line integral of (x + y + z) over segment between pts

[mindy88]

1. Evaluate the integral of xy+y+z ds along the curve r(t)=2ti +tj+ (2-2t)k, 0<t<1

i'm not sure what to do


2. find the line integral of f(x,y,z)=x+y+z over the straight line segment from (1,2,3) to (0,-1,1)

i did:
(0-1)i+(-1-2)j+(1-3)k = -i-3j-2k
x= 1-t
y=2-3t
z= 3-2t
0<t<1
square root of 14

then i took the integral from 0 to 1 of (3t)(square root of 14) dt and i got [3(square root of 14)]/2

the answer in the book is 3 (square root 14)


i don't know why i divided it by 2 and they didn't.

 

[galactus]

Here's a relatively easy way to perform a line integration.

\(\displaystyle \L\\x(t)=2t, \;\ y(t)=t, \;\ z=2-2t\)

Sub into integrand, xy+y+z:

\(\displaystyle \L\\\left[(2t)(t), \;\ t, \;\ 2-2t\right]=\left[2t^{2}, \;\ t, \;\ 2-2t\right]\)

Find the derivative of \(\displaystyle \L\\\frac{d}{dt}[2t, \;\ t, \;\ (2-2t)]=[2, \;\ 1, \;\ -2]\)

Take the dot product of the two:

\(\displaystyle \L\\DotP\left[2t^{2} \;\ t \;\ 2-2t\right]\left[2 \;\ 1 \;\ -2\right]=

4t^{2}+5t-4\)

Now integrate:

\(\displaystyle \L\\\int_{0}^{1}[4t^{2}+5t-4]dt\)

I'll leave you do that.


For the second one:

\(\displaystyle \L\\x+y+z=(1-t)+(2-3t)+(3-2t)=6-6t\)

\(\displaystyle \L\\6\sqrt{14}\int_{0}^{1}(1-t)dt=\fbox{3\sqrt{14}}\)

 

[mindy88]

thanks for the help

so, the answer to the first one is

(4/3)t^3 + (5/2) t^2 -4t from 0 to 1 is -1/6 ?

 

[galactus]

Yep