c Dr Oksana Shatalov, Spring 2014
1
Spring 2014 Math 251
Week in Review 12
courtesy: Oksana Shatalov
(covering Section 14.9 )
14.9: The Divergence Theorem
Key Points
• The Divergence Theorem: Let E be a simple solid region whose boundary surface S has positive (outward) orientation. Let F be a continuous vector field on an open region that contains E. Then
ZZ
ZZZ
F · dS =
divF dV.
S
E
1. Verify the Divergence Theorem for the vector field F~ = x3~i + y 3~j + z 3~k and S, which is the
surface of the region enclosed by x2 + y 2 = 1 and the planes z = 0, z = 2.
c Dr Oksana Shatalov, Spring 2014
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2. Use the Divergence Theorem to find flux of the vector field F = hx, y, 1i across the surface
S which is the boundary of the region enclosed by the cylinder y 2 + z 2 = 1 and the planes
x = 0 and x + y = 5.
3. Apply the Divergence Theorem to compute
RR
S
F · dS for the vector field
F(x, y, z) = hx3 + sin(yz), y 3 , y + z 3 i
over the complete boundary S of the solid hemisphere {(x, y, z) : x2 + y 2 + z 2 ≤ 1, z ≥ 0}
with outward normal.
c Dr Oksana Shatalov, Spring 2014
4. Verify the Divergence Theorem for the region
E = {(x, y, z) : 0 ≤ z ≤ 9 − x2 − y 2 } and the vector field F~ = x~i + y~j + z~k
3
c Dr Oksana Shatalov, Spring 2014
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5. Let F(x, y, z) = hz 3 ey , z 3 ln(x4 + x2 + 2014), zi. Find the flux of F across the part of the
paraboloid x2 + y 2 + z = 5 that lies above the plane z = 4 and is oriented upward.
c Dr Oksana Shatalov, Spring 2014
6. Apply the Divergence Theorem to compute I =
5
RR
S
F · dS, where
1
F(x, y, z) = hxz 2 + cos(y + z), y 3 + ez , x2 z + y + x3 i
3
and S is the top half of the sphere x2 + y 2 + z 2 = 1 with outward normal.
c Dr Oksana Shatalov, Spring 2014
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7. Prove each identity, assuming that S is the boundary surface of a simple solid region E
and the scalar functions and components of the vector fields have continuous second-order
partial derivatives.
ZZ
a · dS, where a is a constant vector.
(a)
S
1
(b) V (E) =
3
ZZ
F · dS, where F = hx, y, zi .
S
ZZ
curlF · dS = 0.
(c)
S