Explanation Given Info: ∫ c 2 x y z d x + x 2 z d y + x 2 y d z Over the path r ( t ) = ( t 2 , Sin ( π t 4 ) , e t 2 − 2 t ) for 0 ≤ t ≤ 2 Formula used: Integral can be converted inform of path integral by following way. ∫ c 2 x y z d x + x 2 z d y + x 2 y d z = ∫ c ( a , b , c ) ( d x , d y , d z ) ∫ c x n d x = x n + 1 n + 1 + c d d t ( u v w ) = v w d d t u + u w d d t v + u v d w d t ∫ d d t [ f ( t ) ] d t = f ( t ) Calculation: The path is as follows: r ( t ) = ( t 2 , Sin ( π t 4 ) , e t 2 − 2 t ) 0 ≤ t ≤ 2 The integral can be written in form of path in tegral, ∫ c 2 x y z d x + x 2 z d y + x 2 y d z = ∫ c ( 2 x y z d x + x 2 z d y + x 2 y d z ) ( d x , d y , d z ) Substitute x with t 2 , y with Sin ( π t 4 ) , z with e t 2 − t 2 ∫ c ( 2 x y z , x 2 z , x 2 y ) ( d x , d y , d z ) = ∫ c ( 2 t 2 sin ( π t 4 ) e t 2 − 2 t , t 4 − e t 2 − 2 t , t 4 Sin ( π t 4 ) ) ( d x , d y , d z ) = ∫ c 2 t 2 Sin ( π t 4 ) e t 2 − 2 t d x + t 4 e t 2 − 2 t d y + t 4 sin ( π t 4 ) d z Since, x = t 2 d x d t = 2 t [ differentiate w .r .t ] d x = 2 t d t y = Sin ( π t 4 ) d y d t = π 4 Cos ( π t 4 ) d y = π 4 Cos ( π 4 t ) d t z = e t 2 − 2 t d z d t = e t 2 − 2 t × ( 2 t − 2 ) d z d t = ( 2 t − 2 ) e t 2 − 2 t d z = ( 2 t − 2 ) e t 2 − 2 t d t Substitute the value of d x 1 , d y 1 d z to get the function in one form = ∫ c 2 t 2 Sin ( π t 4 ) ...

4th Edition

ISBN: 9781319434526

Author: Rogawski

Publisher: MAC HIGHER

See similar textbooks

Concept explainers

Statements and Logic

Logic is the analysis of general patterns of thoughts without regard to any specific sense of context.

Videos

Question

Chapter 17.3, Problem 19E

To determine

The value of the integral $\int_{c} 2 x y z d x + x^{2} z d y + x^{2} y d z$ over the path

$r (t) = (t^{2}, Sin (\frac{π t}{4}), e^{t^{2} - 2 t}) for 0 \leq t \leq 2$

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 17.3, Problem 18E

Chapter 17.3, Problem 20E

Students have asked these similar questions

Which of the following is the general solution to y′′ + 4y = e^2t + 12 sin(2t) ?A. y(t) = c1 cos(2t) + c2 sin(2t) + 1/8 e^2t − 3t cos(2t)B. y(t) = c1e^2t + c2e^−2t + 1/4 te^2t − 3t cos(2t)C. y(t) = c1 + c2e^−4t + 1/12 te^2t − 3t cos(2t)D. y(t) = c1 cos(2t) + c2 sin(2t) + 1/8 e^2t + 3 sin(2t)E. None of the above. Please include all steps! Thank you!

