review2-solabloch/255/review2-sol.pdf · 2012. 3. 20. · title: review2-sol.dvi created date:...

Math 255-003 -- Math 255-003 -- Math 255-003 -- Math 255-003 --

Winter 2012 Math 255

Review Sheet for Second Midterm Exam

Solutions

1. Let r(t) = 〈cos t, sin t, t〉. Find T(π), N(π), and B(π).. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

r′(t) = 〈− sin t, cos t, 1〉,

T(t) =r′

|r′| =1√2〈− sin t, cos t, 1〉,

N(t) =T′

|T′| = 〈− cos t,− sin t, 0〉,

B(t) = T×N =1√2〈sin t,− cos t, 1〉.

We obtain

T(π) =

⟨

0,−1√2,1√2

⟩

, N(π) = 〈1, 0, 0〉, B(π) =

⟨

0,1√2,1√2

⟩

.

2. At what point do the curves r1(t) = 〈t, t2, t3〉 and r2(s) = 〈2s+ 1, s2 +1, es〉 intersect? Find cos θ, where θ is the angle of intersection.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

By solvingt = 2s+ 1, t2 = s2 + 1, t3 = es,

we obtaint = 1, s = 0. (1)

Therefore, the point of intersection is (1, 1, 1). We obtain

r′1(t = 1) = 〈1, 2, 3〉, r′2(s = 0) = 〈2, 0, 1〉.The angle of intersection is the angle θ between these two vectors. We obtain

cos θ =〈1, 2, 3〉 · 〈2, 0, 1〉|〈1, 2, 3〉| |〈2, 0, 1〉| =

√

5

14.

1


3. Consider the Frenet-Serret formulas dT/ds = κN, dN/ds = −κT+τB,and dB/ds = −τN. Here, τ(s) is the torsion. (a) Using the definition ofthe curvature κ = |dT/ds|, derive the first formula. (b) Deduce the secondformula from the first and third formulas.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

(a) We have

κ =

∣

∣

∣

∣

dT

ds

∣

∣

∣

∣

=

∣

∣

∣

∣

dT/dt

ds/dt

∣

∣

∣

∣

, N =T′

|T′| .

Therefore,dT

ds=

T′

s′=

|T′|s′

N = κN,

where we used s′ = |r′(t)| > 0.(b) Using N = B×T, we calculate as

dN

ds=

d

ds(B×T) =

dB

ds×T+B× dT

ds= −τN×T+ κB×N.

Noting that B = T×N and T = N×B, we obtain

dN

ds= τB− κT.

4. Prove (a) r′′ = s′′T + κ(s′)2N, (b) r′ × r′′ = κ(s′)3B, (c) r′′′ = [s′′′ −κ2(s′)3]T + [3κs′s′′ + κ′(s′)2]N + κτ(s′)3B, and finally (d) τ = (r′ × r′′) ·r′′′/|r′ × r′′|2.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

(a) By definition, we have

r′ = |r′|T = s′T.

We obtainr′′ = s′′T+ s′T′.

By Formula 1 of Prob. 3 dT/ds = T′/s′ = κN, we have

r′′ = s′′T+ κ(s′)2N.

(b) Using (a), we have

r′ × r′′ = s′T×[

s′′T+ κ(s′)2N]

.

2


Sine T×T = 0 and T×N = B, we obtain

r′ × r′′ = κ(s′)3N.

(c) Using (a), we obtain

r′′′ = s′′′T+ s′′T′ + κ′(s′)2N+ 2κs′s′′N+ κ(s′)2N′.

Formulas 1 and 2 of Prob. 3 read

dT

ds=

T′

s′= κN,

dN

ds=

N′

s′= −κT+ τB.

Therefore,

r′′′ = s′′′T+ s′′(

κs′N)

+ κ′(s′)2N+ 2κs′s′′N+ κ(s′)2[

s′ (−κT+ τB)]

= [s′′′ − κ2(s′)3]T+ [3κs′s′′ + κ′(s′)2]N+ κτ(s′)3B.

(d) Using (b), we have

|r′ × r′′| = κ(s′)3|B| = κ(s′)3.

Using (b) and (c), we obtain

(r′ × r′′) · r′′′ = κ(s′)3B ·{

[s′′′ − κ2(s′)3]T+ [3κs′s′′ + κ′(s′)2]N+ κτ(s′)3B}

= κ(s′)3B · κτ(s′)3B=

[

κ(s′)3]2

τ.

Therefore, we obtain

τ =(r′ × r′′) · r′′′|r′ × r′′|2 .

