c Dr Oksana Shatalov, Spring 2014
1
Spring 2014 Math 251
Week in Review 10
courtesy: Oksana Shatalov
(covering Sections 14.4-14.6 )
14.4: Green’s Theorem
Key Points
• GREEN’s THEOREM: Let C be a positively oriented, piecewise-smooth, simple closed curve in the plane and let D be the
region bounded by C. If P (x, y) and Q(x, y) have continuous partial derivatives on an open region that contains D, then
I
ZZ ∂Q
∂P
P x + Qy =
−
A.
∂x
∂y
∂D
D
I
1. Compute the integral I =
(cos x4 + xy)dx + (y 2 ey + x2 )dy , where C is the triangular curve consisting of
C
the line segments from (0, 0) to (1, 0), from (1, 0) to (1, 3), and from (1, 3) to (0, 0).
2. Compute the integral along the given positively oriented curve C:
Z
(y 2 − arctan x) dx + (3x + sin y) dy,
C
where C is the boundary of the region enclosed by the parabola y = x2 and the line y = 4.
c Dr Oksana Shatalov, Spring 2014
2
3. Compute the integral
Z
(12 − x2 y − y 3 + tan x) dx + (xy 2 + x3 − ey ) dy
C
where C is positively oriented boundary of the region enclosed by the circle x2 + y 2 = 4. Sketch the curve C
indicating the positive direction.
R
4. Given the line integral I = C 4x2 yx − (2 + x)y where C consists of the line segment from (0, 0) to (2, −2),
the line segment from (2, −2) to (2, 4), and the part of the parabola y = x2 from (2, 4) to (0, 0). Use Green’s
theorem to evaluate the given integral and sketch the curve C indicating the positive direction.
c Dr Oksana Shatalov, Spring 2014
3
14.5: Curl and Divergence
Key Points
i
∂
• curlF = ∇ × F = ∂x
P
j
∂
∂y
Q
k ∂ ∂z R • If F is conservative, then curlF = 0.
• divF = ∇ · F
• div curl F = 0
1. Given the vector field F = zi + 2yzj + (x + y 2 )k.
(a) Find the divergence of the field.
(b) Find the curl of the field.
(c) Is the given field conservative? If it is, find a potential function.
c Dr Oksana Shatalov, Spring 2014
(d) Compute
R
C
4
z dx + 2yz dy + (x + y 2 ) dz where C is the positively oriented curve
y 2 + z 2 = 4, x = 2013
R
(e) Compute C z dx + 2yz dy + (x + y 2 ) dz where C consists of the three line segments: from (0, 0, 0) to
(2013, 0, 0), from (2013, 0, 0) to (2, 3, 1), and from (2, 3, 1) to (1, 1, 1).
2. Is there a vector field such that curlF = −2xi + 3yzj − xz 2 k?
14.6: Parametric surfaces and their areas
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,
(u, v) ∈ D, is a surface.
• 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.
3. Identify the surface x =
√
u cos v, y = 5u, z =
√
u sin v.
4. Identify the surface which is the graph of the vector-function r(u, v) = hu + v, u − v, ui.
c Dr Oksana Shatalov, Spring 2014
5. Find a parametric representation of the following surfaces:
(a) x + 2y + 3z = 0;
(b) the portion of the plane x + 2y + 3z = 0 in the first octant;
(c) the portion of the plane x + 2y + 3z = 0 inside the cylinder x2 + y 2 = 9;
(d) z + zx2 − y = 0;
(e) y = x2 ;
(f) the portion of the cylinder x2 + z 2 = 25 that extends between the planes y = −1 and y = 15
5
c Dr Oksana Shatalov, Spring 2014
6
6. Find an equation of the plane tangent to the surface x = u, y = 2v, z = u2 + v 2 at the point (1, 4, 5).
7. Given a sphere of radius 2 centered at the origin, find an equation for the plane tangent to it at the point
√
(1, 1, 2) considering the sphere as
p
(a) the graph of g(x, y) = 4 − x2 − y 2 (Note that the graph of g(x, y) is the upper half-sphere );
c Dr Oksana Shatalov, Spring 2014
7
(b) a level surface of f (x, y, z) = x2 + y 2 + z 2 ;
(c) a surface parametrized by the spherical coordinates.
8. 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
9. Find the area of the portion of the cone x2 = y 2 + z 2 between the planes x = 0 and x = 2.
8