AMath 231 ASSIGNMENT # 1 Solutions: Review [15 Total] Fall 2014

Due Monday, September 15, 2014 at 2pm in box 7, slot 11 (A-M) and 12 (N-Z), locatedacross from MC4066. Late assignments or assignments submitted to the incorrect dropboxwill receive a grade of zero. Write your solutions clearly and concisely. Marks will bededucted for poor presentation and incorrect notation.

1. Evaluate

x7

1 + x2 dx. [0 marks]

Solution: We substitute u = x2 and du = 2xdx to rewrite this asx7

1 + x2 dx =

u3

1 + u du.

Next substitute v = 1 + u, expand then integrate,u3

1 + u du =

(1 v)3v dv,

=

(1 3v + 3v2 v3)v dv,

=

[2

3v3/2 6

5v5/2 +

6

7v7/2 2

9v9/2)

]+ c

=

[2

3[(1 + x2)3/2 6

5(1 + x2)5/2 +

6

7(1 + x2)7/2 2

9(1 + x2)9/2)

]+ c.

2. Evaluate

sin(log x) dx. [0 marks]

Solution:

Substitute u = log x and du = dx/x or eudu = dxsin(log x) dx =

eu sin(u) du.

There are several ways to integrate this. The standard way you would have learned inAMATH 250. A more clever way is to use Eulers formula realize that the above is theimaginary part of the following, which is easier to evaluate

Imag

[e(1+i)u du

]= Imag

[1

1 + ie(1+i)u

]+ c,

= Imag

[1 i

2e(1+i)u

]+ c,

=1

2eu [sinu cosu] + c,

=1

2x [sin log x cos log x] + c.

1

yx

2

0

Rx2 + (y 1)2 = 1

Figure 1: Region R.

3. Evaluate

(log x)2 dx. [3 marks]

Solution:

Substute u = log x and du = dx/x or eudu = dx(log x)2 dx =

euu2 du.

This we integrate by parts twice,u2eu du = u2eu 2

ueu du,

= u2eu 2ueu + 2eu du,

= u2eu 2ueu + 2eu,= x

((log x)2 2 log x+ 2) .

4. Evaluate

R

(9 x2 y2)dA where R is the region x2 + (y 1)2 1. [4 marks]

Solution:

The region of integration is depicted in Figure 1, which is a circle of radius 1 centredat (0, 1) and hence suggests that a transformation into polar coordinates is an idealchoice for integration. Let x = r cos() and y = r sin(). In polar coordinates theregion R is bounded by

(r cos())2 + (r sin() 1)2 = 1 r(r 2 sin()) = 0.

Either r = 0, which is not a possible choice, or r = 2 sin(). It follows to generate theregion R that r = 2 sin() for from 0 to pi. Furthermore, dA = rdrd (this is from

2

the Jacobian; convince yourself by trying it out as an exercise). Therefore,R

(9 x2 y2)dA =

pi0

2 sin()0

(9 r2)rdrd

=

pi0

(9

2r2 1

4r4) 2 sin()

0

d

=

pi0

(18 sin2() 4 sin4()) d

=

pi0

(18

2(1 cos(2)) 4

4(1 cos(2))2

)d

= 9( 12

sin(2))

pi0

pi0

(1 2 cos(2) + cos2(2))d

= 9pi pi0

(1 2 cos(2) + cos(4) + 1

2

)d

= 9pi ( sin(2) + sin(4)

8+

2

) pi0

= 9pi pi pi2

=15pi

2

where we have applied the double angle formula

cos2() =cos(2) + 1

2and sin2() =

1 cos(2)2

to aid in the integration.

5. Evaluate

G

xyzdV where G is the solid in the first octant bounded by the sphere

x2 + y2 + z2 = 4 and the coordinate planes. [0 marks]Solution:

The region of integration, G, is a sphere constrained to x 0, y 0, z 0, x2 + y2 +z2 22 (see Figure 2). Apply spherical coordinates,

x = r sin() cos()

y = r sin() sin()

z = r cos().

For the first octant, we must have r 2, 0 pi/2 and 0 pi/2. Furthermore,dV = r2 sin()drdd (this is from the Jacobian; convince yourself by trying it out as

3

zy

x

Figure 2: Region G.

an exercise). Therefore, G

xyzdV =

pi2

0

pi2

0

20

(r sin() cos()

)(r sin() sin()

)(r cos()

)r2 sin()drdd

=

20

r5dr

pi2

0

sin3() cos()d

pi2

0

sin() cos()d

=1

6r620

pi2

0

sin3() cos()d

pi2

0

sin() cos()d

=26

6

10

u3du

10

wdw, (where u = sin(), w = sin())

=26

6

u4

4

10

1

2w210

=4

3.

6. The angular momentum of a particle of mass m with position vector r is

H = r (mu).

Let u = r. Using the result

a (b c) = (a c) b (a b) c,

show that H = mr2 when r is perpendicular to . Here r2 = r r. [2 marks]

Solution:

4

We begin by computing H:

H = r (mu),= mr ( r),= m(r r) (r ) r= mr2 ,

as we expected.

7. (a) Solve the differential equationdy

dx= y

x. [0 marks]

Solution: The differential equation is separable and hence through integrationwe get

1

ydy =

1

xdx,

log y = log x+ c,

Solving leads to

xy = constant.

Therefore, xy = c where c is a constant.

(b) Find the solution to (a) that passes through the point (1, 2). [0 mark]Solution: Substituting y(1) = 2 into the solution in (a), we obtain 2 = c andhence c = 2. Therefore, xy = 2.

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