5. Find the tangential and normal components of the acceleration vectorfor the position vector r(t) = (1 + t)i+ sin 3tj+ cos 3tk.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

We calculate v and a as

v = r′ = i+ 3 cos 3tj− 3 sin 3tk, a = r′′ = −9 sin 3tj− 9 cos 3tk.

By definition, we havev = vT,

3


where v = |v|. Hence,a = v′ = v′T+ vT′.

We have T′ = κvN because κ = |T′|/v and N = T′/|T′|. We obtain

a = v′ = v′T+ κv2N.

We can calculate the tangential component v′ as

v′ = T · a =v · av

=(i+ 3 cos 3tj− 3 sin 3tk) · (−9 sin 3tj− 9 cos 3tk)

|i+ 3 cos 3tj− 3 sin 3tk|= 0.

We can calculate the normal component κv2 as

κv2 =|r′ × r′′||r′|3 v2 =

|v × a|v

=|(i+ 3 cos 3tj− 3 sin 3tk)× (−9 sin 3tj− 9 cos 3tk)|

v

=| − 27i+ 9 cos 3tj− 9 sin 3tk|

|i+ 3 cos 3tj− 3 sin 3tk|= 9.

6. Find the limit lim(x,y)→(1,2) xy cos(2x− y) using the ε-δ definition.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

We begin to suspect that the limit is (1)(2) cos(2(1)− 2) = 2. Let ε > 0.We want to find δ > 0 such that

|xy cos(2x− y)− 2| < ε whenever 0 <√

(x− 1)2 + (y − 2)2 < δ. (2)

When√

(x− 1)2 + (y − 2)2 < δ, we have

|x− 1| < δ, |y − 2| < δ.

Let us use

cos θ = 1− 2 sin2θ

2, θ = 2x− y.

4


We obtain

|xy cos(2x− y)− 2| =

∣

∣

∣

∣

xy

(

1− 2 sin2θ

2

)

− 2

∣

∣

∣

∣

=

∣

∣

∣

∣

xy − 2− 2xy sin2θ

2

∣

∣

∣

∣

≤ |xy − 2|+ 2|xy| sin2 θ2.

If 3δ/2 ≤ π/2, we have

sinθ

2= sin

(

2(x− 1)− (y − 2)

2

)

< sin3δ

2.

We will choose δ ≤ π/3. Note that

xy = [(x− 1) + 1][(y − 2) + 2] = (x− 1)(y − 2) + 2(x− 1) + (y − 2) + 2.

Therefore,

|xy cos(2x− y)− 2| < δ2 + 3δ + 2(2 + 3δ + δ2) sin23δ

2≡ f(δ).

Hence, we can choose δ such that f(δ) = ε is satisfied if f(π/3) > ε. Wecan choose δ = π/3 if f(π/3) ≤ ε. With this δ, Eq. (2) is satisfied.

7. Find the limit lim(x,y)→(0,0)(x2 + y2)3 ln(x2 + y2)3.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

lim(x,y)→(0,0)

(x2 + y2)3 ln(x2 + y2)3 = limr→0+

r6 ln r6 = lims→∞

− ln s

s,

where s = 1/r6. We can use L’Hospital’s Rule.

lim(x,y)→(0,0)

(x2 + y2)3 ln(x2 + y2)3 = − lims→∞

ln s

s= − lim

s→∞

1/s

1= 0.

5


8. Find ∂z/∂x and ∂z/∂y for yz = x ln(y + z).. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Note that x and y are independent variables, and z is a function of xand y. By differentiating the equation implicitly with respect to x treatingy as a constant, we obtain

y∂z

∂x= ln(y + z) +

x

y + z

∂z

∂x. (3)

Therefore, we obtain∂z

∂x=

(y + z) ln(y + z)

y(y + z)− x.

By differentiating the equation implicitly with respect to y treating x as aconstant, we obtain

z + y∂z

∂y=

x

y + z

(

1 +∂z

∂y

)

.

Therefore, we obtain∂z

∂y=

x− z(y + z)

y(y + z)− x.

9. Find partial derivative ftrsrs for f(r, s, t) = r2t ln(

rs2et)

.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

We obtain

ft = r2 ln(

rs2et)

+ r2t, ftr = 2r ln(

rs2et)

+ r + 2rt,

then

ftrs =4r

s, ftrsr =

4

s.

Therefore,

ftrsrs = − 4

s2.

10. Determine whether each of the following advection diffusion equationut−uxx+ux = 0. (a) u = et+x, (b) u = e2t+2x, (c) u = e2t−x, (d) u = et−2x.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

(a) not solution, (b) solution , (c) solution, (d) not solution.

