1. No category

Solutions to sample final, MATH 2A03/2X03, Spring 2014
Ruipeng Shen
June 13, 2014
1. Consider the function f (x, y, z) = x2 + y 2 + z 2 + exy .
[1] (a) Find the gradient ∇f of the function f .
[2] (b) Find the Hessian Matrix of the function f .
[1] (c) Calculate div (∇f ).
[1] (d) Evaluate the limit
lim
f (x, y, z).
(x,y,z)→(0,0,1)
Solution
Basic calculation shows
(a) ∇f = (2x + yexy , 2y + xexy , 2z);


 
fxx fxy fxz
2 + y 2 exy (1 + xy)exy 0
(b) H(f ) = fyx fyy fyz  = (1 + xy)exy 2 + x2 exy 0 ;
0
0
2
fzx fzy fzz
(c)div (∇f ) = fxx + fyy + fzz = 6 + (x2 + y 2 )exy .
(d) Since the function f (x, y, z) is continuous, we have
lim
f (x, y, z) = f (0, 0, 1) = 2.
(x,y,z)→(0,0,1)
1
RR
2. Consider the double integral D xdA. Here D is the region bounded by the curves
x2 + y 2 = 1, x2 + y 2 = 4, y = x and y = 0 in the first quadrant.
[1] (a) Sketch the region D in a coordinate system.
[3] (b) Evaluate the double integral
ZZ
(x + y)dA.
D
Solution (a) Please see figure 1.
(b) Rewriting this integral into polar coordinates
y
0
y=x
T
/4
x2+y2=4
D
D*
r
2
1
1
x
2
Figure 1: The region D and D?
ZZ
ZZ
(r cos θ + r sin θ)rdA?
(x + y)dA =
D?
π/4 Z 2
D
Z
=
0
Z
1
π/4
r2 (cos θ + sin θ)drdθ
! Z
2
(cos θ + sin θ)dθ
=
0
r dr
1
=1 × (7/3) = 7/3.
2
2
3. Let D be the region bounded by the lines x − y = 1, x − y = 5 and the coordinate axes,
as shown in the figure below.
[3] (a) Evaluate the double integral
ZZ
x+y
sin
dA.
x−y
D
[1] (b) Determine the positive orientation
of ∂D and mark it in the figure.
R
[2] (c) Evaluate the line integral ∂D xdy.
v
y
v=u
D* 5
1
5
x
D
u
v=-u
Figure 2: The region D
Solution
(a) We need to apply change of variables. Let
u = x − y;
v = x + y.
As a result, the map T is chosen as
(x, y) = T(u, v) =
The Jacobian is
u+v v−u
,
2
2
.
∂(x, y) 1/2 1/2
=
= 1/2.
∂(u, v) −1/2 1/2
Now we need to seek the region D? in uv-plane so that D = T(D? ). The boundary of
D? can be found by plugging (x, y) = ((u + v)/2, (v − u)/2) into the equations x − y = 1,
x − y = 5, x = 0 and y = 0, which defines the boundary of D.
x−y =1
⇒
u = 1;
x−y =5
⇒
u = 5;
x=0
⇒
u + v = 0;
y=0
⇒
v − u = 0.
3
Please see figure 2. Thus we have
ZZ
ZZ x+y
v ∂(x, y) ?
sin
dA =
dA
sin
x−y
u ∂(u, v) D
D?
Z Z u
v
1 5
sin dv du
=
2 1
u
−u
Z 5h
1
v iu
=
du
−u cos
2 1
u v=−u
=0.
(b) Please see the figure 2.
(c) By Green’s theorem, we have
Z
Z
∂
1
xdy =
(x)dA = Area(D) = (52 − 12 ) = 12.
2
D ∂x
∂D
4
4. Consider the solid W bounded by the surfaces z = x2 + y 2 and z = 8 − x2 − y 2
[2] (a) Find the volume of the solid W .
[3] (b) Find the triple integral
ZZZ
(x2 + y 2 )2 dV.
W
[2] (c) Find the flux of the vector field F = ey i + zk out of the solid W .
n
z = 8 - x2 -y2
z = x 2 + y2
n
D
Figure 3: The Solid in problem 4
Solution The solid is bounded from above by the surface z = 8 − x2 − y 2 and from below
by the surface z = x2 +y 2 over the region D = {(x, y)|x2 +y 2 ≤ 4} in the xy-plane, as shown
in figure 3. This region D can be determined by solving the inequality x2 + y 2 ≤ 8 − x2 − y 2 .
(a) The volume can be calculated as the double integral
ZZ
V =
[(8 − x2 − y 2 ) − (x2 + y 2 )]dA
Z ZD
=
(8 − 2x2 − 2y 2 )dA.
D
5
Rewriting this into polar coordinates, we have
Z 2π Z 2
(8 − 2r2 )rdrdθ
V =
0
0
Z 2
=2π
(8r − 2r3 )dr
0
1 4 2
2
=2π 4r − r
2
r=0
=16π.
