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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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1 Answers 1

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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 $u=1+4x^2$.