6


11. Find an equation of the tangent plane to the surface z = e2−x2−y2 ln(1+x2) at the point (1,−1, ln 2).. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Let us define f(x, y) = e2−x2−y2 ln(1 + x2). The tangent plane is ex-pressed as

z − ln 2 = fx(1,−1)(x− 1) + fy(1,−1)(y + 1).

Since

fx(x, y) = e2−x2−y2[

−2x ln(1 + x2) +2x

1 + x2

]

,

fy(x, y) = e2−x2−y2(−2y) ln(1 + x2),

the equation is obtained as

(1− 2 ln 2)x+ (2 ln 2)y − z + 5 ln 2− 1 = 0.

12. The dimensions of a quarter (United States coin) are measured to be24.4mm (diameter) and 1.8mm (thickness). Each measurement is correct towithin 0.2mm. Use differentials to estimate the largest possible error whenthe volume of the quarter is calculated from these measurements. (Useπ ≃ 3.14.). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Let d and w be the diameter and thickness. The volume is V = πd2w/4.The differential is calculated as

dV =∂V

∂ddd+

∂V

∂wdw =

π

2dw dd+

π

4d2 dw. (4)

We are given that |∆d| ≤ 0.2 and |∆w| ≤ 0.2. To find the largest error inthe volume, we therefore use dd = 0.2 and dw = 0.2 together with d = 24.4and w = 1.8.

∆V ≃ dV =3.14

2(24.4)(1.8)(0.2) +

3.14

4(24.4)2(0.2) = (3.14)(24.4)(1.4)

= 107.2624 ≃ 107.

7


13. Prove that if f(x, y) is differentiable at (a, b), then f(x, y) is continuousat (a, b).. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Since f is differentiable at (a, b), we have

f(a+∆x, b+∆y) = f(a, b) + fx(a, b)∆x+ fy(a, b)∆y + ε1∆x+ ε2∆y, (5)

where ε1 and ε2 → 0 as (∆x,∆y) → (0, 0). Hence,

lim(∆x,∆y)→(0,0)

f(a+∆x, b+∆y) = f(a, b).

That is, f is continuous at (a, b).

14. Let f(x, y) = x2y/(x4 + y2) if (x, y) 6= (0, 0) and f(0, 0) = 0. Is thefunction f differentiable at (0, 0)? Explain.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

We let (x, y) approach the origin along the parabola y = x2. We have

limx→0

f(x, x2) = limx→0

x4

x4 + x4=

1

26= 0 = f(0, 0).

Therefore, f is not continuous at the origin and is not differentiable.

15. Find dw/dt for w = xyez, x = t, y = t2, z = ln t.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

dw

dt=

∂w

∂x

dx

dt+

∂w

∂y

dy

dt+

∂w

∂z

dz

dt

= yez + xez(2t) + xyez1

t= 4t3.

16. Finddy

dxfor y4

x + x cos y = y + lnx.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Let F (x, y) be F (x, y) = y4

x +x cos y−y−lnx. Then we have F (x, y) = 0.

dy

dx= −∂F/∂x

∂F/∂y=

y4

x2 − cos y + 1x

4y3

x − x sin y − 1.

8


17. Suppose that the equation F (x, y, z) = 0 implicitly defines each ofthe three variables x, y, and z as functions of the order two: z = f(x, y),y = g(x, z), x = h(y, z). If F is differentiable and Fx, Fy, and Fz are all

nonzero, show that∂z

∂x

∂x

∂y

∂y

∂z= −1.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .By differentiating F (x, y, f(x, y)) = 0 with respect to x, we obtain ∂F

∂x +∂F∂z

∂z∂x = 0. By differentiating F (x, g(x, z), z)) = 0 with respect to z, we

obtain ∂F∂y

∂y∂z + ∂F

∂z = 0. By differentiating F (h(y, z), y, z) = 0 with respect

to y, we obtain ∂F∂x

∂x∂y +

∂F∂y = 0. Thus, by the implicit function theorem, we

obtain∂z

∂x

∂x

∂y

∂y

∂z=

(

−Fx

Fz

)(

−Fy

Fx

)(

−Fz

Fy

)

= −1.

18. Near a buoy, the depth of a lake at the point with coordinates (x, y)is z = 10 + x2 − y3 + 20y, where x, y, and z are measured in meters. Afisherman in a small boat starts at the point (4, 3) and moves toward thebuoy, which is located at (0, 0). Is the water under the boat getting deeperor shallower when he departs? Explain.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Let us define f(x, y) = 10 + x2 − y3 + 20y. Note that we have fx = 2xand fy = −3y2 + 20. We obtain the directional derivative in the direction

of u =

⟨

−4

5,−3

5

⟩

∝ 〈−4,−3〉 as

Duf(x, y) = fx(x, y)

(

−4

5

)

+ fy(x, y)

(

−3

5

)

= −1.6x+ 1.8y2 − 12.

