebra again in college again next semester and wanted to ask about radical solutions.

Okay, go get an apron, a butcher's knife and some intel on the nearest hog lot. You do know where your prof lives, right?

In the meantime I'll try to answer your question:

I have an example in my book where it feels that the numbers simply come out of the blue for some reason and I don't know why they exist. "What do you mean?" you might be asking, well its this.

Imagine a radical stretching over (X-1) and it continues to "equal X-7". From here I understand to get rid of the radical you do the opposite which is exponents.

Which nets me X-1=X(to the second) - 14x +49.  (Note: I don't know how to do exponents on my keyboard, I apologize in advance).

At this point I have to ask why 14x and 49 are existing simultaneously. I guess it has to do with the X being multiplied by the 2 but if thats true then there should only be 2 X's and not 3.

The problem is sqrt(x-1) = x - 7, right? Your approach is perfectly valid, it seems you're just having problems with squaring the x - 7 term.
Remember: x behaves just like a number most of the time. You just multiply like you'd multiply any sum:
(x - 7)^2
= (x - 7)*(x - 7)
= x*x - 7*x - x*7 + 49 (Is this bit clear? You really just dissolve the parentheses.)
= x^2 - 7*x - 7*x + 49 (Since x^2 is defined as x*x and multiplication is commutative )
= x^2 - (7 + 7)*x + 49 (We reintroduce some parentheses to simplify)
= x^2 - 14*x + 49 Voila!

Couple of style tips: Use lower case letters for variables - uppercase ones are usually reserved for sets or probability distributions (or something like that, I don't quite remember). And (as you've seen above) roots of something are usually written as sqrt(something) when writing plain text.





A question of my own: On p 410 of Hatcher's Algebraic Topology he says that

To the extent that fibrations can be regarded as twisted products, up to homotopy equivalence, the spaces X_n in a Postnikov tower for X can be thought of as twisted products of Eilenberg-MacLane spaces K(\pi_n(X),n).

What does he mean by 'twisted products'? Spaces that locally look like products but globally have a different structure? Then I'd need to know whether the twisted product of two finite CW complexes is still finite...

[3man75]

1. I cracked up really well their thank Helgoland.

2. x*x - 7*x - x*7 + 49 (Is this bit clear? You really just dissolve the parentheses.)

x*x=X to the second. Right?

...-7x-x*7 + 49. Again where is this 49 coming from though?

(x-7)*(x-7) turns into ----> ((x*x - 7*x - x*7 + 49)) but why? Why are their 4 x's now and again where is this 49 coming from when both of the 7's are still on the equation? Is this fast forwarding?


3. Just out of curiosity how advanced are you in mathematics?

EDIT:

In my book after squaring I understand that both sides are squared. An they are. However, (x-7) after being squared in my mind should become X (to the second) and 49 but instead I see that part of the equation become X(to the second) - 14x +49. It's that -14x that bothers me because in my understanding the x's and 49 should be alone. That's the part I need clarification on. Sorry about the misunderstanding.

[Helgoland]

Sixth semester, slowly losing my ability to calculate

The 49 is just 7*7, and yeah, x*x=x^2 ('x to the second', the ^2 is just a shorthand) by definition. Lemme give you an example with concrete numbers:

(3+4)*(3+4)
= 3*3 + 4*3 + 3*4 + 4*4 (again, just dissolving parentheses - this you need to pretty much learn by heart, dissolving parentheses is important)
= 3*3 + 4*3 + 4*3 + 4*4
= 3*3 + (4 + 4)*3 + 4*4
= 3^2 + 8*3 + 16
= 49 which is the same as if we'd calculated 7*7 from the beginning.

Now let's say we want to do the same calculation ~fifty times, just with 3 replaced by other numbers. Instead of doing these steps fifty times, we use a placeholder - x - that behaves just like any ordinary number. The 'additional x's' that appear aren't any more additional than the 'additional' threes you see in the example; they just appear because we no longer have the parentheses keeping them bottled up. Kinda like this:
5*7 = 7 + 7 + 7 + 7 + 7
All we're doing is translating from a shorthand version into a more explicit one.

[Graknorke]

Huh, up until now I didn't even know there was a phrase specifically to describe radical equations. That's interesting,
And I agree with Helgo on learning how to multiply out parentheses, it's something that you'll need to do a lot when you're working with equations. Same with factorising too.

[Reelya]

3man75

(x-7) * (x-7) is not equal to  x^2 + 49, because that fails to multiple the x's and -7's together. There is no specific reason to only multiply the x's together and the 7's together. everything has to be multiplied together, hence why the answer is not x^2 + 49

here is another way to break it down. you can split the first (x-7) up:

(x-7) * (x-7)  = x (x-7) - 7(x-7) <= this is just from realizing that you can split up the first x-7 and multiply each component of that by the remaining (x-7)

(x-7) * (x-7)  = x*x - 7*x - 7(x - 7)  <= multiplying out the first term
(x-7) * (x-7)  = x*x - 7*x - 7*x -7*-7  <= multiplying out the second term
(x-7) * (x-7)  = x^2 -(7*x + 7*x) + 7^2  <= grouping the 7*x's and putting squares in
(x-7) * (x-7)  = x^2 -14*x +49  <= simplifying everything

[Helgoland]

Oh c'mon, use parentheses when multiplying negative numbers - you'll confuse the poor guy if you don't.

[3man75]

Actually I figured it out. I've been meaning to post that but other things on meatspace have gotten in the way.

Huh, up until now I didn't even know there was a phrase specifically to describe radical equations. That's interesting,
And I agree with Helgo on learning how to multiply out parentheses, it's something that you'll need to do a lot when you're working with equations. Same with factorising too.

Huh, up until now I didn't even know there was a phrase specifically to describe radical equations. That's interesting,
And I agree with Helgo on learning how to multiply out parentheses, it's something that you'll need to do a lot when you're working with equations. Same with factorising too.

I think I know how to multiply out of parenthesis but then again I might not simply because I need someone to physically show me what is being said sometimes. For example (An a bad one at that)

I saw that (X-7)^= (X-7) (X-7).

Because X-7 * X-7 is true I can distribute it so that I can get X^-14X+49. This is what Helgoland was saying but I missed because of the symbolism online being different. ((2(square) and 2^ just looks so different that I basically didn't know how to interpret it.))


EDIT: Does that make sense to the readers?

[Graknorke]

The caret (^) is a power symbol.
So x^2 is x squared, x^3 is x cubed, and so on.

[Arx]

Slightly better-looking but requiring more effort are the [sup][/sup] ^superscript tags.

[3man75]

Slightly better-looking but requiring more effort are the [sup][/sup] ^superscript tags.

wha-? Que?