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Old 08-26-2007, 12:32 AM   #21
darkstar2466
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I got (sin(4)/2 - 1/2) - (sin(-4)/2 + 1/2). I think it's right.

EDIT: F*ck, I forgot to change the bounds of integration. Whatever, it would end up right if I redid it.
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Old 08-26-2007, 12:32 AM   #22
guywithguitar
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Bingo.


Haha, guys, it's just the area under half of a unit circle, no math involved.
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Old 08-26-2007, 12:34 AM   #23
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I'll be needing this a lot...


You better be telling the truth on the answers!

Cus I sure as hell wouldn't be able to tell if you're lying
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Old 08-26-2007, 12:34 AM   #24
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Pi/2?


Which also equals 1.571. Which is what I said.
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Old 08-26-2007, 12:34 AM   #25
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Ok can someone tell me how to plot those graphs that are like "y=x+2" except they dont tell you x?

im not quite sure what u mean, if u mean as in y=2, then you would just draw a straight line on the y=2 spaces, i.e. (0,2)(1,2)(-1,2) all the way across, so no matter what the x=?, the y=2 always
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Old 08-26-2007, 12:35 AM   #26
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Originally Posted by bigwillie
Which also equals 1.571. Which is what I said.


Yeah, but you actually did the integral .
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Old 08-26-2007, 12:36 AM   #27
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Which also equals 1.571. Which is what I said.


I know, but keep it in radians. Sorry I didn't mention it.
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Old 08-26-2007, 12:38 AM   #28
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Anyone done double/triple or flux/circulation integrals here?
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Old 08-26-2007, 12:39 AM   #29
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Yeah, but you actually did the integral .


But the long way was much more fun.

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I know, but keep it in radians. Sorry I didn't mention it.


Ah well, it's all good.
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Old 08-26-2007, 12:41 AM   #30
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Originally Posted by myself101
Ok can someone tell me how to plot those graphs that are like "y=x+2" except they dont tell you x?


y = x+2

your slope is 1/1, and your y-intercept is (0,2). So start at that point and move up one and over one. Plot that point. then move up one and over one. and plot that point. you'll soon see what the line is supposed to look like.
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Old 08-26-2007, 12:41 AM   #31
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Originally Posted by darkstar2466
Anyone done double/triple or flux/circulation integrals here?

I've had one double integral problem ever, and was kind of walked through it, but I think I understand how to do them. But do elaborate on the flux and circulation integrals
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Old 08-26-2007, 12:44 AM   #32
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Anyone done double/triple or flux/circulation integrals here?


I've done a few triples, but I've never heard of the flux/circulation integrals before either.
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Old 08-26-2007, 12:44 AM   #33
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Ok here's another easy one. This is trig.

Give me the area of a triangle with sides a-9.12, b-4.67, c-10.83 and tell me how you did it.
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Old 08-26-2007, 12:49 AM   #34
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Originally Posted by guywithguitar
I've had one double integral problem ever, and was kind of walked through it, but I think I understand how to do them. But do elaborate on the flux and circulation integrals
.


Flux is the amount of flow over a certain space, so you're usually given a scalar quantity and you integrate over that region to the flow from it. It's usually used for electricity and magnetism. Here's the basic form:



Circulation is similar, except it's over a closed loop and you figure out the integral of the velocity of a fluid or something circulating. Here's the basic form for that:



But before you do that, you must learn line integrals, which is the basis of these.
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Old 08-26-2007, 12:51 AM   #35
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Ah, I'm taking E&M next semester, a.k.a. about a week from now, so I'll be familiar with those soon .
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Old 08-26-2007, 12:54 AM   #36
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Ah, I'm taking E&M next semester, a.k.a. about a week from now, so I'll be familiar with those soon .


Yea, you should hit these pretty soon. The math series I'm taking does this first, to prepare for E&M, I guess.

What math are you taking?
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Old 08-26-2007, 12:56 AM   #37
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im a dumbass when it comes to math

n science


n all school stuff
















i like music class tho that count?
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Old 08-26-2007, 12:58 AM   #38
Zackie EL
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No answers?
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Old 08-26-2007, 01:03 AM   #39
darkstar2466
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Originally Posted by Zackie EL
Ok here's another easy one. This is trig.

Give me the area of a triangle with sides a-9.12, b-4.67, c-10.83 and tell me how you did it.


Uh, I think I would use law of cosines to get the angle C, facing the hypotenuse c, then I would use the general formula of the triangle area A = 1/2 a * b * sin(C). I don't feel like plugging in the numbers.
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Old 08-26-2007, 01:03 AM   #40
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Ok here's another easy one. This is trig.

Give me the area of a triangle with sides a-9.12, b-4.67, c-10.83 and tell me how you did it.


21.07

I used the Law of Cosines to solve for angle A, (56.436), then solved for the height using sine equals opposite over hypotenuse. Then just basic A=b*h/2.

EDIT: I wrote 27.07, but meant 21.07. I can't even read my own writing.

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