When we start at (x, y) = (4, 3), we have Duf < 0. Therefore, the waterunder the boat is getting shallower.

19. If g(x, y) = x2+y2+2xy+2x−2y, find the gradient vector ∇g(1,−1)and use it to find the tangent line to the level curve g(x, y) = 4 at the point(1,−1). Sketch the level curve, the tangent line, and the gradient vector.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

The gradient vector is calculated as ∇g = 〈2x+2y+2, 2y+2x− 2〉 andso ∇g(1,−1) = 〈2,−2〉. Noting that the tangent line is perpendicular to thegradient vector, we write a vector equation of the tangent line as

〈x− 1, y + 1〉 · 〈2,−2〉 = 0.

9


We obtainy = x− 2.

To sketch the level curve, let us first understand what g(x, y) = 4 lookslike by defining X = (x + y)/

√2 and Y = (x − y)/

√2. We have Y =

−(1/√2)X2 +

√2, so g(x, y) = 4 is a parabola. The level curve, tangent

line, and gradient vector are sketched as follows.

20. Show that the function f(x, y) = (xy)1/3 is continuous and the partialderivatives fx and fy exist at the origin but the directional derivatives in allother directions do not exist.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

The function f(x, y) is continuous at (0, 0) if

lim(x,y)→(0,0)

f(x, y) = f(0, 0) = 0.

That is, (xy)1/3 is continuous at the origin if for every ε > 0 there existsδ > 0 such that

∣

∣

∣(xy)1/3

∣

∣

∣< ε whenever 0 <

√

x2 + y2 < δ.

10


Since 0 ≤ (x− y)2, we have 2xy ≤ x2 + y2 < δ2. Thus,

∣

∣

∣(xy)1/3

∣

∣

∣<

∣

∣

∣

∣

∣

(

δ2

2

)1/3∣

∣

∣

∣

∣

.

For every ε, we can choose δ such that ε = (δ2/2)1/3. Therefore, (xy)1/3 iscontinuous at the origin.

The partial derivatives are calculated as

fx(0, 0) = fy(0, 0) = 0.

Let us calculate the directional derivative in the direction of u = 〈a, b〉(a 6= 0, b 6= 0).

Duf(0, 0) = limh→0

[ha(hb)]1/3 − [0(0)]1/3

h= lim

h→0

(ab)1/3

h1/3= ∞.

Therefore, the directional derivatives in the direction of such u do not exist.

21. Find the local maximum and minimum values and saddle point(s) ofthe function f(x, y) = xye−x−2y.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

We find

fx = y(1− x)e−x−2y,

fy = x(1− 2y)e−x−2y,

fxx = y(x− 2)e−x−2y,

fyy = 4x(y − 1)e−x−2y,

fxy = fyx = (1− x)(1− 2y)e−x−2y.

The function has critical points at (0, 0) and

(

1,1

2

)

. Let us use the second

derivatives test. We obtain

D = fxxfyy − (fxy)2

=[

4xy(x− 2)(y − 1)− (1− x)2(1− 2y)2]

e−2x−4y.

Therefore, we have

D(0, 0) = −1, D

(

1,1

2

)

= e−4.

11


Note that

fxx

(

1,1

2

)

= −1

2e−2.

By the second derivatives test, we see that the point (0, 0) is a saddle point

and the point

(

1,1

2

)

is a local maximum.

22. A rectangular house is designed to minimize heat loss. The east andwest walls lose heat at a rate of 9 units/m2 per day, the north and southwalls at a rate of 8 units/m2 per day, the floor at a rate of 1 units/m2 perday, and the roof at a rate of 5 units/m2 per day. Each wall must be atleast 20m long, the height must be at least 2m, and the volume must beexactly 1000m3. Find the dimensions that minimize heat loss. Is it possibleto design a house with even less heat loss if the restrictions on the lengthsof the walls were removed?. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Let L denote the heat loss.

L = (9 + 9)xz + (8 + 8)yz + (1 + 5)xy.

We have the following conditions.

x ≥ 20, y ≥ 20, z ≥ 4, xyz = 1000.

Since the volume is fixed, we can eliminate z.

L(x, y) = 6xy +16000

x+

18000

y.

Let us find the critical point(s).

∂L

∂x= 6y − 16000

x2= 0,

∂L

∂y= 6x− 18000

y2= 0.

