I have some doubts whether the result I obtained is correct. As the topic title says - I am looking for the mass of a curve with a density of $$\sigma(x,y)= \sqrt{x}$$ The curve K is described as follows: $$x=y^2$$ $$y \in [-1,1]$$
Alright, so I begin with substituting $$y=t$$ I know that to calculate the mass of such a curve I need to use following equation $$M=\int_{K}^{}\sigma dl$$ $$K(t)=(t^2,t)$$ $$K'(t)=(2t,1)$$ $$I=\int_{K}\sigma dl=\int_{-1}^{1}\sqrt{t^2}\sqrt{1+4t^2} dt$$ $$I=\frac{1}{8}\int_{-1}^{1}8t\sqrt{1+4t^2} dt$$ Which, due to the function being even is equal to: $$I=2\frac{1}{8}\int_{0}^{1}8t\sqrt{1+4t^2} dt$$ Using substitution $$1+4t^2=s$$ $$8tdt=ds$$ $$I=\frac{1}{4}\int_{1}^{5}s^{\frac{1}{2}} ds$$ $$I=\frac{1}{6}s^{\frac{3}{2}} dt$$ Which after substituting the limits is equal to $$\frac{1}{6}(\sqrt{125}-1)$$
However, Wolfram after giving it the first integral to solve - just prints that the result is 0.
What have I done wrong?
I haven't checked the entire computation, but you've done at least two things wrong. First, $$\int_{-1}^{1}\sqrt{t^2}\sqrt{1+4t^2}\, dt = \int_{-1}^1 |t|\sqrt{1+4t^2}\,dt,$$ and then $$\int_{-1}^1 |t|\sqrt{1+4t^2}\,dt$$ is an even function (the function you posted is an odd function, and indeed its integral is zero), so you can write $$\int_{-1}^1 |t|\sqrt{1+4t^2}\,dt = 2\int_0^1 t\sqrt{1+4t^2}\,dt.$$