Question

Evaluate the triple integral. X dv, where e is bounded by the paraboloid x = 4y2 4z2 and the plane x = 4. E

Answer 1

The triple integral that is bounded by a paraboloid x = 4y2 4z2 given as

$16pi / 3$

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

limts of the integral

0 ≤ θ ≤ 2π

4r^2 ≤ x ≤ 4

0 ≤ r ≤ 1

$\int\limits\int\limits\int\limits {x} \, dV = \int\limits^1_0\int\limits^a_b\int\limits^c_d {x} \, dx ( rdrdz)$

where

a = 4

b = 4r^2

c = 2r

d = 0

The first integral using limits c and d gives:

$2pi\int\limits^1_0\int\limits^a_b {xr} \, dx$

The second integral using limits a and b

$pi\int\limits^1_0 {16 } } \, rdr - pi\int\limits^1_0 {16r^{5} \, dx$

$16pi\int\limits^1_0 { } } \, rdr - 16pi\int\limits^1_0 {r^{5} \, dx$

$16pi\int\limits^1_0 { } } \, [r-r^{5}]dr$

The third integral using limits 1 and 0 gives:

$16pi / 3$

Read more on Triple integral here: brainly.com/question/27171802