Using the above two equations and noting that 1 = 4000/xy, we obtain

x =40

3, y = 15, z = 5.

The obtained x and y are less than 20. Let us find the extreme values of Lon the boundary.

L(x, 20) = 120x+16000

x+ 900, x ≥ 20.

12


This function L(x, 20) (x ≥ 20) has the minimum value at x = 20.

L(20, y) = 120y + 800 +18000

y, y ≥ 20.

This function L(20, y) (y ≥ 20) has the minimum value at y = 20. Whenx = y = 20, we have z = 2.5. Therefore, the heat loss is minimized whenx = y = 20m, z = 2.5m, and the minimum heat loss is L(20, 20) = 4100.

At the critical point, we obtain

L(40

3, 15) = 6

(

40

3

)

(15) +16000

40/3+

18000

15

= 1200 + 1200 + 1200

= 3600.

This number is smaller than 4100.

23. The plane

√5

2x− y +

11

2z = 2 intersects the cone z2 = x2 + y2 in an

ellipse. Find the highest and lowest points on the ellipse.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Let us define

f(x, y, z) = z, g(x, y, z) =

√5

2x− y +

11

2z, h(x, y, z) = z2 − x2 − y2.

With the Lagrange condition ∇f = λ∇g + µ∇h and two constraints g = 2and h = 0, we have

0 =

√5

2λ− 2xµ, (6)

0 = −λ− 2yµ, (7)

1 =11

2λ+ 2zµ, (8)

√5

2x− y +

11

2z = 2, (9)

z2 − x2 − y2 = 0. (10)

From Eqs. (6), (7), and (8), we obtain

x =

√5λ

4µ, y = − λ

2µ, z =

2− 11λ

4µ.

13


By plugging them into Eq. (9), we obtain

µ =11− 56λ

8.

Therefore,

x =2√5λ

11− 56λ, y = − 4λ

11− 56λ, z =

4− 22λ

11− 56λ.

By using these expressions in Eq. (10), we obtain

28λ2 − 11λ+ 1 = 0.

We obtain

λ =1

4,

1

7.

If λ = 1/4, we have

x = −√5

6, y =

1

3, z =

1

2.

If λ = 1/7, we have

x =2√5

21, y = − 4

21, z =

2

7.

Therefore, the highest point is

(

−√5

6,1

3,1

2

)

and the lowest point is

(

2√5

21,− 4

21,2

7

)

.

24. Evaluate the double integral

∫∫

D

(

1−√

x2 + y2)

dA, D = {(x, y)|−

1 ≤ x ≤ 1, −√

1− x2 ≤ y ≤√

1− x2} by using an iterated integral.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

The double integral is expressed as

V =

∫∫

D

(

1−√

x2 + y2)

dA =

∫ 1

−1

∫

√1−x2

−√1−x2

(

1−√

x2 + y2)

dydx

= 4

∫ 1

0

∫

√1−x2

0

(

1−√

x2 + y2)

dydx.

14


We introduce y = x sinh t, dy = x cosh tdt. We have sinh−1(a) = ln(

a+√a2 + 1

)

for a > 0 because if we put sinh−1 a = lnu, then a = (u − 1u)/2 and

u = a+√a2 + 1. Thus,

v =

∫

√1−x2

0

√

x2 + y2dy

=

∫ sinh−1(√1−x2/x)

0x2√

1 + sinh2 t cosh tdt

= x2∫ ln

(

1+

√1−x2

x

)

0

1 + cosh 2t

2dt.

Note that sinh 2t = 2 sinh t√

1 + sinh2 t.

v =x2

2

[

ln

(

1 +√1− x2

x

)

+

√1− x2

x2

]

.

We obtain

V = 2

∫ 1

0

[

√

1− x2 − x2 ln

(

1 +√1− x2

x

)]

dx.

Let us first calculate the first integral.

∫ 1

0

√

1− x2dx =

∫ π/2

0

√

1− sin2 θ cos θ dθ

=

∫ π/2

0

1 + cos 2θ

2, dθ

=1

2

[

θ − sin 2θ

2

]π/2

0

=π

4.

To calculate the second integral, note that

d

dx

[

x3

3ln

(

1 +√1− x2

x

)]

= x2 ln

(

1 +√1− x2

x

)

− x2

3√1− x2

.

Therefore,[

x3

3ln

(

1 +√1− x2

x

)]1

0

=

∫ 1

0x2 ln

(

1 +√1− x2

x

)

dx−∫ 1

0

x2

3√1− x2

dx.

15


Since,∫ 1

0

x2

3√1− x2

dx =[

−x

3

√

1− x2]1

0+

1

3

∫ 1

0

√

1− x2dx =π

12,

we have∫ 1

0x2 ln

(

1 +√1− x2

x

)

dx =π

12.

