Let 9 be the function f: R³ → R, and let C denote the region f(x, y, z) Sketch the region C. Compute the triple integral = 2 x+y - 625 C = {(x, y, z) = R³ | -(5-z) ≤ x ≤ (5-z), (5-z) ≤ y ≤ (5-z), 0≤z≤ 5}. 5 (x+y), ffff dv.
Let 9 be the function f: R³ → R, and let C denote the region f(x, y, z) Sketch the region C. Compute the triple integral = 2 x+y - 625 C = {(x, y, z) = R³ | -(5-z) ≤ x ≤ (5-z), (5-z) ≤ y ≤ (5-z), 0≤z≤ 5}. 5 (x+y), ffff dv.
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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Transcribed Image Text:Let
9
be the function
f: R³ → R,
f(x, y, z)
-
2
625
x+y
es (x+y),
and let C denote the region
C = {(x, y, z) € R³ | −(5-z) ≤ x ≤ (5-z), − (5-z) ≤ y ≤ (5 - z), 0 ≤ z ≤ 5}.
Sketch the region C.
Compute the triple integral
Sff fav.
C

Transcribed Image Text:Consider the given region C defined as follows.
C = {(x,y.z) ER ³1 -(5-z) ≤x≤(5-z), (5-z) ≤y < (5-z), 0≤z <5}.
We need to sketch the region C.
-
Step 2: Sketching region C
Consider the given region C.
C = {(x,y.z) E R³ - (5-z) ≤x≤ (5-z), (5-z) ≤y ≤ (5-z), 0≤z<5}.
Given that 0 ≤z ≤5.
Since -(5-z) ≤x≤ (5-z), therefore there will be two plane X = -(5-z), x=5-z inersecting at x = 0, z = 5.
Similarly - (5-z) ≤ y ≤ (5-z) implies that the planes y = -(5-z), y=5-z intersecting at y = 0, z = 5.
Hence the region C is drawn as follows.
Z
5
x=0, z=5
y=0, z=5
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