## anonymous 5 years ago Evaluate the line integral with respect to arc length (intergral)(3*(x^2)*y*z)ds Where c is given by X=t, y=t^2, z=(2/3)*(t^3), 0<=t<=1

1. watchmath

$$\int_0^1 3t^2\cdot t^2\cdot \frac{2}{3}t^3\sqrt{1^2+(2t)^2+(2t^2)^2}\,dt$$ $$\int_0^12t^7\sqrt{((2t^2)+1)^2}\,dt$$ $$\int_0^1 2t^7\cdot ( (2t)^2+1)\, dt$$

2. anonymous

thanks mate, do you reckon you could explain how you got from $\sqrt{1^{2}+2t^2+(2t^2)^2}$ to $\sqrt{((2t^2)+1)^2 }$

3. watchmath

no, i didn't say that :) I said that $$1^2+(2t)^2+(2t^2)^2=4t^4+4t^2+1=((2t^2)+1)^2$$ You can easily check that if you expand $$((2t^2)+1)^2$$.