Next, we need to parameterize the curve.

$$\vec r\left( t \right) = \left( {1 - t} \right)\left\langle {4, - 1,2} \right\rangle + t\left\langle {1,7, - 1} \right\rangle = \left\langle {4 - 3t, - 1 + 8t,2 - 3t} \right\rangle \hspace{0.25in}0 \le t \le 1$$ Show Step 3

Now we need to evaluate the line integral. Be careful with this type of lines integral. In this case we have both a $dy$ and a $dz$ in the integrand. Recall that this is just a simplified notation for,

$$\int\limits_{C}{{{x^2}\,dy - yz\,dz}} = \int\limits_{C}{{{x^2}\,dy}} - \int\limits_{C}{{yz\,dz}}$$

Then all we need to do is recall that $dy = y'\,dt$ and $dz = z'\,dt$ when we convert the line integral into a “standard” integral.

So, let’s evaluate the line integral. Just remember to “plug in” the parameterization into the integrand (i.e. replace the $x$, $y$ and $z$ in the integrand with the $x$, $y$ and $z$ components of the parameterization) and to convert the differentials properly.

Here is the line integral.

$$\begin{align*}\int\limits_{C}{{{x^2}\,dy - yz\,dz}} & = \int\limits_{C}{{{x^2}\,dy}} - \int\limits_{C}{{yz\,dz}}\\ & = \int_{0}^{1}{{{{\left( {4 - 3t} \right)}^2}\left( 8 \right)\,dt}} - \int_{0}^{1}{{\left( { - 1 + 8t} \right)\,\,\left( {2 - 3t} \right)\left( { - 3} \right)\,dt}}\\ & = \int_{0}^{1}{{8{{\left( {4 - 3t} \right)}^2} - 3\left( {24{t^2} - 19t + 2\,dt} \right)\,dt}}\\ & = \left. {\left[ { - \frac{8}{9}{{\left( {4 - 3t} \right)}^3} - 3\left( {8{t^3} - \frac{{19}}{2}t^{2} + 2t} \right)} \right]} \right|_0^1 = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{109}}{2}}}\end{align*}$$

Note that, in this case, we combined the two integrals into a single integral prior to actually evaluating the integral. This doesn’t need to be done but can, on occasion, simplify the integrand and hence the evaluation of the integral.