Parametrizing a curve between surfaces with boundary values

Question

I'm evaluating a line integral of the function $T= x^2 + 4xy + 2yz^3$ from $a = (0,0,0)$ to $b=(1,1,1)$ on the path $z = x^2$, and $y = x$ without using the fundamental theorem.

My question is how to factor in the boundaries of the integral when I parameterize the path in terms of $t$

So far I have:
Let $x = t$
so $r= \langle t,t,t^2\rangle$ and $dr = \langle 1,1,2t\rangle$

how do I factor in the boundaries $a=(0,0,0)$ and $b=(1,1,1)$ for my integral? After I have the boundaries, solving the line integral is not a problem

Thank you

Answer 1

The path must be contained in the curve $$\gamma(t)=(t,t,t^2), \, t \in \mathbb{R}$$ It is easily seen that $\gamma$ is a regular curve and, moreover, that $a=\gamma(0)$ and $b=\gamma(1)$.

We know that $\gamma'(t)=(1,1,2t) \Rightarrow ||\gamma'(t)||=\sqrt{2+4t^2}$. So, the required integral can be written as follows $$\int_0^1 T(x,y,z) \sqrt{2+4t^2} \, \text{d} t$$

Remark: I suppose that you did mean $T$ is a scalar field, not a vector field.

Answer 2

From the formula here

$r=(t,t,t^2);\;r'(t)=(1,1,2t)\to |r'(t)|=\sqrt{1^2+1^2+(2t)^2}=\sqrt{2+4t^2}$

$$\int_r(x^2+4 x y+2 y z^3)\,ds=\int_0^1 \left(t^2+4t\cdot t+2t\cdot(t^2)^3\right)\sqrt{2+4t^2}\,dt=$$ $$=\int_0^1 \left(2 t^7+5 t^2\right) \sqrt{4 t^2+2}\, dt=$$ $$\small=\left[\frac{\sqrt{4 t^2+2} \left(1120 t^8+80 t^6-48 t^4+6300 t^3+32 t^2+1575 t-32\right)}{5040}-\frac{5}{16} \text{arcsinh}\left(\sqrt{2} t\right)\right]_0^1=$$ $$=\frac{1}{5040}\left(32 \sqrt{2}+9027 \sqrt{6}-1575 \text{arcsinh}\sqrt{2}\right)\approx 4.038$$