Show that i cote +1 = cosec 20 tan 20+1 = sec² O २ cos² + sin 20 = 1 using pythagon's theorem

Find the general solution to the differential equation

Chapter 17 Solutions

CALCULUS 4E (LL) W/ ACHIEVE ACCESS

Ch. 17.1 - Prob. 1PQ Ch. 17.1 - Prob. 2PQ Ch. 17.1 - Prob. 3PQ Ch. 17.1 - Prob. 4PQ Ch. 17.1 - Prob. 1E Ch. 17.1 - Prob. 2E Ch. 17.1 - Prob. 3E Ch. 17.1 - Prob. 4E Ch. 17.1 - Prob. 5E Ch. 17.1 - Prob. 6E

Ch. 17.1 - Prob. 7E Ch. 17.1 - Prob. 8E Ch. 17.1 - Prob. 9E Ch. 17.1 - Prob. 10E Ch. 17.1 - Prob. 11E Ch. 17.1 - Prob. 12E Ch. 17.1 - Prob. 13E Ch. 17.1 - Prob. 14E Ch. 17.1 - Prob. 15E Ch. 17.1 - Prob. 16E Ch. 17.1 - Prob. 17E Ch. 17.1 - Prob. 18E Ch. 17.1 - Prob. 19E Ch. 17.1 - Prob. 20E Ch. 17.1 - Prob. 21E Ch. 17.1 - Prob. 22E Ch. 17.1 - Prob. 23E Ch. 17.1 - Prob. 24E Ch. 17.1 - Prob. 25E Ch. 17.1 - Prob. 26E Ch. 17.1 - Prob. 27E Ch. 17.1 - Prob. 28E Ch. 17.1 - Prob. 29E Ch. 17.1 - Prob. 30E Ch. 17.1 - Prob. 31E Ch. 17.1 - Prob. 32E Ch. 17.1 - Prob. 33E Ch. 17.1 - Prob. 34E Ch. 17.1 - Prob. 35E Ch. 17.1 - Prob. 36E Ch. 17.1 - Prob. 37E Ch. 17.1 - Prob. 38E Ch. 17.1 - Prob. 39E Ch. 17.1 - Prob. 40E Ch. 17.1 - Prob. 41E Ch. 17.1 - Prob. 42E Ch. 17.1 - Prob. 43E Ch. 17.1 - Prob. 44E Ch. 17.1 - Prob. 45E Ch. 17.1 - Prob. 46E Ch. 17.1 - Prob. 47E Ch. 17.1 - Prob. 48E Ch. 17.1 - Prob. 49E Ch. 17.1 - Prob. 50E Ch. 17.1 - Prob. 51E Ch. 17.1 - Prob. 52E Ch. 17.1 - Prob. 53E Ch. 17.1 - Prob. 54E Ch. 17.1 - Prob. 55E Ch. 17.1 - Prob. 56E Ch. 17.1 - Prob. 57E Ch. 17.2 - Prob. 1PQ Ch. 17.2 - Prob. 2PQ Ch. 17.2 - Prob. 3PQ Ch. 17.2 - Prob. 4PQ Ch. 17.2 - Prob. 1E Ch. 17.2 - Prob. 2E Ch. 17.2 - Prob. 3E Ch. 17.2 - Prob. 4E Ch. 17.2 - Prob. 5E Ch. 17.2 - Prob. 6E Ch. 17.2 - Prob. 7E Ch. 17.2 - Prob. 8E Ch. 17.2 - Prob. 9E Ch. 17.2 - Prob. 10E Ch. 17.2 - Prob. 11E Ch. 17.2 - Prob. 12E Ch. 17.2 - Prob. 13E Ch. 17.2 - Prob. 14E Ch. 17.2 - Prob. 15E Ch. 17.2 - Prob. 16E Ch. 17.2 - Prob. 17E Ch. 17.2 - Prob. 18E Ch. 17.2 - Prob. 19E Ch. 17.2 - Prob. 20E Ch. 17.2 - Prob. 21E Ch. 17.2 - Prob. 22E Ch. 17.2 - Prob. 23E Ch. 17.2 - Prob. 24E Ch. 17.2 - Prob. 25E Ch. 17.2 - Prob. 26E Ch. 17.2 - Prob. 27E Ch. 17.2 - Prob. 28E Ch. 17.2 - Prob. 29E Ch. 17.2 - Prob. 30E Ch. 17.2 - Prob. 31E Ch. 17.2 - Prob. 32E Ch. 17.2 - Prob. 33E Ch. 17.2 - Prob. 34E Ch. 17.2 - Prob. 35E Ch. 17.2 - Prob. 36E Ch. 17.2 - Prob. 37E Ch. 17.2 - Prob. 38E Ch. 17.2 - Prob. 39E Ch. 17.2 - Prob. 40E Ch. 17.2 - Prob. 41E Ch. 17.2 - Prob. 42E Ch. 17.2 - Prob. 43E Ch. 17.2 - Prob. 44E Ch. 17.2 - Prob. 45E Ch. 17.2 - Prob. 46E Ch. 17.2 - Prob. 47E Ch. 17.2 - Prob. 48E Ch. 17.2 - Prob. 49E Ch. 17.2 - Prob. 50E Ch. 17.2 - Prob. 51E Ch. 17.2 - Prob. 52E Ch. 17.2 - Prob. 53E Ch. 17.2 - Prob. 54E Ch. 17.2 - Prob. 55E Ch. 17.2 - Prob. 56E Ch. 17.2 - Prob. 57E Ch. 17.2 - Prob. 58E Ch. 17.2 - Prob. 59E Ch. 17.2 - Prob. 60E Ch. 17.2 - Prob. 61E Ch. 17.2 - Prob. 62E Ch. 17.2 - Prob. 63E Ch. 17.2 - Prob. 64E Ch. 17.2 - Prob. 65E Ch. 17.2 - Prob. 66E Ch. 17.2 - Prob. 67E Ch. 17.2 - Prob. 68E Ch. 17.2 - Prob. 69E Ch. 17.2 - Prob. 70E Ch. 17.2 - Prob. 71E Ch. 17.2 - Prob. 72E Ch. 17.2 - Prob. 73E Ch. 17.2 - Prob. 74E Ch. 17.2 - Prob. 75E Ch. 17.3 - Prob. 1PQ Ch. 17.3 - Prob. 2PQ Ch. 17.3 - Prob. 3PQ Ch. 17.3 - Prob. 4PQ Ch. 17.3 - Prob. 1E Ch. 17.3 - Prob. 2E Ch. 17.3 - Prob. 3E Ch. 17.3 - Prob. 4E Ch. 17.3 - Prob. 5E Ch. 17.3 - Prob. 6E Ch. 17.3 - Prob. 7E Ch. 17.3 - Prob. 8E Ch. 17.3 - Prob. 9E Ch. 17.3 - Prob. 10E Ch. 17.3 - Prob. 11E Ch. 17.3 - Prob. 12E Ch. 17.3 - Prob. 13E Ch. 17.3 - Prob. 14E Ch. 17.3 - Prob. 15E Ch. 17.3 - Prob. 16E Ch. 17.3 - Prob. 17E Ch. 17.3 - Prob. 18E Ch. 17.3 - Prob. 19E Ch. 17.3 - Prob. 20E Ch. 17.3 - Prob. 21E Ch. 17.3 - Prob. 22E Ch. 17.3 - Prob. 23E Ch. 17.3 - Prob. 24E Ch. 17.3 - Prob. 25E Ch. 17.3 - Prob. 26E Ch. 17.3 - Prob. 27E Ch. 17.3 - Prob. 28E Ch. 17.3 - Prob. 29E Ch. 17.3 - Prob. 30E Ch. 17.3 - Prob. 31E Ch. 17.3 - Prob. 32E Ch. 17.3 - Prob. 33E Ch. 17.3 - Prob. 34E Ch. 17.3 - Prob. 35E Ch. 17.4 - Prob. 1PQ Ch. 17.4 - Prob. 2PQ Ch. 17.4 - Prob. 3PQ Ch. 17.4 - Prob. 4PQ Ch. 17.4 - Prob. 5PQ Ch. 17.4 - Prob. 6PQ Ch. 17.4 - Prob. 1E Ch. 17.4 - Prob. 2E Ch. 17.4 - Prob. 3E Ch. 17.4 - Prob. 4E Ch. 17.4 - Prob. 5E Ch. 17.4 - Prob. 6E Ch. 17.4 - Prob. 7E Ch. 17.4 - Prob. 8E Ch. 17.4 - Prob. 9E Ch. 17.4 - Prob. 10E Ch. 17.4 - Prob. 11E Ch. 17.4 - Prob. 12E Ch. 17.4 - Prob. 13E Ch. 17.4 - Prob. 14E Ch. 17.4 - Prob. 15E Ch. 17.4 - Prob. 16E Ch. 17.4 - Prob. 17E Ch. 17.4 - Prob. 18E Ch. 17.4 - Prob. 19E Ch. 17.4 - Prob. 20E Ch. 17.4 - Prob. 21E Ch. 17.4 - Prob. 22E Ch. 17.4 - Prob. 23E Ch. 17.4 - Prob. 24E Ch. 17.4 - Prob. 25E Ch. 17.4 - Prob. 26E Ch. 17.4 - Prob. 27E Ch. 17.4 - Prob. 28E Ch. 17.4 - Prob. 29E Ch. 17.4 - Prob. 30E Ch. 17.4 - Prob. 31E Ch. 17.4 - Prob. 32E Ch. 17.4 - Prob. 33E Ch. 17.4 - Prob. 34E Ch. 17.4 - Prob. 35E Ch. 17.4 - Prob. 36E Ch. 17.4 - Prob. 37E Ch. 17.4 - Prob. 38E Ch. 17.4 - Prob. 39E Ch. 17.4 - Prob. 