[T] The transformation T k , 1 , 1 : R 3 → R 3 , T k , 1 , 1 ( u , v , w ) = ( x , y , z ) of the form x = ku, y = v, z = w. where k ≠ 1 is a positive real number is called a stretch if k > 1 and a compression if 0 < k < 1 in the x-direclion. Use a CAS to evaluate the integral ∭ s e − ( 4 x 2 + 9 y 2 + 25 z 2 ) d x d y d z on the solid S = { ( s , y , z ) | 4 x 2 + 9 y 2 + 25 z 2 ≤ 1 } by considering the compression T 2 , 3 , 5 ( u , v , w ) = ( x , y , z ) defined by x = u 2 , y = v 3 and z = w 5 Round your answer to four decimal places.
[T] The transformation T k , 1 , 1 : R 3 → R 3 , T k , 1 , 1 ( u , v , w ) = ( x , y , z ) of the form x = ku, y = v, z = w. where k ≠ 1 is a positive real number is called a stretch if k > 1 and a compression if 0 < k < 1 in the x-direclion. Use a CAS to evaluate the integral ∭ s e − ( 4 x 2 + 9 y 2 + 25 z 2 ) d x d y d z on the solid S = { ( s , y , z ) | 4 x 2 + 9 y 2 + 25 z 2 ≤ 1 } by considering the compression T 2 , 3 , 5 ( u , v , w ) = ( x , y , z ) defined by x = u 2 , y = v 3 and z = w 5 Round your answer to four decimal places.
[T] The transformation
T
k
,
1
,
1
:
R
3
→
R
3
,
T
k
,
1
,
1
(
u
,
v
,
w
)
=
(
x
,
y
,
z
)
of the form x = ku, y = v, z = w. where
k
≠
1
is a positive real number is called a stretch if k >1 and a compression if 0 < k < 1 in the x-direclion. Use a CAS to evaluate the integral
∭
s
e
−
(
4
x
2
+
9
y
2
+
25
z
2
)
d
x
d
y
d
z
on the solid
S
=
{
(
s
,
y
,
z
)
|
4
x
2
+
9
y
2
+
25
z
2
≤
1
}
by considering the compression
T
2
,
3
,
5
(
u
,
v
,
w
)
=
(
x
,
y
,
z
)
defined by
x
=
u
2
,
y
=
v
3
and
z
=
w
5
Round your answer to four decimal places.
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Linear Equation | Solving Linear Equations | What is Linear Equation in one variable ?; Author: Najam Academy;https://www.youtube.com/watch?v=tHm3X_Ta_iE;License: Standard YouTube License, CC-BY