Download Midterm Solution | Advanced Multivariable Calculus 1 | MATH 324 and more Exams Calculus in PDF only on Docsity!
1 (10 points)^ Consider the integral:
0
y^2 / 4
y cos(x
2 ) dx dy
(a) Sketch the region in the x, y plane corresponding to this integral.
(b) Reverse the order of integration by treating the region as the other type.
(c) Compute the integral.
Solution:
(a) Since x = y
2 /4 meets x = 4 at (4, 4), the region corresponding to the integral is the
region in the x, y plane in the first quadrant bounded by the curves x = y
2 /4, x = 4 and
y = 0.
(b) The integral as written is type II. To write it as type I we want to write x = y
2 /4 with
y being a function of x; that is,
y
2 = 4x
y = 2
x
Then the integral as type I becomes,
0
∫ 2 √x
0
y cos(x
2 ) dy dx
(c) This integral is now possible to evaluate,
4
0
2
√ x
0
y cos(x
2 ) dy dx =
4
0
y
2 cos(x
2 )
2
√ x
0
dx
0
2 x cos(x
2 )dx
= sin(x
2 )
4 = sin 16
2 (10 points)^ In polar coordinates^ r^ =^ θ, θ^ ≥^ 0 represents a spiral. The first ”revolution” of
this spiral can be parameterized as, x(t) = t cos t, y(t) = t sin t, 0 ≤ t ≤ 2 π. Call this portion
of the spiral, C.
Find, ∫
C
x
2
2 ds
Solution: Our parameterization for the curve C is,
r(t) = 〈t cos t, t sin t〉, 0 ≤ t ≤ 2 π
To compute the line integral we need to write the correct limits, rewrite the function in terms
of t and rewrite ds in terms of t and dt. We have,
x
2
2
t
2 cos
2 t + t
2 sin
2 t = t
ds =
x
′ (t)
2
′ (t)
2 dt
with,
x
′ (t) = cos t − t sin t
y
′ (t) = sin t + t cos t
then,
ds =
(cos t − t sin t)^2 + (sin t + t cos t)^2 dt
1 + t
2 dt
Thus, ∫
C
x
2
2 ds =
∫ (^2) π
0
t
1 + t
2 dt
making a substitution, u = 1 + t
2 , allows the integral to be evaluated as,
(1 + t
2 )
3 / 2
2 π
0
[
(1 + 4π
2 )
3 / 2 − 1
]
4 (10 points)^ Let^ R^ be the region in the first quadrant of the^ x, y^ plane bounded by the
curves, xy = 1, y = 4x, y = x/4.
Consider the change of variables, x = u/v, y = uv, u, v ≥ 0.
(a) Sketch the region R in the x, y plane and the corresponding region D in the u, v plane.
(notice that the line u = 0 corresponds to (0, 0) in the x, y plane.)
(b) Find the Jacobian,
∂(x,y) ∂(u,v)
(c) Use the change of variables to find the area of R by evaluating
R
1 dxdy.
Solution:
(a) One way to find D is to find the curves in the u, v plane that correspond to the bound-
ary curves of R. Notice that we are restricting both u and v to be non-negative. This
corresponds to x and y also being non-negative.
xy = 1 −→ u
2 = 1 −→ u = 1
y = 4x −→ uv = 4u/v −→ v
2 = 4 −→ v = 2
y = x/ 4 −→ uv = u/ 4 v −→ v
2 = 1/ 4 −→ v = 1/ 2
We may also notice, as in the hint, that the origin in the x, y plane corresponds to u = 0.
Thus the corresponding region, D in the u, v plane is the rectangle,
D = {(u, v)| 0 ≤ u ≤ 1 ,
≤ v ≤ 2 }
(b)
∂(x, y)
∂(u, v)
xu yu
yu yv
1 v
u v^2 v u
u
v
(c)
Area(R) =
R
dxdy =
D
∂(x, y)
∂(u, v)
dudv
1 / 2
0
u
v
dudv
1 / 2
v
dv
= ln |v|
2
1 / 2
= 2 ln 2
