Math 227
Carter Sample for test 3
By popular request, I am preparing a sample study guide. The first question is familar,
but now more questions are attached. A problem such as this will be a substantial portion
of the test. The remaining integrals should be as straight forward, but not identical to those
listed here.
1. Consider the quadratic surface f (x, y) = (x − 5)2 − (y − 12)2 .
(a) Sketch the z = 0, z = −1, and z = 1 levels of the surface.
This is the output of the mathematica function, contour plot. The point is that
the level curves here are hyperbolas. The z = 0 level appears below that.
30
20
10
0
-10
-5
-10
5
0
10
15
25
20
15
10
5
-10
-5
5
1
10
15
16
14
12
10
8
20
0
-20
0
2
4
6
8
10
~ .
(b) Compute the gradient ∇f
~ = 2(x − 5)i + 2(y − 12)j]
[ ∇f
(c) Find the critical point(s) of f (x, y).
∂f
= 0 ⇒ x = 5.
∂x
∂f
∂y
= 0 ⇒ y = 12.
The critical point is (5, 12).

(d) Compute the determinant of the Hessian H = 
∂2f
∂x2
∂2f
∂y∂x
point.


∂2f
∂x2
∂2f
∂y∂x
∂2f
∂x∂y
∂2f
∂y 2
2


=
2 0
0 −2
!
∂2f
∂x∂y
∂2f
∂y 2


at the critical
Therefore det H = −4 and the critical point is a saddle point.
(e) Is the critical point a local maximum, minimum, or neither? Explain why?
(f) Sketch the gradient vector field at appropriate points.
(g) Compute the work done in moving a particle once around a circle x2 + y 2 = 4 in
the gradient field. The gradient field is conservative, so W = 0. Alternatively,
you can compute
Z
2pi
2(2 cos t − 5)i − 2(2 sin t − 12)j) · (−2 sin ti + 2 cos tj) dt.
0
2. Calculate the volume that is enclosed by the sphere x2 + y 2 + z 2 = 25 and the cylinder
x2 + y 2 ≤ 1.
V
Z
=
Z
2π
Z
0
1
2(25 − r2 )1/2r drdθ
0
−2
=
(25 − r2 )3/2|10 dθ
3
0
√
4π
[125 − ( 24)3]
=
3
2π
3. Set up an integral that computes the surface area of the region of a sphere x2 +y 2 +z 2 =
a2 that lies above the circle r = a cos (θ).
Sol’n: The area form is dA =
comes dA =
q
Z
a2
rdrdθ.
a2 −r2
π/2
−π/2
Z
a
2
r
1+
∂f 2
∂x
+
∂f 2
dxdy
∂y
In polar coordinates this be-
We compute
2 −1/2
a(a − r )
rdrdθ = −a
0
Z
π/2
−π/2
4. Compute the line integral I =
z = t2 for t ∈ [0, 1].
R
Sol’n:
I=
C
(a2 − r2 )1/2|a0 dθ = a2π
zdx + xdy + ydz over the curve x = t2 , y = t3, and
Z
1
(2t3 + 3t4 + 2t4)dt =
0
3
2
~ is conservative. If so, find a potential function.
5. Determine whether or not F
~ (x, y) = (6x + 5y)i + (5x + 6y)j Sol’n
F
F = P i + Qj,
R
and ∂P
= 5 = ∂Q
. Therefore the field is conservative. Let f (x, y) = (6x + 5y) dx =
∂y
∂x
= 5x + g 0 (y) = 5x + 6y So g(y) = 3y 2 + K for some
3x2 + 5xy + g(y). We compute, ∂f
∂y
constant K. Any potential function will be of the form:
f (x, y) = 3x2 + 5xy + 3y 2 + k.
Notice that (0, 0) is a critical point for this function, and that it is a local minimum.
3