Question

Evaluate the line integral 4z ds, where C is the line segment from (1, 0, 1) to (2, -2, 2).

I got (1, -2, 1) for the vector so ds=sqrt(6). And x(t)=1+t, y(t)=-2t and z(t)=1+t and t=0 but how do I set up the integral? The answer should be 6sqrt(6).

Answer 1

The line segment from (1,0,1) to (2,-2,2) is given by the vector-valued function \(\vec{r}(t)=(1,0,1)(1-t)+(2,-2,2)t\) subject to \(0\le t\le1\).

Another way to express the vector function is \(\vec{r}(t)=x(t)\vec{i}+y(t)\vec{j}+z(t)\vec{k}.\) From this, you have \(\vec{r}(t)=(1+t)\vec{i}-2t\vec{j}+(1+t)\vec{k}\).

In this problem, \(dS=\sqrt{(x'(t))^2+(y'(t))^2+(z'(t))^2}\;dt\).
The integral itself is
\[\int_C4z\;dS=\int_0^14(1+t) \sqrt{(1)^2+(-2)^2+(1)^2}\;dt=4\sqrt6\int_0^1(1+t)\;dt\]
To answer your question specifically, you replace any x's, y's, and z's with their vector representations as part of r(t).