40E Ch. 17.4 - Prob. 41E Ch. 17.4 - Prob. 42E Ch. 17.4 - Prob. 43E Ch. 17.4 - Prob. 44E Ch. 17.4 - Prob. 45E Ch. 17.4 - Prob. 46E Ch. 17.4 - Prob. 47E Ch. 17.4 - Prob. 48E Ch. 17.4 - Prob. 49E Ch. 17.4 - Prob. 50E Ch. 17.4 - Prob. 51E Ch. 17.5 - Prob. 1PQ Ch. 17.5 - Prob. 2PQ Ch. 17.5 - Prob. 3PQ Ch. 17.5 - Prob. 4PQ Ch. 17.5 - Prob. 5PQ Ch. 17.5 - Prob. 6PQ Ch. 17.5 - Prob. 7PQ Ch. 17.5 - Prob. 1E Ch. 17.5 - Prob. 2E Ch. 17.5 - Prob. 3E Ch. 17.5 - Prob. 4E Ch. 17.5 - Prob. 5E Ch. 17.5 - Prob. 6E Ch. 17.5 - Prob. 7E Ch. 17.5 - Prob. 8E Ch. 17.5 - Prob. 9E Ch. 17.5 - Prob. 10E Ch. 17.5 - Prob. 11E Ch. 17.5 - Prob. 12E Ch. 17.5 - Prob. 13E Ch. 17.5 - Prob. 14E Ch. 17.5 - Prob. 15E Ch. 17.5 - Prob. 16E Ch. 17.5 - Prob. 17E Ch. 17.5 - Prob. 18E Ch. 17.5 - Prob. 19E Ch. 17.5 - Prob. 20E Ch. 17.5 - Prob. 21E Ch. 17.5 - Prob. 22E Ch. 17.5 - Prob. 23E Ch. 17.5 - Prob. 24E Ch. 17.5 - Prob. 25E Ch. 17.5 - Prob. 26E Ch. 17.5 - Prob. 27E Ch. 17.5 - Prob. 28E Ch. 17.5 - Prob. 29E Ch. 17.5 - Prob. 30E Ch. 17.5 - Prob. 31E Ch. 17.5 - Prob. 32E Ch. 17.5 - Prob. 33E Ch. 17.5 - Prob. 34E Ch. 17.5 - Prob. 35E Ch. 17.5 - Prob. 36E Ch. 17.5 - Prob. 37E Ch. 17.5 - Prob. 38E Ch. 17 - Prob. 1CRE Ch. 17 - Prob. 2CRE Ch. 17 - Prob. 3CRE Ch. 17 - Prob. 4CRE Ch. 17 - Prob. 5CRE Ch. 17 - Prob. 6CRE Ch. 17 - Prob. 7CRE Ch. 17 - Prob. 8CRE Ch. 17 - Prob. 9CRE Ch. 17 - Prob. 10CRE Ch. 17 - Prob. 11CRE Ch. 17 - Prob. 12CRE Ch. 17 - Prob. 13CRE Ch. 17 - Prob. 14CRE Ch. 17 - Prob. 15CRE Ch. 17 - Prob. 16CRE Ch. 17 - Prob. 17CRE Ch. 17 - Prob. 18CRE Ch. 17 - Prob. 19CRE Ch. 17 - Prob. 20CRE Ch. 17 - Prob. 21CRE Ch. 17 - Prob. 22CRE Ch. 17 - Prob. 23CRE Ch. 17 - Prob. 24CRE Ch. 17 - Prob. 25CRE Ch. 17 - Prob. 26CRE Ch. 17 - Prob. 27CRE Ch. 17 - Prob. 28CRE Ch. 17 - Prob. 29CRE Ch. 17 - Prob. 30CRE Ch. 17 - Prob. 31CRE Ch. 17 - Prob. 32CRE Ch. 17 - Prob. 33CRE Ch. 17 - Prob. 34CRE Ch. 17 - Prob. 35CRE Ch. 17 - Prob. 36CRE Ch. 17 - Prob. 37CRE Ch. 17 - Prob. 38CRE Ch. 17 - Prob. 39CRE Ch. 17 - Prob. 40CRE Ch. 17 - Prob. 41CRE Ch. 17 - Prob. 42CRE Ch. 17 - Prob. 43CRE Ch. 17 - Prob. 44CRE Ch. 17 - Prob. 45CRE Ch. 17 - Prob. 46CRE Ch. 17 - Prob. 47CRE Ch. 17 - Prob. 48CRE Ch. 17 - Prob. 49CRE Ch. 17 - Prob. 50CRE Ch. 17 - Prob. 51CRE Ch. 17 - Prob. 52CRE Ch. 17 - Prob. 53CRE Ch. 17 - Prob. 54CRE Ch. 17 - Prob. 55CRE Ch. 17 - Prob. 56CRE Ch. 17 - Prob. 57CRE Ch. 17 - Prob. 58CRE Ch. 17 - Prob. 59CRE Ch. 17 - Prob. 60CRE Ch. 17 - Prob. 61CRE