(b) This triple integral can be writen into the iterated integral
ZZZ
(x2 + y 2 )2 dV =
W
ZZ
Z
(x2 + y 2 )2 dz
(8 − 2x2 − 2y 2 )(x2 + y 2 )2 dA.
=
D
As in part (a), we can rewrite this into polar coordinates
ZZZ
(x2 + y 2 )2 dV =
W
dA
x2 +y 2
D
ZZ
!
8−x2 −y 2
2π
Z
Z
2
(8 − 2r2 )r4 · rdrdθ
0
0
Z
2
(8r5 − 2r7 )dr
0
4 6 1 8 2
=2π r − r
3
4
r=0
=2π
=128π/3.
(c) By the divergence theorem, the flux can be calculated as
ZZ
ZZZ
Flux =
F · dS =
div FdV
∂W
Z Z ZW ∂ y
∂
=
(e ) +
(z) dV
∂z
W ∂x
ZZZ
=
1dV
W
=V (W ) = 16π.
6
5. Let S be a surface as shown in figure 4. The blue and red shapes show the positive and
negative sides of the surface, respectively.
[1] (a) Mark the positive orientation of the boundary ∂S of the surface S in the figure.
[2] (b) Suppose that r(u, v) = (u sin v, v, u cos v), (u, v) ∈ [−1, 1]×[0, π] is a parametrization
of the surface S, Calculate the tangent vectors Tu = ∂r/∂u, Tv = ∂r/∂v and the normal
vector N(u, v) = Tu × Tv .
[1] (c) Determine whether the parametrization r is orientation-preserving or orientationreversing.
RR √
[2] (d) Calculate the surface integral S x2 + z 2 dS.
z
n
y
x
Figure 4: The Surface S
Solution (a) Please see figure 4.
(b) This is a basic calculation
∂r
= (sin v, 0, cos v);
∂u
∂r
= (u cos v, 1, −u sin v);
Tv =
∂v
N =Tu × Tv
i
j
k
cos v = sin v 0
u cos v 1 −u sin v Tu =
=(− cos v, u, sin v).
(c) The normal vector N(u, v) points to the negative side of the surface. (This can be easily
tested at the point r(0, 0) = (0, 0, 0)) Thus the parametrization r is orientation-reversing.
7
(d) By definition we have
ZZ p
Z
2
2
x + z dS =
S
π
Z
π
Z
0
Z
1
−1
1
|u|(u2 + 1)1/2 dudv
=
0
Z
p
(u sin v)2 + (u cos v)2 kN(u, v)kdudv
−1
1
u(u2 + 1)1/2 du
0
1
1 2
3/2
=2π (u + 1)
3
0
2π 3/2
= (2 − 1).
3
=2π
6. Let S be the ellipsoid x2 /9 + y 2 /25 + z 2 /25 = 1.
[2] (a) Give a parametrization of S.
[3] (b) Calculate the surface integral
ZZ
(x2 i + zk) · dS.
S
[2] (c) Find an equation of the tangent plane to S at the point (0, 3, 4).
Solution
(a) The parametrization can be given by “modified spherical coordiates”
r(u, v) = (3 sin u cos v, 5 sin u sin v, 5 cos u), (u, v) ∈ [0, π] × [0, 2π].
(b) A basic calculation shows
∂r
= (3 cos u cos v, 5 cos u sin v, −5 sin u);
∂u
∂r
Tv =
= (−3 sin u sin v, 5 sin u cos v, 0);
∂v
N =Tu × Tv
i
j
k = 3 cos u cos v 5 cos u sin v −5 sin u
−3 sin u sin v 5 sin u cos v
0 Tu =
=(25 sin2 u cos v, 15 sin2 u sin v, 15 sin u cos u)
8
As a result, the surface integral can be calculated by
ZZ
(x2 i + zk) · dS
S
Z 2π Z π
[(9 sin2 u cos2 v)i + (5 cos u)k] · N(u, v)dudv
=
0
0
Z 2π Z π
=
(9 sin2 u cos2 v, 0, 5 cos u) · (25 sin2 u cos v, 15 sin2 u sin v, 15 sin u cos u)dudv
0
0
Z 2π Z π
(225 sin4 u cos3 v + 75 sin u cos2 u)dudv
=
0
Z0 π
Z 2π
Z π
Z 2π
4
3
2
=225
sin udu
cos vdv + 75
sin u cos udu
1dv
0
0
0
0
π
1
=0 + 150π × − cos3 u
3
u=0
=100π.
(c) This ellipsoid is actually a level surface of the function
f (x, y, z) =
y2
z2
x2
+
+
9
25 25
A normal vector at the point (0, 3, 4) can be given by the gradient
2x 2y 2z
∇f (0, 3, 4) =
, ,
|
9 25 25 (x,y,z)=(0,3,4)
2
= (0, 3, 4) .
25
Thus the tangent plane is
2
(0, 3, 4) · (x − 0, y − 3, z − 4) = 0 ⇒ 3y + 4z = 25.
