Evaluate the line integral ∫C 𝐅 ·d 𝐫, where C is given by the vector function 𝐫(t) . 𝐅(x, y, z)=

Evaluate the line integral ∫C 𝐅 ·d 𝐫, where C is given by...

Key Concepts

Line Integral of a Vector Fields

A line integral of a vector field involves calculating the cumulative effect of the field along a curve. This is typically interpreted as the work done by a force field when moving along a path, and it is computed by taking the dot product of the vector field with the tangent (derivative) of the curve and integrating with respect to the parameter.

Curve Parameterization

Parameterization of a curve involves describing the curve using one or more parameters, usually by expressing the coordinates as functions of a single parameter. This approach converts a multi-dimensional integration problem into a single-variable problem, which simplifies the evaluation of line integrals.

Dot Product

The dot product is a fundamental operation in vector calculus that combines two vectors to yield a scalar. In the context of line integrals, the dot product is used to extract the component of the vector field that is tangent to the curve, which is essential for calculating work or other related quantities.

Integration in One Variable

After the vector field is evaluated along the parameterized curve and the dot product is taken with the derivative of the parameterization, the line integral reduces to a single-variable integral. This integral is then solved using standard techniques of calculus to obtain the final result.

Transcript

00:01 All right, so in this question we're asked to determine the line integral, and we're given that the big f or the force or the field is sine x -i -hat plus cosine y -j -hat plus x -z -k -hat, and we're also given that the position vector is t -cubbed i -hat minus t -squared j -hat plus t -khat, as t goes from zero to one.

00:27 So now we want to write f in terms of t, so instead of x, we're going to plug in t -cube, instead of y, we're going to plug in negative t squared, and instead of z, we're going to plug in t.

00:43 So, um, sine of x becomes sine t -cube, uh, cosine of uh, of y becomes cosine of negative t squared, and then, uh, x -z becomes t cubed times t.

01:02 All right, so there's something we can simplify.

01:07 So, for example, we know that cosine is an even function.

01:11 So cosine of negative t squared is just cosine of t squared, the negative curves.

01:15 And t -cube times t is t -to -the -power of four.

01:19 So now we can write f in terms of t -sign t -cub, cosine, cosine, t -squared, and t -to -the -power of four.

01:28 And now r -t, we can write r -ttcubed, comma, negative t -cq.

01:33 Squared comma t and r prime of t you just take the derivative of r with respect to t so t q becomes three t squared um negative t squared becomes negative two t and t becomes one so that's our r prime all right so now the line integral is just we can write it as the integral from a to b of f of r of t dot product with r prime of t and um our upper limit our lower limit is zero and our upper limit is one, just like it's specified above.

02:09 And now we find the dot product of these two vectors.

02:14 And this is simple, so we just multiply the first term here by the first term here, second term here by the second term here, third term here, third term here.

02:25 And then we get this integral right over here.

02:29 All right, so now we have three terms, and two of these terms involve sine and cosine.

02:35 What we notice is that we have sine of t cubed, and on the outside we have 3t squared.

02:42 So we realize that we can use u substitution.

02:45 So if we let u be t -q, then d -u is just 3 -t -square d -t.

02:50 And for the cosine t -square, we can let w be t -squared, then d -w is just 2t -t.

02:58 Again, we're going to use u -s substitution in both.

03:02 All right, so now we have the integral from 0 .0.

03:06 0 to 1 of sine d u.

03:08 And we have the integral, like also from 0 to 1 of cosine w dw.

03:13 Now if you pay attention, these, the limit, so the limit for the first one, the upper and the lower limit of the integral, remains from 0 to 1...

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