Knowledge Booster

$Calculus$

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, calculus and related others by exploring similar questions and additional content below.

The value of the integral ∫ c 2 x y z d x + x 2 z d y + x 2 y d z over the path r ( t ) = ( t 2 , Sin ( π t 4 ) , e t 2 − 2 t ) for 0 ≤ t ≤ 2

Solution Summary: The author explains that the integral can be written in form of path in t.

Concept explainers

Videos

The value of the integral $\int_{c} 2 x y z d x + x^{2} z d y + x^{2} y d z$ over the path

$r (t) = (t^{2}, Sin (\frac{π t}{4}), e^{t^{2} - 2 t}) for 0 \leq t \leq 2$

Want to see the full answer?

Chapter 17 Solutions

The value of the integral ∫ c 2 x y z d x + x 2 z d y + x 2 y d z over the path r ( t ) = ( t 2 , Sin ( π t 4 ) , e t 2 − 2 t ) for 0 ≤ t ≤ 2

Solution Summary: The author explains that the integral can be written in form of path in t.

Concept explainers

Videos

The value of the integral ∫c2xyzdx+x2zdy+x2ydz over the path r(t)=(t2,Sin(πt4),et2−2t) for 0≤t≤2

Want to see the full answer?

Chapter 17 Solutions

The value of the integral $\int_{c} 2 x y z d x + x^{2} z d y + x^{2} y d z$ over the path

$r (t) = (t^{2}, Sin (\frac{π t}{4}), e^{t^{2} - 2 t}) for 0 \leq t \leq 2$