How do I calculate this line integral? The answer to this porblem is 61/6 here but I am not very confident in line integration. My attempt is below the problem.
Problem:
Attempt: $$ \int_CB \cdot dl = \int_C (-x^2,y,3) \cdot (dx,dy,dz) = $$ $$ = \int_C -x^2 dx + \int_C y dy + \int_C 3 dz = \int_0^1 -x^2 dx + \int_2^5 y dy + \int_5^5 3 dz $$
So the last term (involving z) is 0. But what is the strategy for the two first terms? I am trying to rewrite the functions using $y=3x^2+2$.
For the first term: $$y=3x^2+2 \longrightarrow x^2 = (y-2)/3 \longrightarrow dx = ?$$ $x$ from 0 to 1 on C given by: $$\int_0^1 -x^2 dx = \int_0^1 -\frac{(y-2)}{3} dx $$
For the second term: y from 2 to 5 on C given by: $$y=3x^2+2 \longrightarrow dy = dx 6x$$ $$\int_2^5 y dy = \int_2^5 6x dx $$
What I am unsure about is the limits of integration (if it should be from 0 to 1 when I have dx in 2nd term and how to get the differential line element dx (like if you can get it from same function $y=3x^2+2$ or need to derive $x^2 = (y-2)/3 \longrightarrow dx = ?$.
I would think you should parameterize the path of integration, and rewrite the integral so you are integrating with respect to that parameter. So the path $C$ is parameterized as $$(x(t),y(t),z(t)) = (t, 3t^2+2, 5), 0\leq t \leq 1$$ and you have $$B = (-x(t)^2, y(t), 3) = (-t^2, 3t^2+2, 3)$$ $$dl = (x'(t)dt, y'(t)dt, z'(t)dt) = (1, 6t, 0)\, dt$$ so $$ B\cdot dl = (18t^3 - t^2 + 12t)\, dt$$ and $$\int_C B\cdot dl = \int_{t=0}^1(18t^3 - t^2 + 12t)\, dt$$ $$=\left(\tfrac92t^4 - \tfrac13t^3 + 6t^2\right)\Biggr\rvert_{t=0}^1$$ $$= \tfrac92 - \tfrac13 + 6 = \tfrac{61}{6}$$
[Note: I corrected the minus sign I missed in the first component of $B$ in my original post]