Problem

Set up an integral that represents the length of …

00:57

Question

Answered step-by-step

Problem 2 Medium Difficulty

Use the arc length formula to find the length of the curve $ y = \sqrt{2 - x^2} $, $ 0 \le x \le 1 $. Check your answer by noting that the curve is part of a circle.

Video Answer

Solved by verified expert

Textbook Answer

Official textbook answer

Video by Harriet O'Brien

Numerade Educator

This textbook answer is only visible when subscribed! Please subscribe to view the answer

Related Courses

Calculus 2 / BC

Calculus: Early Transcendentals

Chapter 8

Further Applications of Integration

Section 1

Arc Length

Discussion

You must be signed in to discuss.

Top Calculus 2 / BC Educators

Watch More Solved Questions in Chapter 8

Problem 1

Problem 2

Problem 3

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15

Problem 16

Problem 17

Problem 18

Problem 19

Problem 20

Problem 21

Problem 22

Problem 23

Problem 24

Problem 25

Problem 26

Problem 27

Problem 28

Problem 29

Problem 30

Problem 31

Problem 32

Problem 33

Problem 34

Problem 35

Problem 36

Problem 37

Problem 38

Problem 39

Problem 40

Problem 41

Problem 42

Problem 43

Problem 44

Problem 45

Problem 46

Video Transcript

we'll be looking at Why equals radical to minus X squared will want to use the arc length formula, which has a why prime in it. So we'll want to differentiate, what with respect to X. So we get negative X over radical to minus X squared. So we're on the interval. X is greater than or equal to zero, but less than or equal to one. So we will use that in arc length formula, putting in zero and one for our A and B one plus plugging in. Why prime here squared t X and then we can simplify this. Let's go to the next page. Um, we can see that X squared er negative X squared is going to be X squared over to minus X squared because squaring the radical cancels out and then plus one next will come over here and adding the one we can just say one. This is another way we can write one so that we can adhere. Um, thes cancel out and we're left with two over two minus X squared under the radical still D X equal. Come to the next page here so we can pull out the radical, too. And we're left with the integral 0 to 11 over radical to minus X squared DX. And we can see that this corresponds to a substitution we can make with Mark Sein. You over a So once we make the substitution, we can see that it comes down to radical, too. Sign our ark Sign X over radical too from 120 So we can then plug these in which I did on the next page, we can see that, um, we'll plug in one over radical too. And we can pull out the radical too first and then subtract arc sine one of a radical to minus arc science zero, which is just zero. So arc sine won over. Radical to is high over four minus euro. Not forgetting about that radical, too. On this equals pi ride too over four. And we'll go to the next page here to verify that this is actually an eighth of a circle. So the why equals rad to minus um X squared is the graph of the upper 70 circles under dhe at the origin with a radius of Brad, too. So, on this interval, um, dark length is 1/4 of the earth clink of the semi circle. So it is an eighth of the circumference of the whole circle. Um, which would be to pie are so we can verify here that's plugging in a radius of red two will just get us the answer we had before.

Get More Help with this Textbook

James Stewart

Calculus: Early Transcendentals

View More Answers From This Book

Find Another Textbook

Study Groups

Study with other students and unlock Numerade solutions for free.

Problem

Set up an integral that represents the length of …

This textbook answer is only visible when subscribed! Please subscribe to view the answer

Related Courses

Calculus: Early Transcendentals

Related Topics

Discussion

Top Calculus 2 / BC Educators

Anna Marie Vagnozzi

Kristen Karbon

Samuel Hannah

Michael Jacobsen

Calculus 2 / BC Courses

Recommended Videos

Watch More Solved Questions in Chapter 8

Video Transcript

James Stewart

Calculus: Early Transcendentals

Anna Marie Vagnozzi

Kristen Karbon

Samuel Hannah

Michael Jacobsen

This textbook answer is only visible when subscribed! Please subscribe to view the answer

Calculus: Early Transcendentals

Anna Marie Vagnozzi

Kristen Karbon

Samuel Hannah

Michael Jacobsen

James StewartCalculus: Early Transcendentals

Anna Marie Vagnozzi

Kristen Karbon

Samuel Hannah

Michael Jacobsen

James Stewart

Calculus: Early Transcendentals