Evaluate the line integral, where C is the given curve. ∫C x y z d s C : x=2 sin t, y=t

Evaluate the line integral, where C is the given curve. ...

Key Concepts

Arc Length Differential

The arc length differential, denoted by ds, represents an infinitesimal segment of the curve's length. When a curve is parameterized, ds is obtained by computing the magnitude of the derivative of the position vector with respect to the parameter, i.e., ds = |r'(t)| dt. This conversion is crucial for accurately translating the geometric properties of the curve into the integral setup.

Line Integral of a Scalar Field

A line integral of a scalar field involves integrating a scalar function along a curve. It extends the concept of ordinary integration to situations where the values of a function are accumulated along a specific path in space, taking into account the curve's geometry through the arc length element. This concept is fundamental in areas such as physics and engineering, where it is used to calculate quantities like work or mass distributed along a curve.

Curve Parameterization

Curve parameterization is the process of expressing a curve in space using a set of equations that define each coordinate as a function of a single parameter. This method simplifies the evaluation of line integrals by converting the multi-variable function into a single-variable function, allowing standard techniques of integration to be applied effectively.

Transcript

00:01 So in this question, we're asked to determine the line integral, and we're given a line integral of x, y, z, ds, and we're given the parameterization of the curve.

00:15 X is 2 sine t, y is t, z is negative 2 cosine t, and z goes from 0 to 5.

00:26 All right, so the first thing we have to do, so our x is given in terms of t or y is t, is given in terms of t and or z is given in terms of t.

00:35 However, ds we need to determine it also in terms of t.

00:39 So we're going to use to determine ds.

00:42 We're going to use this formula, which is a square root of the derivative of x with respect to t squared, plus a derivative of y with respect to t squared, plus the derivative of z with respect to t squared.

00:55 And then we take the square root of that, multiply it by dt.

00:59 All right, now the derivative, so x is 2 sine t.

01:02 The derivative of that with respect to t is two cosine t.

01:06 The derivative of y, which is t with respect to t, is just one.

01:11 The derivative of z, which is negative two cosine t with respect to t, is just two sine t.

01:18 All right, so now we have two cosine t squared plus one squared is two sine t squared.

01:25 Two cosine t squared is four cosine square t, and two sine t squared is four sine square t.

01:35 So we can pull that 4 out as a common factor, and we're left with 4 times cosine square plus sine square.

01:43 And we know that cosine square plus sine squared is simply...

01:49 So we have the square root of 4 times 1 plus 1, which is a square root of 5, and then dt.

01:58 All right, so d .s is just a square root of 5d .t.

02:02 All right, great.

02:03 So now we have everything in terms of t.

02:05 So we're going to plug everything back in.

02:07 And then we have, now we have to deal with this integral right here.

02:12 And our lower limit of the integral is zero, the upper limit is pi.

02:17 All right, what we can do is we can pull out a negative 2 square root of 5, and we're going to leave these two inside, and now we're going to see one...

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