Problem

Sketch the region enclosed by the given curves an…

04:48

Question

Answered step-by-step

Problem 17 Easy Difficulty

Sketch the region enclosed by the given curves and find its area.

$ x = 2y^2 $ , $ x = 4 + y^2 $

Video Answer

Solved by verified expert

Textbook Answer

Official textbook answer

Video by Sky Li

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 6

Applications of Integration

Section 1

Areas Between Curves

Discussion

You must be signed in to discuss.

Top Calculus 2 / BC Educators

Watch More Solved Questions in Chapter 6

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

Problem 47

Problem 48

Problem 49

Problem 50

Problem 51

Problem 52

Problem 53

Problem 54

Problem 55

Problem 56

Problem 57

Problem 58

Problem 59

Problem 60

Problem 61

Video Transcript

All right, let's sketch the region cost by those 2 curve, serves first, so here you saw our x y plane and now x, equals to 2 equals to 2 y square. So it's a problem open to the right. It looks like this. This is our x goes through 2 y square and the second 1. Also a problem is some x equals to 4 plus y square, so is also open to the open to the right. So here is 4, since the coefficient is 1 instead of 2, which means this first curve you'll catch up with the second curve. So you have the intersection right. So the intersection intersect here- and this is symmetric- the x axis- all right. So eventually we'll get this. Some moon, ship, enclosed area and you observe that everything is represented by y. It'S like is our independent variable here. So to find this area we take the integral integral with respect to y okay. So then we need to find a boundary for y and as we see this, i call this y 1. This is our y 2, so this is our y 1. This is our y of the intersection point. So how do we find this 1 way to? Basically, we just plug in those 2 functions. You know this intersection happens when 2 y square equals to 4 plus y square. So we can solve this. We know this is y equals to. We move this y square to the other side, so we will get 1 y squared equals to 4. So we get y equals to positive or negative 2 point. So since our 1 is negative to 1 equal to negative 2 and our 2 will go to positive point, so our integral goes from negative 2 to 2 point and the thing inside the integral is just the right curve: minus the left curve. So our right curve, where is 4 plus y square minus our last curve, our left curve, is some left curve, is 2 y square okay. So basically, this is just 4 minus y square and anti derivative for 4 minus y squared 4 y minus 1 cube 1. Third y cube and you mietheboundary y equals to 2 and y e goes to negative 2. So this goes to the first. 1 is 8 minus 8 over 3 minus active 8 plus 8 over 3 point right. So this as to 32 over 3.

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