25
9
7. Consider the function f (x, y) = x2 + xy + y 2 defined in the disk D = {(x, y)|x2 + y 2 ≤ 1}.
[3] (a) Find the absolute extreme values of the function f .
[2] (b) Find an upper bound of the double integral
ZZ
(x2 + xy + y 2 )1/3 dA.
D
Solution
(a) A Critical point (x, y) in the interior of the domain satisfies
∇f = (2x + y, x + 2y) = 0.
Thus there is only one critical point, f (0, 0) = 0. Now let us seek the extreme values on
the boundary by Lagrange multipliers. We have the equation system (The constraint is
g(x, y) = x2 + y 2 = 1)

 2x + y = 2λx;
∇f (x, y) = λ∇g(x, y);
2y + x = 2λy;
⇒
g(x, y) = 1
 2
x + y2 = 1
This can be rewritten into

 y = (2λ − 2)x;
x = (2λ − 2)y;
 2
x + y2 = 1
First of all, both x and y are nonzero. Otherwise they must be both zero by the first two
equations above, but it is a contradiction with the third equation. Now by the first two
equations we have
y/x = 2λ − 2 = x/y
Thus y = ±x. Plugging this into the constraint, we have four critical points
√ √
√
√
√ √
√
√
2 2
2
2
2
2
3
− 2 2
1
f(
,
) = f (−
,−
) = ; f(
,
) = f(
,−
)= .
2 2
2
2
2
2
2
2
2
2
By a basic comparison, we obtain two absolute maxima
√
√
√
√
f (1/ 2, 1/ 2) = f (−1/ 2, −1/ 2) = 3/2
and a single absolute minimum f (0, 0) = 0.
(b) By part (a), we have
(x2 + xy + y 2 )1/3 ≤ (3/2)1/3
for all (x, y) ∈ D. Thus we have
ZZ
ZZ
2
2 1/3
(x + xy + y ) dA ≤
(3/2)1/3 dA = (3/2)1/3 π.
D
D
10
8. Let c(t) = (et cos t, et sin t) with t ∈ [0, 2π] be a spiral .
[2] (a) Calculate the length of the spiral.
[2] (b) Calculate the following path integral
Z
p
cos( x2 + y 2 )ds.
c
Solution
(a) We have
c0 (t) =(et (cos t − sin t), et (sin t + cos t));
p
kc0 (t)k = e2t (cos t − sin t)2 + e2t (sin t + cos t)2 ;
√
= 2et .
Thus
Z
2π
0
Z
2π
kc (t)kdt =
l=
0
√ t
√
2e dt = 2(e2π − 1).
0
(b) By definition, we have
Z 2π
p
p
√
2
2
cos( x + y )ds =
cos( (et cos t)2 + (et sin t)2 ) · 2et dt
c
0
Z 2π √
2(cos et )et dt
=
0
i2π
h√
= 2 sin et
0
√
= 2(sin e2π − sin 1).
Z
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9. [1 mark each] Determine whether the following statements are true or false. Simply give
your answer in one word. No argument, reasoning or proof is required.
(a) The tangent planes of the cone x2 + y 2 = z 2 always contain the origin.
Solution: True Let (x0 , y0 , z0 ) be an arbitrary point on the cone, then a normal vector
to the cone at that point is given by the gradient of the function f (x, y, z) = x2 + y 2 − z 2 ,
since the cone is a level surface of this function. More precisely, a normal vector can be
given as
∇f (x0 , y0 , z0 ) = (2x0 , 2y0 , −2z0 ).
Thus an equation of the tangent plane is
2x0 (x − x0 ) + 2y0 (y − z0 ) − 2z0 (z − z0 ) = 0,
that is (Use the assumption that (x0 , y0 , z0 ) is on the cone)
x0 x + y0 y − z0 z = x20 + y02 − z02 = 0.
This is clear that the origin is on the tangent plane.
(b) The line integral
R
c [(e
x
+ y + z 2 )i + xj + (2xz + z 2 )k] · ds is independent of path in R3 .
Solution: True This is because the curl of the vector field is zero and the domain R3 is
simple-connected.
(c) The integral
RR
S
x2 dS over any surface S is positive.
Solution: False If the surface is part of the plane x = 0, then x2 = 0 holds on the whole
surface. Thus the integral is zero in this case.
(d) The parametrization r(u, v) = (u + v, u + v, u + sin v), (u, v) ∈ [−1, 1] × [−1, 1] is a
smooth parametrization.
Solution: False
The normal vector F(0, 0) = 0. This is not smooth.
(e) The function f (x, y) = |xy| is differentiable at the origin.
Solution: True
Both partial derivatives exist fx (0, 0) = fy (0, 0) = 0. we need to check
|xy|
p
= 0.
(x,y)→(0,0)
x2 + y 2
lim
This is true by the inequality
|xy|
0≤ p
≤ |x|.
x2 + y 2
12
(f) The curl of a gradient vector field is still a gradient vector field.
Solution: True The curl of a gradient vector field is zero, thus it is still a gradient vector
field. (It is the gradient of zero function.)
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