Homework Help is Here – Start Your Trial Now!
Math
Calculus
(x - z)j + (x – y)k tegr S: z = 9 – x² – y², z 2 0 3. F(x, | 4 F(x. y, z) = (-y + z)i + (x – 2)j + (x - y)k S: z = /1 – x² – y² 5. F(x, y, z) = xyzi + yj + zk - | %3D S: 6x + 6y + z = 12, first octant 6. F(x, y, z) = z²i + x²j + y°k %3D S: z = y², 0 < x< a, 0 s y < a

(x - z)j + (x – y)k tegr S: z = 9 – x² – y², z 2 0 3. F(x, | 4 F(x. y, z) = (-y + z)i + (x – 2)j + (x - y)k S: z = /1 – x² – y² 5. F(x, y, z) = xyzi + yj + zk - | %3D S: 6x + 6y + z = 12, first octant 6. F(x, y, z) = z²i + x²j + y°k %3D S: z = y², 0 < x< a, 0 s y < a

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
See similar textbooks
icon
Related questions
Question

I need help figuring out number 6. Thanks! 

### Verifying Stokes's Theorem In Exercises 3–6, verify **Stokes's Theorem** by evaluating the integral \(\oint_C \mathbf{F} \cdot d\mathbf{r}\) as a line integral and as a double integral. #### Exercise 3 \(\mathbf{F}(x, y, z) = (-y + z)\mathbf{i} + (x - z)\mathbf{j} + (x - y)\mathbf{k}\) \[ S: z = 9 - x^2 - y^2, \quad z \geq 0 \] #### Exercise 4 \(\mathbf{F}(x, y, z) = (-y + z)\mathbf{i} + (x - z)\mathbf{j} + (x - y)\mathbf{k}\) \[ S: z = \sqrt{1 - x^2 - y^2} \] #### Exercise 5 \(\mathbf{F}(x, y, z) = xyz\mathbf{i} + y\mathbf{j} + z\mathbf{k}\) \[ S: 6x + 6y + z = 12, \text{ first octant} \] #### Exercise 6 \(\mathbf{F}(x, y, z) = z^2\mathbf{i} + x^2\mathbf{j} + y^2\mathbf{k}\) \[ S: z = y^2, \quad 0 \leq x \leq a, \quad 0 \leq y \leq a \]
Transcribed Image Text:### Verifying Stokes's Theorem In Exercises 3–6, verify **Stokes's Theorem** by evaluating the integral \(\oint_C \mathbf{F} \cdot d\mathbf{r}\) as a line integral and as a double integral. #### Exercise 3 \(\mathbf{F}(x, y, z) = (-y + z)\mathbf{i} + (x - z)\mathbf{j} + (x - y)\mathbf{k}\) \[ S: z = 9 - x^2 - y^2, \quad z \geq 0 \] #### Exercise 4 \(\mathbf{F}(x, y, z) = (-y + z)\mathbf{i} + (x - z)\mathbf{j} + (x - y)\mathbf{k}\) \[ S: z = \sqrt{1 - x^2 - y^2} \] #### Exercise 5 \(\mathbf{F}(x, y, z) = xyz\mathbf{i} + y\mathbf{j} + z\mathbf{k}\) \[ S: 6x + 6y + z = 12, \text{ first octant} \] #### Exercise 6 \(\mathbf{F}(x, y, z) = z^2\mathbf{i} + x^2\mathbf{j} + y^2\mathbf{k}\) \[ S: z = y^2, \quad 0 \leq x \leq a, \quad 0 \leq y \leq a \]
Verify Stokes’s Theorem for #6, page 1123 **Line Integral:** From the figure you see that **Explanation of Diagram:** The diagram illustrates a three-dimensional coordinate system with axes labeled \(x\), \(y\), and \(z\). It shows a curved surface bounded by a closed path, which is divided into four segments labeled \(C_1\), \(C_2\), \(C_3\), and \(C_4\). - The surface appears to be curved and spans within the dashed rectangular prism, indicating the region of interest in the space. - The path segments are oriented such that they form a loop around the surface. In the context of Stokes's Theorem, this diagram represents the line integral along the closed path \(C\) enclosing the surface. The theorem relates this line integral to a surface integral over the surface bounded by \(C\).
Transcribed Image Text:Verify Stokes’s Theorem for #6, page 1123 **Line Integral:** From the figure you see that **Explanation of Diagram:** The diagram illustrates a three-dimensional coordinate system with axes labeled \(x\), \(y\), and \(z\). It shows a curved surface bounded by a closed path, which is divided into four segments labeled \(C_1\), \(C_2\), \(C_3\), and \(C_4\). - The surface appears to be curved and spans within the dashed rectangular prism, indicating the region of interest in the space. - The path segments are oriented such that they form a loop around the surface. In the context of Stokes's Theorem, this diagram represents the line integral along the closed path \(C\) enclosing the surface. The theorem relates this line integral to a surface integral over the surface bounded by \(C\).
Expert Solution
trending now

Trending now

This is a popular solution!

steps

Step by step

Solved in 2 steps with 2 images

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer
Recommended textbooks for you
Calculus: Early Transcendentals
Calculus: Early Transcendentals
Calculus
ISBN:
9781285741550
Author:
James Stewart
Publisher:
Cengage Learning
Thomas' Calculus (14th Edition)
Thomas' Calculus (14th Edition)
Calculus
ISBN:
9780134438986
Author:
Joel R. Hass, Christopher E. Heil, Maurice D. Weir
Publisher:
PEARSON
Calculus: Early Transcendentals (3rd Edition)
Calculus: Early Transcendentals (3rd Edition)
Calculus
ISBN:
9780134763644
Author:
William L. Briggs, Lyle Cochran, Bernard Gillett, Eric Schulz
Publisher:
PEARSON
Calculus: Early Transcendentals
Calculus: Early Transcendentals
Calculus
ISBN:
9781319050740
Author:
Jon Rogawski, Colin Adams, Robert Franzosa
Publisher:
W. H. Freeman
Precalculus
Precalculus
Calculus
ISBN:
9780135189405
Author:
Michael Sullivan
Publisher:
PEARSON
Calculus: Early Transcendental Functions
Calculus: Early Transcendental Functions
Calculus
ISBN:
9781337552516
Author:
Ron Larson, Bruce H. Edwards
Publisher:
Cengage Learning
SEE MORE TEXTBOOKS