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The triple integral that is bounded by a paraboloid x = 4y2 4z2 given as 16.762
Parabloid, x = 4y² + 4z²
plane x = 4
x = 4y² + 4z²
x = 4
4 = 4y² + 4z²
4 = 4 (y² + z² )
1 = y² + z²
from polar coordinates
y = r cos θ
z = r sin θ
r² = y² + z²
0 ≤ θ ≤ 2π
4r² ≤ x ≤ 4
0 ≤ r ≤ 1
[tex]\int\limits\int\limits\int\limits {x} \, dV = \int\limits^1_0\int\limits^a_b\int\limits^c_d {x} \, dx ( rdrdz)[/tex]
where
a = 4
b = 4r²
c = 2r
d = 0
The first integral using limits c and d gives:
[tex]2pi\int\limits^1_0\int\limits^a_b {xr} \, dx[/tex]
The second integral using limits a and b
[tex]pi\int\limits^1_0 {16 } } \, rdr - pi\int\limits^1_0 {16r^{5} \, dx[/tex]
[tex]16pi\int\limits^1_0 { } } \, rdr - 16pi\int\limits^1_0 {r^{5} \, dx[/tex]
[tex]16pi\int\limits^1_0 { } } \, [r-r^{5}]dr[/tex]
The third integral using limits 1 and 0 gives: 16.762
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The triple integral that is bounded by a paraboloid x = 4y2 4z2 given as 16.762
Integration is defined as adding small parts to form a new significant part.
Parabloid, x = 4y² + 4z²
plane x = 4
x = 4y² + 4z²
x = 4
4 = 4y² + 4z²
4 = 4 (y² + z² )
1 = y² + z²
from polar coordinates
y = r cos θ
z = r sin θ
r² = y² + z²
The limits of the integral
0 ≤ θ ≤ 2π
4r² ≤ x ≤ 4
0 ≤ r ≤ 1
[tex]\int\int\intxdV = \int_0_1\int_b_a\int_d_cxdx(rdrdz)[/tex]
where
a = 4
b = 4r²
c = 2r
d = 0
The first integral using limits c and d gives:
[tex]2\pi\int_0^1\int_b^axydx[/tex]
The second integral using limits a and b
[tex]\pi \int_0^116rdr-\pi\int_0^116r^5dx[/tex]
[tex]16\pi \int_0^1rdr-16\pi\int_0^1 r^5dx[/tex]
[tex]16\pi\int_0^1[r-r^5]dr[/tex]
The third integral using limits 1 and 0 gives: 16.762
Read more on Triple integral here:
brainly.com/question/27171802
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