We wanted to evaluate the line integral
along the curve C
where the curve C is a line segment
from the point zero comma six comma two
to the point negative one comma eight comma seven.
So our first step is to parameterize
the curve C of this line segment,
which means we need to write parametric equations
for this line segment.
And if we can just look at these coordinates
and write the parametric equations, that's great.
But if we struggle with this,
we can always use these equations here
for the parametric equations for a line segment in space,
where x sub one, y sub one and z sub one
would be the coordinates of the first point
and x sub two, y sub two, z sub two
would be the coordinates of the second point.
So here we have x sub one, y sub one, z sub one,
and x sub two, y sub two and z sub two.
When we define a line segment using these equations,
the interval for t will always be the closed interval
from zero to one.
So for x or x sub t,
we would have the quantity one minus t
times x sub one, which is zero,
plus t times x sub two, which is negative one.
Simplifying, we just get x sub t
equals negative t.
For y of t,
we would have the quantity one minus t
times y sub one, which is six,
plus t times y sub two, which is eighth.
Simplifying, we get y of t equals,
we'll have six,
and then negative six t plus eight t,
that's plus two t.
And finally, for z of t,
we'd have the quantity one minus t
times z sub one, which is two,
plus t times z sub two, which is seven.
So we have z of t equals
two.
And then negative two t plus seven t,
that's plus five t.
So the parameterization of the curve C
is x or x of t
equals negative t.
y of t or y
is equal to six plus two t.
z or z of t
is equal to two plus five t.
Again, t is on the closed interval from zero to one.
Now that we've parameterized the curve C
looking at our notes below,
to evaluate the line integral,
we're going to write the integrand function
as a function of t
using the parametric equations.
And then differential s is equal to the square root
of x prime of t squared plus y prime of t squared
plus z prime of t squared dt.
Let's go ahead and find x prime of t,
y prime of t and z prime of t.
So x prime of t
is going to be equal to,
the derivative of negative t with respect to t
is negative one.
y prime of t
is the derivative of six plus two t,
which would be two.
And z prime of t
is equal to the derivative of two plus five t,
which would be five.
Now, for our next step, let's go ahead and write
the given integrand function as a function of t.
We're given f of x comma y comma z equals x squared plus z,
which means f of x of t
comma y of t
comma z of t
would be equal to,
again, x squared would be negative t squared,
times z which would be times
the quantity two plus five t.
Now let's find the square root
of x prime of t squared
plus y prime of t squared
plus z prime of t squared.
So we'd have the square root,
x prime of t squared would be negative one squared,
plus y prime of t squared would be two squared,
plus z prime of t squared would be five squared.
So here, we have the square root of one plus four plus 25
which is the square root of 30.
And now we have all the information we need.
So the line integral along the curve C of x squared z
integrated with respect to s or arc length
is equal to the integral of,
f expresses the function of t which we already found here,
let's go ahead and distribute.
We'd have t squared times two, that's two t squared.
And then plus,
t squared times five t, that's plus five t to the third.
So we have the quantity two t squared plus
five t to the third.
And we already found the value of the square root
to be the square root of 30,
so we have times the square root of 30
dt.
And now the limits of integration for t
are from zero to one
because we set these parametric equations up
so that it traces out this line segment
from t equals zero to t equals one.
Now, let's evaluate this on the next slide.
Let's rewrite this as the square root of 30
times integral from zero to one
of two t squared
plus five t to the third.
Integrating with respect to t,
we're going to have the square root of 30
times
the antiderivative of,
two t to the second with respect to t is going to be
two times t to the third divided by three
or 2/3 t to the third
plus the antiderivative of,
five t to the third is going to be
five times t to the fourth divided by four
or 5/4 t to the fourth.
Evaluating here, we're going to have the square root of 30
times,
when t is equal to one, we'll just have
2/3 times one to the third
plus 5/4 times one to the fourth
minus, when t is zero,
notice both terms contain a factor of t
so both terms are going to be zero.
So we have the square root of 30
times the quantity 2/3
plus 5/4.
Appending a common denominator which would be 12,
we'll multiple by four over four here
and three over three here.
So we have the square root of 30 times,
here, we're going to have 8/12 plus 15/12,
which is 23/12.
Let's go ahead and write this as 23 square root of 30
divided by 12,
which as a decimal would be approximately
10.4980.
I hope you found this helpful.