Our Discord hit 10K members! 🎉 Meet students and ask top educators your questions.Join Here!

Like

Report

Numerade Educator

Like

Report

Problem 9 Easy Difficulty

Evaluate $\int _ { C } ( x + y ) d s$ where $C$ is the straight-line segment $x = t , y = ( 1 - t ) , z = 0 ,$ from $( 0,1,0 )$ to $( 1,0,0 )$

Answer

$\sqrt{2}$

Discussion

You must be signed in to discuss.

Video Transcript

okay. What we want to dio is we want to step through the process of being able to evaluate, um, the line integral of, um X plus y with respect to s, um where, um see is the straight line, um x equal to t. And, um, Yeah, why? Equal to one minus t and C equal to zero. And we're going from, um 010210 comma zero. Okay. And so what we need to do is we need to find a vector equation. So we're gonna let our of t equal to T I plus T j um and, um, T is going to go from 0 to 1. And so we know that, um, the derivative of that victory equation is going to give May V of tea, which is I plus J. And the magnitude of that V of tea is going to be the square root of one squared plus one squared, which is the square root of two. And so D s is equal to that magnitude of e of t times D t. And that is really why I needed to find that V of tea. And so this is gonna be the square root of two d t. Which is what I need right here. OK? And now what we want to do is, um, determine that integral. And so now that integral that get over over that straight line segment of eggs Plus why d s is now equal to the integral from 0 to 1. Because we're changing everything in terms of tea, Um, and of x is t Why is one minus T and D s is the square root of to t. T. So this becomes the integral from zero to a one a the square root of two DT. And so that is going to become, um, the square root of two tea evaluated at the upper limit of one and the lower limit of zero. So this is gonna be this square root of two.

University of Central Arkansas
Top Calculus 3 Educators
Kayleah T.

Harvey Mudd College

Caleb E.

Baylor University

Michael J.

Idaho State University

Joseph L.

Boston College