Thus we obtain

V = 2

∫ 1

0

[

√

1− x2 − x2 ln

(

1 +√1− x2

x

)]

dx

= 2(π

4− π

12

)

=π

3.


∫∫

D

(

1−√

x2 + y2)

dA, D = {(x, y)|0 ≤√

x2 + y2 ≤ 1} by first identifying it as the volume of a solid.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

The solid is a cone. The volume is 12π(1)/3 = π/3.

26. Evaluate the integral

∫∫

D

(

1−√

x2 + y2)

dA where D = {(x, y)|0 ≤x2 + y2 ≤ 1} by changing to polar coordinates.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

The region D is expressed as D = {(r, θ)|0 ≤ θ ≤ 2π, 0 ≤ r ≤ 1}.

V =

∫∫

D

(

1−√

x2 + y2)

dA

=

∫ 2π

0

∫ 1

0(1− r)rdrdθ

=

∫ 2π

0

[

r2

2− r3

3

]1

0

dθ

=1

6

∫ 2π

0dθ

=π

3.

16


27. Calculate the double integral

∫∫

Rsin(3x + 2y) dA, R = {(x, y)|0 ≤

x ≤ π/2, 0 ≤ y ≤ π/2}. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

∫∫

Rsin(3x+ 2y) dA =

∫ π/2

0

∫ π/2

0sin(3x+ 2y)dydx

=

∫ π/2

0

[

−1

2cos(3x+ 2y)

]y=π/2

y=0

dx

= −1

2

∫ π/2

0[cos(3x+ π)− cos(3x)] dx

= −1

2

[

sin(3x+ π)

3− sin(3x)

3

]π/2

0

= −1

3.

28. Find the volume of the solid that lies under the hyperbolic paraboloidz = 2 + x2 − 3y2 and above the square R = [1, 2]× [−1, 1].. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

[volume] =

∫ 2

1

∫ 1

−1(2 + x2 − 3y2)dydx

=

∫ 2

1

[

2y + x2y − y3]1

−1dx

=

∫ 2

1(2 + 2x2)dx

=

[

2x+2

3x3]2

1

=20

3.

29. Find the average value of f(x, y) = ln(x + y + 1) over the rectangleR = [0, 2]× [0, 3].. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

17


The average value fave is calculated as

fave =1

6

∫ 2

0

∫ 3

0ln(x+ y + 1)dydx.

Hence,

fave =1

6

∫ 2

0[(x+ y + 1) ln(x+ y + 1)− y]y=3

y=0 dx

=1

6

∫ 2

0[(x+ 4) ln(x+ 4)− (x+ 1) ln(x+ 1)− 3] dx

=1

6

[

(x+ 4)2

2ln(x+ 4)− (x+ 4)2

4− (x+ 1)2

2ln(x+ 1) +

(x+ 1)2

4− 3x

]2

0

=ln 2

3+

9 ln 3

4− 3

2.

30. Let f(x, y) = xy(x2 − y2)/(x2 + y2)3 for (x, y) 6= (0, 0) and f(0, 0) =

0. Show that Fubini’s theorem does not hold and

∫ 1

0

∫ 2

0f(x, y) dxdy 6=

∫ 2

0

∫ 1

0f(x, y) dydx.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .We have

∫ 1

0

∫ 2

0f(x, y) dxdy =

∫ 1

0

[

− x2y

2(x2 + y2)2

]x=2

x=0

dy

= −∫ 1

0

2y

(y2 + 4)2dy

=

[

1

y2 + 4

]1

0

= − 1

20.

18


On the other hand,

∫ 2

0

∫ 1

0f(x, y) dydx =

∫ 2

0

[

xy2

2(x2 + y2)2

]y=1

y=0

dx

=

∫ 2

0

x

2(x2 + 1)2dx

=

[

− 1

4(x2 + 1)

]2

0

=1

5.

Therefore, Fubini’s theorem does not hold.

31. Evaluate the iterated integral

∫ 1

0

∫ 1−x

x(x− y2) dydx

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

∫ 1

0

∫ 1−x

x(x− y2) dydx =

∫ 1

0

[

xy − y3

3

]y=1−x

y=x

dx

=

∫ 1

0

(

−1

3+ 2x− 3x2 +

2

3x3)

dx

=

[

−x

3+ x2 − x3 +

1

6x4]1

0

= −1

6.


