Let 9 be the function f: R³ → R, f(x, y, z) = 2 -e (x+y), 625 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 f av. C

Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter9: Systems Of Equations And Inequalities
Section: Chapter Questions
Problem 12T
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Hello, can you please help me compute the triple integral, I've attatched a sketch of the reigon C, I just need help computing the triple integral, many thanks.

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: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
Consider the given region C defined as follows.
C = {(x,y,z) ER³| -(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) ER³| -(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
Transcribed Image Text:Consider the given region C defined as follows. C = {(x,y,z) ER³| -(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) ER³| -(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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