Find the mass M of a fluid with a constant mass density flowing across the paraboloid z = 36-x² - y², z ≥ 0, in a unit of time in the direction of the outer unit normal if the velocity of the fluid at any point on the paraboloid is F = F(x, y, z) = xi + yj + 13k. (Express numbers in exact form. Use symbolic notation and fractions where needed.) M = Incorrect 31740π. Feedback Use A Surface Integral Expressed as a Double Integral Theorem. Let S be a surface defined on a closed bounded region R with a smooth parametrization r(r, 0) = x(r, 0) i+ y(r, 0)j + z(r, 0) k. Also, suppose that F is continuous on a solid containing the surface S. Then, the surface integral of F over S is given by F(x, F(x, y, z) ds = ´= ا³. Recall that the mass of fluid flowing across the surface S in a unit of time in the direction of the unit normal n is the flux of the velocity of the fluid F across S. M = 0₂² F-ndS F(x(r, 0), y(r, 0), z(r, 0))||I, X Io|| dr dº

Advanced Engineering Mathematics
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ISBN:9780470458365
Author:Erwin Kreyszig
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Chapter2: Second-order Linear Odes
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Find the mass M of a fluid with a constant mass density flowing across the paraboloid z = 36 - x² - y², z ≥ 0, in a unit of
time in the direction of the outer unit normal if the velocity of the fluid at any point on the paraboloid
is F = F(x, y, z) = xi + yj + 13k.
(Express numbers in exact form. Use symbolic notation and fractions where needed.)
M =
Incorrect
31740πσ
> Feedback
Use A Surface Integral Expressed as a Double Integral Theorem.
Let S be a surface defined on a closed bounded region R with a smooth
parametrization r(r, 0) = x(r, 0) i + y(r, 0) j + z(r, 0) k. Also, suppose that F is
continuous on a solid containing the surface S. Then, the surface integral of F over
S is given by
[[F(x, y, z) ds = = D₁²
Recall that the mass of fluid flowing across the surface S in a unit of time in the
direction of the unit normal n is the flux of the velocity of the fluid F across S.
M =
F(x(r, 0), y(r, 0), z(r, 0))||1r × 1e|| dr de
F.ndS
X
Transcribed Image Text:Find the mass M of a fluid with a constant mass density flowing across the paraboloid z = 36 - x² - y², z ≥ 0, in a unit of time in the direction of the outer unit normal if the velocity of the fluid at any point on the paraboloid is F = F(x, y, z) = xi + yj + 13k. (Express numbers in exact form. Use symbolic notation and fractions where needed.) M = Incorrect 31740πσ > Feedback Use A Surface Integral Expressed as a Double Integral Theorem. Let S be a surface defined on a closed bounded region R with a smooth parametrization r(r, 0) = x(r, 0) i + y(r, 0) j + z(r, 0) k. Also, suppose that F is continuous on a solid containing the surface S. Then, the surface integral of F over S is given by [[F(x, y, z) ds = = D₁² Recall that the mass of fluid flowing across the surface S in a unit of time in the direction of the unit normal n is the flux of the velocity of the fluid F across S. M = F(x(r, 0), y(r, 0), z(r, 0))||1r × 1e|| dr de F.ndS X
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