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|dw:1443350101709:dw|

Here, the average distance turns out to be \(2\cdot 2/3 + 4\cdot 1/3 = 8/3\) which makes sense.

By unit equilateral triangle do you mean the side length is 1 ?

Yes.

Then simply setup a double integral for the distance expression

How?!

|dw:1443350378977:dw|

\[\text{Average distance} = \dfrac{1}{A}\iint \sqrt{x^2+y^2}dA\]

\(s\) is the displacement function

yes it does

That's great - you understand my love for physics.\[x_{CM} = \frac{\displaystyle \int x dm}{ M}\]

Area of the whole region: \(\displaystyle \int x^2dx/4 \) from 0 to ???

Mass = density*Area
lets just assume that the object shown in above diagram has uniform density of 1

Wow I definitely know how to solve this problem - just tell me the upper limit.

|dw:1443352602540:dw|

I'm not looking at the original problem yet

\[\int ^{+\sqrt{3}/4}_{-\sqrt{3}/4}\int^{-y/\sqrt{3}+1/4 }_{y/\sqrt{3}-1/4 } \sqrt{x^2 + y^2}dx dy\]

Not so fast, do you see why the double integral gives the averge value ?

Yeah, we've covered a lot of center of mass and moment of inertia in the class. :)

but you said earlier that you don't know about double integrals

So I'm confused how you're able to setup that double integral in two minutes hmm

I realised that the inner integral is really just a function of the outer integral.

Hmm that doesn't make sense to me
but its okay, if you're good, im good.

just evaluate the integral and we're done!
there isn't much to it

|dw:1443353241031:dw|

oops

Yes, but again, I don't really know what exactly you're asking... I feel you're rushing

maybe find the center of mass of that parabolic region
you will see whats going on with the bounds..

wolfram understands most latex

Interesting, that worked. Can I have the closed form though?

try evaluating it manually
\(\sqrt{a^2+x^2}\) has a closed form

nooooooooooooooooooooooooooooooooooooo

really ugly form, that.

http://www.wolframalpha.com/input/?i=%5Cint+%5Csqrt%7Bx%5E2+%2B+y%5E2%7D+dx+

you may try changing it to polar but then the bounds will get ugly

Maybe try evaluating just the inner thing?

notice that the integrand is simply \(r\) in polar form

what if it returns the answers in terms of logs and inverse hyperbolic tangnets ?

The answer is supposed to be in a really simple form.

:(

but it's a calculus problem

yeah i mean there are geometrical methods in calc too but

wolfram is not cooperating, so can't do much with that double integral

need to think of somethign else..

never mind, it really was dumb