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[Darvi]

I get to correct somebody who should know better than I do? Feels like school all over again.

By the way, PH, it's called multiplying by the conjugate or something.

[PsyberianHusky]

Part of the problem is I do not understand myself
can I add the square root of Y to both the top and bottom ?
or can I only do multiplication to the top and bottom

Y+ [√y]
_______
Y-√y [+√r]



y+√y
______
y





√y

[Darvi]

Hell fucking no. Only multiplications, and only by the same factor.

[PsyberianHusky]

so


Y*Y-√y
_______
Y-√y *Y-√y 
Because as long as multiply the top and the bottom evenly the ratio remains unchanged


is that what Vector was trying to suggest

[Virex]

Part of the problem is I do not understand myself
can I add the square root of Y to both the top and bottom ?
or can I only do multiplication to the top and bottom

Y+ [√y]
_______
Y-√y [+√r]



y+√y
______
y





√y

Technically you can, as long as you also subtract it:

Code: [Select]

y      Y + √Y - √Y      √y
____ = ____________ = _______ + 1
y-√y       y-√y         y-√y

Alternatively you can divide by √y on both sides of the divisor, giving you:
√y
____
√y-1



Further simplification is as far as I know not possible, unless √Y >> 1 , in which case the equation reduces to:
√Y
__  = 1 because √Y-1 ~ √Y
√Y

[PsyberianHusky]

Thank you so much, I recall it now

[Sowelu]

I found a funny.

Spoiler (click to show/hide)

[Virex]

Well at least the mathematician knows how to read

[Darvi]

Heh. I' the kind of person to give the second answer, and then the school system makes me fail because it's not the answer they wanted.

Then ask unambiguous questions you jackasses!

[Virex]

If they do that, get cross because pi can definitely be written as a fraction, just not as a fraction of 2 integers

[Dr. D]

My matrix is defective. Can I get another one?

[Christes]

At first, I assumed you meant it had determinant 0 xD

[Vector]

At first, I assumed you meant it had determinant 0 xD

I know, I too was sitting there thinking "... is this a mathematical word that I don't know of?  The generally defective linear group?  Whyyy"

[Vector]

Fuck.

Okay, dudes, this is going to make me look dumb, but I'm writing it out anyway.  The problem:

Consider the area in the xy-plane bounded by line y=1 and parabola y=x^2.  Let P be the solid obtained by revolving this area around the line y=1.

Considering P as the sum of its circular cross-sections, apply Cavalieri's results to obtain v(P) = 16pi/15.


Now, the Cavalieri part doesn't matter--that just gives us an integral for any whole-numbered power of x.  What does matter is that I can't quite figure out what I'm doing wrong here.

Observation: this volume is the same as that you'd get from rotating the given figure around the y-axis, by symmetry.  This isn't important, but I thought it was kind of cool.

So okay.  "Remembering" what I learned from high school calculus, we know that we need the integral of (1-x^2)^2 from 0 to 1.  Or something like that.

... Well, that gives us the lovely answer of 8pi/15, not 16pi/15, which is precisely half of what we're "supposed" to get.  Any advice?

[Darvi]

Isn't it pi times the integral of (x^2-1)^2 from 0 to 1 that's asked for?

Wait, due to the square that's the same thing.