∫∫

D(x+y) dA, D is bounded by y = x1/3

and y = x3

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

19


∫ 1

0

∫ x1/3

x3

(x+ y) dydx =

∫ 1

0

[

xy +y2

2

]y=x1/3

y=x3

dx

=

∫ 1

0

(

x4/3 +x2/3

2− x4 − x6

2

)

dx

=

[

3

7x7/3 +

3

10x5/3 − x5

5− x7

14

]1

0

=16

35.


∫∫

Dxy dA, D is the triangular region

with vertices (0, 0), (3, 2), and (0, 4). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

∫ 3

0

∫ 2x/3

0xy dydx+

∫ 4

3

∫ −2x+8

0xy dydx

=

∫ 3

0

[x

2y2]y=2x/3

y=0dx+

∫ 4

3

[x

2y2]y=−2x+8

y=0dx

=

∫ 3

0

2

9x3 dx+

∫ 4

3

(

2x3 − 16x2 + 32x)

dx

=

[

x4

18

]3

0

+

[

x4

2− 16

3x3 + 16x2

]4

3

=20

3.

34. Find the volume of the solid under the surface z = x + y2 and abovethe region bounded by x =

√y and x = y3.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

20


∫ 1

0

∫

√y

y3

(

x+ y2)

dxdy =

∫ 1

0

[

x2

2+ xy2

]x=√y

x=y3dy

=

∫ 1

0

(

y

2+ y5/2 − y6

2− y5

)

dy

=

[

y2

4+

2

7y7/2 − y7

14− y6

6

]1

0

=25

84.

35. Show that

∫∫

D

(

x2 + y2 − 4x− 2y + 5)−3/2

dA ≤ π/2, where D is the

disk with center (2, 1) and radius 2.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Note that for (x, y) ∈ D, we have

(x− 2)2 + (y − 1)2 ≤ 4.

The double integral is estimated as∫∫

D

(

x2 + y2 − 4x− 2y + 5)−3/2

dA =

∫∫

D

[

(x− 2)2 + (y − 1)2]−3/2

dA

≤∫∫

D(4)−3/2 dA

= (4π)4−3/2

=π

2.

36. Find the volume of the solid inside the sphere x2 + y2 + z2 = 9 andoutside the cylinder x2 + z2 = 1.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

We consider polar coordinates in the x-z plane. The volume is obtainedas

2

∫ 2π

0

∫ 3

1

√

9− r2 rdrdθ = 2

∫ 2π

0

[

−1

3

(

9− r2)3/2

]3

1

dθ

=64√2

3π.

21


37. Evaluate the iterated integral

∫ 4

0

∫

√4x−x2

0(x2 + y2) dydx

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .Note that y =

√4x− x2 implies (x− 2)2 + y2 = 4.

∫ 4

0

∫

√4x−x2

0(x2 + y2) dydx =

∫ π/2

0

∫ 4 cos θ

0r2 rdrdθ

=

∫ π/2

0

[

r4

4

]r=4 cos θ

r=0

dθ

= 43∫ π/2

0cos4 θ dθ

= 43∫ π/2

0

(

3

8+

cos 2θ

2+

cos 4θ

8

)

dθ

= 43[

3

8θ +

sin 2θ

4+

sin 4θ

32

]π/2

0

= 12π.

38. Show the Gaussian integral

∫ ∞

−∞e−x2

dx =√π.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

(∫ ∞

−∞e−x2

dx

)2

=

∫ ∞

−∞e−x2

dx

∫ ∞

−∞e−y2dy

=

∫ ∞

−∞

∫ ∞

−∞e−(x2+y2)dydx

=

∫ 2π

0

∫ ∞

0e−r2rdrdθ

= 2π

[

−1

2e−r2

]∞

0

= 2π1

2= π.

Therefore,∫ ∞

−∞e−x2

dx =√π.

22


39. Find the mass m and center of mass (x, y) of the lamina that occupiesthe Cardioid D = {(r, θ)|0 ≤ θ ≤ 2π, 0 ≤ r ≤ 1 + sin θ} when the densityfunction is ρ(x, y) =

√

x2 + y2.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

m =

∫∫

Dρ(x, y) dA

=

∫ 2π

0

∫ 1+sin θ

0r rdrdθ

=

∫ 2π

0

[

r3

3

]r=1+sin θ

r=0

dθ

=1

3

∫ 2π

0

(

5

2+

15

4sin θ − 3

2cos(2θ)− 1

4sin(3θ)

)

dθ

=1

3

[

5

2θ − 15

4cos θ − 3

4sin(2θ) +

1

12cos(3θ)

]2π

0

=5π

3.

x =1

m

∫∫

Dxρ(x, y) dA

=3

5π

∫ 2π

0

∫ 1+sin θ

0r2 cos θ rdrdθ

=3

5π

∫ 2π

0cos θ

[

r4

4

]r=1+sin θ

r=0

dθ

=3

20π

∫ 2π

0cos θ(1 + sin θ)4 dθ

=3

20π

∫ 2π

0cos θ

(

35

8+ 7 sin θ − 7

2cos(2θ)− sin(3θ) +

1

8cos(4θ)

)

dθ

= 0.

