Problem 1 The present problem focuses on the function f defined by Equation (1), f(x, y) = x³ · y³ . 2³, (1) as well as the box B with length L > 0, width W > 0, and height H defined by Equation (3), B = [0, L] × [0, W] × [0, H] {(x, y, z) E R³ : (0

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Chapter2: Second-order Linear Odes
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Problem 1 The present problem focuses on the function f defined by Equation (1),
f (x, y)
= x³ · y³. 2³,
(1)
as well as the box B with length L > 0, width W > 0, and height H defined by Equation (3),
B = [0, L] × [0, W] × [0, H]
= {(x, y, z) E R³ : (0 <x < L) AND (0 < y < W) AND (0 < z < H)}.
(2)
6.
(3)
Thus, the corners of the box B are at (0,0,0), (L,0,0), (L, W,0), and (0, W, 0) counterclock-
wise seen from above on the bottom square face, as well as (0,0, H), (L,0, H), (L, W, H), and
(0, W, H) counterclockwise seen from above on the top square face.
Question: Calculate the integral of the function f over the box B.
Caution: Your correct answer may be different from, but algebraically equivalent to, one of the
multiple choices. Select a choice that is algebraically equivalent to a correct answer, if any exists.
Transcribed Image Text:Problem 1 The present problem focuses on the function f defined by Equation (1), f (x, y) = x³ · y³. 2³, (1) as well as the box B with length L > 0, width W > 0, and height H defined by Equation (3), B = [0, L] × [0, W] × [0, H] = {(x, y, z) E R³ : (0 <x < L) AND (0 < y < W) AND (0 < z < H)}. (2) 6. (3) Thus, the corners of the box B are at (0,0,0), (L,0,0), (L, W,0), and (0, W, 0) counterclock- wise seen from above on the bottom square face, as well as (0,0, H), (L,0, H), (L, W, H), and (0, W, H) counterclockwise seen from above on the top square face. Question: Calculate the integral of the function f over the box B. Caution: Your correct answer may be different from, but algebraically equivalent to, one of the multiple choices. Select a choice that is algebraically equivalent to a correct answer, if any exists.
Expert Solution
Step 1

Given the function fx,y,z=x3·y3·z3

To find the integral over the box B=0,L×0,W×0,H.

The integral of f over B is Bfx,y,zdV=Bfx,y,zdxdydz

Step 2

To find the limits:

Given the box is B=0,L×0,W×0,H

Thus,

0xL0yW0zH

Hence the limit is, x:0  L, y:0  W, z:0  H

 

 

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