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Mass of the wire of a curve described by $$y = x^2 +1 $$ where $0\le x\le 1$ with density $$ \rho(x,y) = 12x $$

I couldn't get the correct answer for this one.

What I did:

$$ m = \int^{0}_{1}\int^{1}_{x^2} 12 x dy dx $$

that gives me $3$, which is not correct.

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Did you look at math.stackexchange.com/questions/77486/… at all? –  user22805 Nov 3 '12 at 7:35
    
Should it not be $\int_0^1 12 x (x^2+1) dx = \int_0^1 \int_0^{x^2+1} 12 x \, dy \, dx$? –  copper.hat Nov 3 '12 at 7:46
    
Is it because it is a "Wire" so you cannot integrate it the way I did above and have to use Line integral? –  40Plot Nov 3 '12 at 7:50
    
Also, I'm not sure if copper.hat's method is valid or not. Can anyone confirm? –  40Plot Nov 3 '12 at 7:52
    
@40Plot: Have you read the post that David Wallace linked to? It describes exactly how to do these kinds of problems. Yes, you have to use a line integral rather than integrating over a 2D region. –  wj32 Nov 3 '12 at 8:08
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1 Answer

We need to work with a modification of the arclength formula. Imagine a tiny segment of the curve, going from $x$ to $x+dx$. This will have length approximately equal to $\sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx$, and therefore mass approximately equal to $12x\sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx$. "Add up," with $x$ going from $0$ to $1$. We find that the mass is equal to $$\int_0^1 12x\sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx.$$ Since $\dfrac{dy}{dx}=2x$, we get an expression that is straightforward to integrate by making the substitution $u=1+4x^2$.

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