23


y =1

m

∫∫

Dyρ(x, y) dA

=3

5π

∫ 2π

0

∫ 1+sin θ

0r2 sin θ rdrdθ

=3

5π

∫ 2π

0sin θ

[

r4

4

]r=1+sin θ

r=0

dθ

=3

20π

∫ 2π

0sin θ(1 + sin θ)4 dθ

=3

20π

∫ 2π

0sin θ

(

35

8+ 7 sin θ − 7

2cos(2θ)− sin(3θ) +

1

8cos(4θ)

)

dθ

=21

20.

Thus, the mass is m = 5π/3 and the center of mass is (x, y) = (0, 1.05).

40. Suppose X and Y are random variables with joint density function

f(x, y) =

√

2

πe−(x−1)2/2e−2y for −∞ < x < ∞, 0 ≤ y < ∞. Find P (Y ≥

ln 3), and the expected values of X and Y .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

P (Y ≥ ln 3) =

√

2

π

∫ ∞

−∞

∫ ∞

ln 3e−(x−1)2/2e−2y dydx

= 2

∫ ∞

ln 3e−2y dy

= 2

[

−e−2y

2

]∞

ln 3

= e−2 ln 3

=1

9.

24


µ1 =

√

2

π

∫ ∞

−∞

∫ ∞

0xe−(x−1)2/2e−2y dydx

=1

2

√

2

π

∫ ∞

−∞xe−(x−1)2/2 dx

=1

2

√

2

π

{∫ ∞

−∞(x− 1)e−(x−1)2/2 dx+

∫ ∞

−∞e−(x−1)2/2 dx

}

=1

2

√

2

π

{∫ ∞

−∞te−t2/2 dt+

∫ ∞

−∞e−t2/2 dt

}

=1

2

√

2

π

{

[

−e−t2/2]∞

−∞+√2π

}

=1

2

√

2

π

√2π

= 1.

µ2 =

√

2

π

∫ ∞

−∞

∫ ∞

0ye−(x−1)2/2e−2y dydx

=

√

2

π

√2π

∫ ∞

0ye−2y dy

= 2

{

[

−y

2e−2y

]∞

0+

1

2

∫ ∞

0e−2y dy

}

= 2

{

1

2

[

−1

2e−2y

]∞

0

}

=1

2.

41. Find the area of the surface which is the part of the sphere x2+y2+z2 =9 that lies within the cylinder x2 + y2 − 3y = 0.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

The domain D is the circle x2 +

(

y − 3

2

)2

=

(

3

2

)2

; we have r = 3 sin θ

in polar coordinates. The sphere is expressed as z = ±√

9− x2 − y2. There-

25


fore, we obtain

∂z

∂x=

∓x√

9− x2 − y2,

∂z

∂y=

∓y√

9− x2 − y2.

The surface area is calculated as

2

∫∫

D

√

1 +

(

∂z

∂x

)2

+

(

∂z

∂y

)2

dA = 2

∫ π

0

∫ 3 sin θ

0

3√9− r2

rdrdθ

= 6

∫ π

0

[

−√

9− r2]3 sin θ

0dθ

= 18

∫ π

0(1− | cos θ|) dθ

= 18

{

∫ π

0dθ −

∫ π/2

0cos θ dθ +

∫ π

π/2cos θ dθ

}

= 18{

[θ]π0 − [sin θ]π/20 + [sin θ]ππ/2

}

= 18(π − 1− 1)

= 18(π − 2).

42. Evaluate the triple integral

∫∫∫

Ex3ey dV , where E is bounded by the

parabolic cylinder z = 1− 2y2 and the planes z = 0, x = 0, and x = 1.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

∫ 1

0

∫ 1/√2

−1/√2

∫ 1−2y2

0x3ey dzdydx =

∫ 1

0

∫ 1/√2

−1/√2x3ey(1− 2y2) dydx

=

[

x4

4

]1

0

[

ey(−2y2 + 4y − 3)]1/

√2

−1/√2

= e1/√2

(

1√2− 1

)

+ e−1/√2

(

1√2+ 1

)

.

26

review2-solabloch/255/review2-sol.pdf · 2012. 3. 20. · title: review2-sol.dvi created date:...

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