Q. 25 In Exercises 21鈥�28, evaluate t... [FREE SOLUTION]

Chapter 14: Q. 25 (page 1107)

In Exercises 21鈥�28, evaluate the multivariate line integral of the given function over the specified curve.
$f (x, y, z) = e^{x + y + z}$ , with C the straight line segment from the origin to (1,2,3).

Short Answer

Expert verified

The multivariate line integral of the given function over the specified curve is ${鈭�}_{C} f (x, y, z) d s = \frac{\sqrt{14}}{6} [e^{6} - 1]$ .

Step by step solution

Step 1. Given Information

In the given exercises we have to evaluate the multivariate line integral of the given function over the specified curve.

$f (x, y, z) = e^{x + y + z}$ , with C the straight line segment from the origin to (1,2,3).

Step 2. The given function is f(x,y,z)=ex+y+z

The points are $(0, 0, 0) to (1, 2, 3)$

We write the line segment as a vector function:

localid="1650989416785" $r = (0, 0, 0) + t (1 - 0, 2 - 0, 3 - 0) 0 鈮� t 鈮� 1, or in parametric form r = (0, 0, 0) + t (1, 2, 3) x = 0 + t, y = 0 + 2 t, z = 0 + 3 t x = t, y = 2 t, z = 3 t$

Step 3. Now the integral is ∫Cf(x,y)ds=∫abf(r(t))(x'(t))2+(y'(t))2+(z'(t))2dt

${鈭�}_{C} f (x, y, z) d s = {鈭�}_{0}^{1} e^{t + 2 t + 3 t} \sqrt{{(1)}^{2} + {(2)}^{2} + {(3)}^{2}} d t {鈭�}_{C} f (x, y, z) d s = {鈭�}_{0}^{1} e^{6 t} \sqrt{1 + 4 + 9} d t {鈭�}_{C} f (x, y, z) d s = {鈭�}_{0}^{1} e^{6 t} \sqrt{14} d t L e t u = 6 t d u = 6 d t \frac{1}{6} d u = d t {鈭�}_{C} f (x, y, z) d s = \frac{\sqrt{14}}{6} {鈭�}_{0}^{1} e^{u} d u {鈭�}_{C} f (x, y, z) d s = \frac{\sqrt{14}}{6} {[e^{u}]}_{0}^{1} {鈭�}_{C} f (x, y, z) d s = \frac{\sqrt{14}}{6} {[e^{6 t}]}_{0}^{1} {鈭�}_{C} f (x, y, z) d s = \frac{\sqrt{14}}{6} [e^{6 路 1} - e^{6 路 0}] {鈭�}_{C} f (x, y, z) d s = \frac{\sqrt{14}}{6} [e^{6} - e^{0}] {鈭�}_{C} f (x, y, z) d s = \frac{\sqrt{14}}{6} [e^{6} - 1]$

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In Exercises 21鈥�28, evaluate the multivariate line integral of the given function over the specified curve.
$f (x, y, z) = e^{x + y + z}$ , with C the straight line segment from the origin to (1,2,3).

Short Answer

Step by step solution

Step 1. Given Information

Step 2. The given function is f(x,y,z)=ex+y+z

Step 3. Now the integral is ∫Cf(x,y)ds=∫abf(r(t))(x'(t))2+(y'(t))2+(z'(t))2dt

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91影视

In Exercises 21鈥�28, evaluate the multivariate line integral of the given function over the specified curve.f(x,y,z)=ex+y+z, with C the straight line segment from the origin to (1,2,3).

Short Answer

Step by step solution

Step 1. Given Information

Step 2. The given function is f(x,y,z)=ex+y+z

Step 3. Now the integral is ∫Cf(x,y)ds=∫abf(r(t))(x'(t))2+(y'(t))2+(z'(t))2dt

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Probability and Statistics

Logic and Functions

Theoretical and Mathematical Physics

Mechanics Maths

Pure Maths

Discrete Mathematics

Study anywhere. Anytime. Across all devices.

In Exercises 21鈥�28, evaluate the multivariate line integral of the given function over the specified curve.
$f (x, y, z) = e^{x + y + z}$ , with C the straight line segment from the origin to (1,2,3).