Find the line integral of f(x, y)=√(y) / x along the curve 𝐫(t)=t^3 𝐢+t^4 𝐣, 1 / 2 ≤t ≤1.

Name: Problem 24, Chapter 15: University Calculus: Early Transcendentals
Uploaded: 2020-05-30T00:16:36-08:00
Duration: 4 min 28 s
Channel: Jack Hou
Description: Find the line integral of f(x, y)=√(y) / x along the curve 𝐫(t)=t^3 𝐢+t^4 𝐣, 1 / 2 ≤t ≤1.

Find the line integral of f(x, y)=√(y) / x along the curve...

Key Concepts

Line Integral

A line integral is a generalization of an integral where a function is integrated along a curve. In the context of a scalar field, it involves summing the function's values weighted by the infinitesimal lengths along the curve. It is a foundational concept in multivariable calculus and is used to compute various physical quantities, such as work done by a force field along a path.

Parameterization of a Curve

Parameterization is the process of expressing a curve in terms of a single variable, often called the parameter. This allows one to translate the problem of integrating along a curve into a single-variable integral by representing the coordinates (such as x and y) as functions of the parameter. This is a critical step in setting up and evaluating line integrals.

Arc Length Differential

When integrating a function along a curve, the differential element of arc length (usually denoted as ds) represents a small segment of the curve's length. It is computed as the magnitude of the derivative of the parameterization of the curve with respect to the parameter. This concept is essential in converting a line integral into an ordinary single-variable integral by incorporating the stretching or compression of the curve as dictated by its parameterization.

Change of Variables in Integration

Changing variables in integration involves substituting the expressions for the coordinates and the differential element resulting from the parameterization into the original integral. This technique exploits the parameter to simplify the integration process, making it possible to evaluate integrals that are originally expressed in multiple coordinates.

Transcript

00:01 Okay, folks, so in this video, we're going to have a integral, a line integral to evaluate.

00:07 So let's write it down here.

00:08 We have a line integral along the curve c.

00:11 Well, not the curve c, but the curve r, which is parameterized by the variable t here.

00:19 We're integrating the function f, which is a function of x and y along the curve.

00:24 Okay.

00:25 So now let's copy down, let's write down the function f, but not as a function of x and y, but as a function of x and y, but as a function.

00:32 Function of t because both x and y are functions of t let's write it down we have square root of y which is really the square root of t fourth divided by the divided by x but x is t cubed okay ds is really just x prime squared plus y prime squared multiplied by d t x prime is 3 t squared squared plus y prime squared but y prime is 4 t to the power of 3 4 t to the power of 3 squared multiplied by d t okay um and then the limits of integration is between 1 half and 1 okay so let's crank this out we have uh 1 over t here for this term because we have t to the power of 2 over t to the power 3 so that's 1 over t multiplied by the uh by this thing right here 9 t to the power of fourth plus 16 t to the power of 6 d t okay um so now i'm going to have 2 to the power of negative 1 because that's what 1 over t is multiplied by t to the power of 2 9 plus 16 t to the power of 2 and now i'm going to multiply these two terms together so now i have t to the power of 1 multiplied by 16 t squared plus 9 d t so now is now is the time now is the time for us to do a little bit of a u substitution here i'm going to define the variable u as this term here inside of the square root 16 t squared plus 9 okay so d u is 30 32 t d t okay um i'm going to substitute this by in here i have square root of u, which is just u to the power of one half, multiplied by it, multiplied by td t, which i'm going to write as d u over 32, as you can clearly see right here.

02:52 So now i have 1 over 32 of u to the power 3 half over 3 half evaluated at the end points...

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