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.
Best 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 $u=1+4x^2$.