Download Calculus IV Homework 3: Line and Surface Integrals - Prof. Michael Fairchild and more Assignments Advanced Calculus in PDF only on Docsity!
Math 2242 (Calculus IV) Graded Homework 3 DUE 04/21/
- (Path integrals) (10 points each):
(a) Let f (x, y, z) = xy++yz and c : [1, 2] → R^2 : t 7 → (t, 23 t^3 /^2 , t). Calculate
c f^ ds. (b) Let f (x, y, z) = |z| and c be the helical path c : [−t 0 , t 0 ] : t 7 → (cos t, sin t, t), where t 0 > 0. Calculate
c f^ ds. [Note: Your answer should depend on^ t^0 .]
- (Line integrals) (10 points each):
(a) Let F(x, y) = (1, y) and c(t) = (2 cos t, 2 t) for 0 ≤ t ≤ 2 π. Find
c F^ ·^ ds. (b) Calculate (^) ∫
C
6 xy dx + 3x^2 dy + ez^ dz
where C is any oriented simple curve connecting (0, 0 , 0) to (3, 2 , 0). [Hint: Theorem 7. on p.440. Can you find a scalar function g such that ∇g = (6xy, 3 x^2 , ez^ )?]
- (Surface area) (10 points each):
(a) Let D ⊂ R^2 be the region 0 ≤ u ≤ 1 and 0 ≤ v ≤ 1 (i.e. the unit square in the uv-plane). Let Φ : D → R^3 : (u, v) 7 → (x(u, v), y(u, v), z(u, v)) = (u + v, u, v). Use the formula
A(S) =
D
||Tu × Tv|| du dv,
where Tu = ∂ ∂uΦ and Tv = ∂ ∂vΦ , to find the area of the surface S = Φ(D). (b) Let S denote the disk of radius R in the xy-plane. Thinking of this disk as a surface in R^3 , find a parameterization of it, i.e. find a region D ⊂ R^2 and a mapping Φ : D → R^3 : (u, v) 7 → (x(u, v), y(u, v), z(u, v)) such that S = Φ(D). Then use the formula given in part (a) to find its surface area.
- (Surface integrals of scalar fields) (10 points each):
(a) Let Φ(u, v) = (2 cos u, 2 sin u, v) for 0 ≤ u ≤ 2 π and − 1 ≤ v ≤ 1. Evaluate ∫ ∫
Φ
|z| dS.
(b) Let S be the triangle with vertices (1, 0 , 0), (0, 2 , 0), and (0, 1 , 1). Evaluate ∫ ∫
S
2 x + y + z dS.
[Hint: Find an equation z = g(x, y) for the plane containing the triangle. Since S is the graph of the function g(x, y), you may use the formula ∫ ∫
S
f (x, y, z) dS =
D
f (x, y, g(x, y)) cos θ
dx dy,
where θ is the angle a unit normal vector ˆn to the plane makes with the z-axis, and D is the region of integration in the xy-plane (i.e the projection of the triangle S onto the xy-plane). Recall that cos θ = ˆn · kˆ, where ˆn is a unit normal vector to the plane S. Be careful! In this problem, the region D should be a certain obtuse triangle in the xy-plane with one edge along the y-axis.]
1
- (Surface integrals of vector fields) (10 points each):
(a) Let F(x, y, z) = (F 1 (x, y, z), F 2 (x, y, z), F 3 (x, y, z)) = (x, y, 0) and S be the “upside down” paraboloid z = g(x, y) = 4 − x^2 − y^2 , 0 ≤ z ≤ 4. Use equation (4) on p.495, namely ∫ ∫
S
F · dS =
D
[
F 1
∂g ∂x
+ F 2
∂g ∂y
+ F 3
]
dx dy,
to calculate the flux of F across S. [Hint. Sketch S and D. To evaluate the integral over D, switch to polar coordinates using the change of variables theorem.] (b) Let S 1 be the lateral side of the cone described by z =
x^2 + y^2 for 0 ≤ z ≤ 3. Let S 2 be the “top” of the cone, described by x^2 + y^2 ≤ 3 , z = 3. Let S = S 1 ∪ S 2 , i.e the lateral side of the cone plus its top (thus S is a closed surface). Put F(x, y, z) = (y, −x, x^2 + y^2 ). Calculuate the flux of F through the cone, i.e.
S F^ ·^ dS.^ Make sure your normal vectors point “outward.” [Hint: You need to do two surface integrals and add the results; one over the lateral part of the cone and one over its top part. You may parameterize the lateral part of the cone by Φ(θ, r) = (r cos θ, r sin θ, r) for 0 ≤ r ≤ 3 and 0 ≤ θ ≤ 2 π. The outward normal vector is Tθ × Tr. You’ll have to figure out how to integrate over the top part of the cone.]
- EXTRA CREDIT (5 points each):
(a) Let c(t) be a path. Recall the unit tangent vector of c is T = c
′(t) ∫^ ||c′(t)||^.^ If you calculate c T^ ·^ ds, you will get a number. Describe in words what this number represents. Briefly justify your answer. (b) (A familiar formula): If f : [a, b] → R is piecewise continuously differentiable, define the length of the graph of f on [a, b] as the arc-length of the path c : [a, b] → R^2 : t 7 → (t, f (t)). Using this definition, show that the length of the graph of f on [a, b] is given by the usual calculus formula (^) ∫ (^) b
a
1 + [f ′(x)]^2 dx.
(c) (Change of variables): Let D?^ be a v-simple region in the uv-plane described by a ≤ u ≤ b and h(u) ≤ v ≤ g(u). Suppose ψ : D?^ → R is a C^1 function that is never zero on D?, and let T : R^2 → R^2 be the transformation T (u, v) = (x(u, v), y(u, v)) = (u, ψ(u, v)). Assume that T (D?) = D is a y-simple region; show that if f : D → R is continuous, then ∫ ∫
D
f (x, y) dx dy =
D?
f (u, ψ(u, v))
∂ψ ∂v
∣ du^ dv.
[Hint: Draw a picture of the situation and use the change of variables theorem. This problem is easier than it looks.] (d) (Closed line integrals of gradient fields): Let c be a path whose image is a closed curve. If F = ∇g, show that (^) ∮
c
F · ds = 0.
[Hint: Theorem 7.3 on p.440. The circle on the integral sign is just a notational reminder to you that the integral is over a closed loop.]
