The triple integral that is bounded by a paraboloid x = 4y2 4z2 given as
Parabloid, x = 4y^2 + 4z^2
plane x = 4
x = 4y^2 + 4z^2
x = 4
4 = 4y^2 + 4z^2
4 = 4 (y^2 + z^2 )
1 = y^2 + z^2
from polar coordinates
y = r cos θ
z = r sin θ
r^2 = y^2 + z^2
<h3>limts of the integral</h3>
0 ≤ θ ≤ 2π
4r^2 ≤ x ≤ 4
0 ≤ r ≤ 1
where
a = 4
b = 4r^2
c = 2r
d = 0
The first integral using limits c and d gives:
The second integral using limits a and b
The third integral using limits 1 and 0 gives:
Read more on Triple integral here: brainly.com/question/27171802
Answer:
22
Step-by-step explanation:
Hope this helps!
Yes it's a function.
the domains and ranges are {(1,6),(2,2),(5,-3),(6,-7)}
the domain are the x numbers so, (1,2,5) are your x's
the rangers are the y numbers so, (6,2,-3)
Answer:
All of the symbols don't make this statement true.
Step-by-step explanation:
The absolute value of -1 is 1
The absolute value of 1 is 1
it should be |-1| = |1|
If you can only choose one symbol, my best bet would be >, because 1 can't be greater than 1.