Let S be the part of the plane x + y+ z = 2 in the first octant oriented up, and let F(x, y, z) = (x + Y, x – y, 2z). If we change the surface integral (f. F · dS into iterated integrals, we get: 2-r 2 * (x + z)dydx o f S (4 – 2y)dydx O 13 *(4 – 2y)dydx 2-x OS S*(4 – 2y)dydx 2-x
Let S be the part of the plane x + y+ z = 2 in the first octant oriented up, and let F(x, y, z) = (x + Y, x – y, 2z). If we change the surface integral (f. F · dS into iterated integrals, we get: 2-r 2 * (x + z)dydx o f S (4 – 2y)dydx O 13 *(4 – 2y)dydx 2-x OS S*(4 – 2y)dydx 2-x
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![Let \( S \) be the part of the plane \( x + y + z = 2 \) in the first octant oriented up, and let
\[
\vec{F}(x, y, z) = \langle x + y, x - y, 2z \rangle.
\]
If we change the surface integral \(\iint_S \vec{F} \cdot d\vec{S}\) into iterated integrals, we get:
- \( \bigcirc \ 2 \int_0^2 \int_0^{2-x} (x + z) \, dy \, dx \)
- \( \bigcirc \ \int_0^2 \int_0^2 (4 - 2y) \, dy \, dx \)
- \( \bigcirc \ \sqrt{3} \int_0^2 \int_0^{2-x} (4 - 2y) \, dy \, dx \)
- \( \bigcirc \ \int_0^2 \int_0^{2-x} (4 - 2y) \, dy \, dx \)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9a885923-ea1b-4069-b3f1-c8a8d20774f8%2F87858496-4ddd-45bd-a8d2-8fd0118cb0ca%2F9y1yl8c_processed.png&w=3840&q=75)
Transcribed Image Text:Let \( S \) be the part of the plane \( x + y + z = 2 \) in the first octant oriented up, and let
\[
\vec{F}(x, y, z) = \langle x + y, x - y, 2z \rangle.
\]
If we change the surface integral \(\iint_S \vec{F} \cdot d\vec{S}\) into iterated integrals, we get:
- \( \bigcirc \ 2 \int_0^2 \int_0^{2-x} (x + z) \, dy \, dx \)
- \( \bigcirc \ \int_0^2 \int_0^2 (4 - 2y) \, dy \, dx \)
- \( \bigcirc \ \sqrt{3} \int_0^2 \int_0^{2-x} (4 - 2y) \, dy \, dx \)
- \( \bigcirc \ \int_0^2 \int_0^{2-x} (4 - 2y) \, dy \, dx \)
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