Solution verification for a line integral

Question

Question: Sami walked from the point $(0,1,2)$ to $(3,2,4)$ in a straight line. Given that $A$ is the line segment from which she walked, determine the value of the following integral(Hint: Use parametric equations.): $$ \int_{A}^{} x^{2}dx + y^{2}dy + xzdz $$

Answer:(Sorry if this seems like more of a discussion than an answer, i'm still learning and it helps when I write down everything going on in my head.)

First we parameterise the curve $A$. First, find the direction vector for the line segment. In order to do this we subtract the two points that we originally have, thus: $(3,2,4)-(0,1,2) = \langle 3,1,2 \rangle$. Now let us define the parametric function $r(t)$, in order to parameterise the curve we must multiply by $t$. Thus: $r(t) = \langle 0,1,2 \rangle + t\langle 3,1,2 \rangle $, simplifying we eventually get $r(t) = \langle 3t,t + 1,2t + 2 \rangle$. In order for us to get a definite integral we must find some sort of interval. To do this let us test the points $0,1$. It follows that: $r(0) = \langle 0,1,2 \rangle$ and that, $r(1) = \langle 3,2,4 \rangle$. Now we can finally get rid of this Line integral and replace it with a definite one with the interval $[0,1]$, However first we must calculate a few derivatives.

$$A: \begin{cases} x(t)=0+3t,\\y(t)=1+t,\\z(t)+2+2t\end{cases},\quad 0\leqslant t\leqslant 1.$$

Therefore, $x'(t)=3,y'(t)=1,z'(t)=2$.

Now that we have these values we can proceed to evaluate the definite integral:

$$\begin{equation}\begin{aligned} \int_{A}^{} x^{2}dx + y^{2}dy + xzdz &= \int_{0}^{1} x^{2}dx + y^{2}dy + xzdz \\ &= \int_{0}^{1} (3t)^{2}(3) + (1 + t)^{2}(1) + (3t)(2t + 2)(2) dt \\ &= \int_{0}^{1} 27t^{2} + t^{2} + 2t + 1 + 12t^{2} + 12t dt \\ &= \int_{0}^{1} 40t^{2} + 1 + 14t dt \end{aligned}\end{equation}$$

Now we can simply evaluate this using the power rule: $$ \int_{0}^{1} \frac{40t^{3}}{3} + t + 7t^{2} dt= \left[\frac{40(1)^{3}}{3} + (1)^{2} + 7(1)^{2}\right] - \left[\frac{40(0)^{3}}{3} + (0)^{2} + 7(0)^{2}\right]=\frac{64}{3}.$$

Answer 1

• $\color{blue}{\text{You wrote $r(t)=(0,1,2)+t(3,1,2)$ that is correct}}$, but you should write the movement of the parameter $t$, that is, $0\leqslant t\leqslant 1$. In general we can parametrize a line segment with initial point $r_{0}$ and endpoint $r_{f}$ as $r(t)=r_{0}+t(r_{f}-r_{0})$ for $0\leqslant t\leqslant 1$.
• $\color{blue}{\text{You wrote $r(0)=(0,1,2)$ and $r(1)=(3,2,4)$ and it is correct}}$.
• $\color{blue}{\text{You wrote $x=3t, y=1+3t,z=2+2t$ that is correct}}$, but you should write on a better way, e.g., $$A: \begin{cases} x(t)=0+3t,\\y(t)=1+t,\\z(t)+2+2t\end{cases},\quad 0\leqslant t\leqslant 1.$$ so you can write it as $x'(t)=3,y'(t)=1,z'(t)=2$ or $\frac{{\rm d}x}{{\rm d}t}=3, \frac{{\rm d}y}{{\rm d}t}=1$ and $\frac{{\rm d}z}{{\rm d}t}=2$ because ${\rm d}x=3, {\rm d}y=1$ and ${\rm d}z=2$ does not make much sense.
• $\color{blue}{\text{You wrote $\int_{0}^{1}x^{2}\,{\rm d}x+y^{2}\, {\rm d}y+xz\, {\rm d}z$ but you should write $\int_{A}x^{2}\,{\rm d}x+y^{2}\, {\rm d}y+xz\, {\rm d}z$}}$ because it is the correct notation for the line integral, so you should write in the first line $$\int_{A}x^{2}\,{\rm d}x+y^{2}\, {\rm d}y+xz\, {\rm d}z=\int_{0}^{1}\left(3(0+3t)^{2}+1(1+t)^{2}+2(3t)(2+2t)\right)\, {\rm d }t$$
• $\color{blue}{\text{Your calculations they are correct}}$, that is, indeed, $$\int_{A}x^{2}\,{\rm d}x+y^{2}\, {\rm d}y+xz\, {\rm d}z=\int_{0}^{1}\left(27t^{2}+12(t+1)t+(t+1)^{2}\right)\, {\rm d}t=\color{blue}{\frac{64}{3}}.$$ when $A$ is a line segment from the point $(0,1,2)$ to $(3,2,4)$.