c Dr Oksana Shatalov, Spring 2014
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Spring 2014 Math 251
Week in Review 11
courtesy: Oksana Shatalov
(covering Sections 14.6(continued),14.7-14.8 )
14.6: Parametric surfaces and their areas (continued)
Key Points
• Parametric representation of surfaces: x = x(u, v), y = y(u, v), z = z(u, v),
(u, v) ∈ D.
• The graph of a vector-function of two variables, r(u, v) = x(u, v)i + y(u, v)j + z(u, v)k,
surface.
(u, v) ∈ D, is a
• A normal vector to the tangent plane to a parametric surface at point (x0 , y0 , z0 ) = (x(u0 , v0 ), y(u0 , v0 ), z(u0 , v0 ))
is N(u0 , v0 ) = ru (u0 , v0 ) × rv (u0 , v0 ).
RR
RR
• Surface area: A(S) = S dS = D |ru × rv |dA.
Examples
9. Find the area of the part of the cylinder x2 + z 2 = 1 which lies between the planes y = 0
and x + y + z = 4.
c Dr Oksana Shatalov, Spring 2014
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10. Find the area of the portion of the cone x2 = y 2 + z 2 between the planes x = 0 and x = 2.
14.7: Surface Integrals
Key Points
• If S is given by r(u, v) = x(u, v)i + y(u, v)j + z(u, v)k, (u, v) ∈ D, then the surface integral of f over the
surface S is:
ZZ
ZZ
ZZ
f (x, y, z) dS =
f (r(u, v))|N(u, v)| dA =
f (r(u, v))|ru × rv | dA
S
D
D
Note that if f (x, y, z) > 0 for all (x, y, z) ∈ D then the above integral is equal to mass of a thin sheet of
shape S and the density f (x, y, z).
• If r(φ, θ) = ha sin φ cos θ, a sin φ sin θ, a cos φi then |rφ × rθ | = a2 sin φ
• Surface integral of vector field F over S, or flux of F across S:
ZZ
ZZ
ZZ
ZZ
F · dS =
F · n̂ dS =
F(r(u, v)) · N(u, v) dA =
F(r(u, v)) · (ru × rv ) dA.
S
S
D
D
Examples
1. Evaluate the surface integral
RR
(x2 z + y 2 z)dS where
S
(a) S is the part of the plane z = 4 + x + y that lies inside the cylinder x2 + y 2 = 4.
c Dr Oksana Shatalov, Spring 2014
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(b) S is the portion of the sphere x2 + y 2 + z 2 = 25 such that z ≤ 0, y ≥ 0.
2. Evaluate the surface integral
RR
S
the inequalities 0 ≤ x, y, z ≤ 1.
(x + y + z)dS where S is the surface of the cube defined by
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3. Find the mass of the lamina that is the portion of the surface z = 2013 − x2 /2 between the
planes x = 0, x = 1, y = 0, y = 1, if the density is ρ(x, y, z) = x.
4. 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. Use positive
(outward) orientation for S.
c Dr Oksana Shatalov, Spring 2014
5. Evaluate the surface integral
5
RR
hx, y, 1i · dS where S is the portion of the paraboloid
S
z = 1 − x2 − y 2 in the first octant, oriented by downward normals.
14.8: Stokes’ Theorem
Key Points
• Let S be an oriented piece-wise-smooth surface that is bounded by a simple, closed, piecewise smooth
boundary curve C with positive orientation. Let F be a vector field whose components have continuous
partial derivatives on an open region in R3 that contains S. Then
I
ZZ
F · dr =
curlF · dS.
C
S
Examples
H
6. Use Stokes’ Theorem to evaluate C F~ · d~r, where F~ (x, y, z) =< 3z, 5x, −2y > and C is
the ellipse in which the plane z = y + 3 intersects the cylinder x2 + y 2 = 4, with positive
orientation as viewed from above.
c Dr Oksana Shatalov, Spring 2014
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7. Find the work performed by the forced field F = h−3y 2 , 4z, 6xi on a particle that traverses
the triangle C in the plane z = 21 y with vertices A(2, 0, 0), B(0, 2, 1), and O(0, 0, 0) with a
counterclockwise orientation looking down the positive z-axis.
I
8. Evaluate I =
C
F · dr if F = h2y + 3ex , z − y 8 , x + ln(z 2 + 1)i and C is the curve of inter-
section of the plane x+y+z = 0 and sphere x2 +y 2 +z 2 = 1. (Orient C to be counterclockwise
when viewed from above).
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9. Verify Stokes’ Theorem for the surface S: x2 + y 2 + 5z = 1, z ≥ −5 (oriented by upward
normal) and the vector field F~ = xz~i + yz~j + (x2 + y 2 )~k.