Module MA1132 (Frolov), Advanced CalculusHomework Sheet 7

Each set of homework questions is worth 100 marks

Due: at the beginning of the tutorial session Thursday/Friday, 17/18 March 2016

Name:

You may use Mathematica to sketch the integration regions and solids, and to check theresults of integration.

1. Evaluate the double integral

I =

∫∫R

1

x+ ydxdy ,

where R is the region enclosed by the lines x = 2, y = x, and the hyperbola xy = 1.

Solution: The region R is shown below

0.0 0.5 1.0 1.5 2.0

0.0

0.5

1.0

1.5

2.0

We consider R as a region of type I. Then I is given by

I =

∫ 2

1

[∫ x

1/x

1

x+ ydy

]dx =

∫ 2

1

ln(x+ y)|y=xy=1/x dx =

∫ 2

1

ln(2x2

1 + x2) dx

= ln(2x2

1 + x2)x

∣∣∣∣21

−∫ 2

1

x d ln(2x2

1 + x2) = 2 ln

8

5−∫ 2

1

2

1 + x2dx

= 2 ln8

5+π

2− 2 arctan 2 ≈ 0.296506 .

(1)

2. Sketch the integration region R and reverse the order of integration

(a) ∫ 7/2

−1/2

∫ 2+√

7+12y−4y2

2−√

7+12y−4y2f(x, y)dxdy (2)

1

Solution: The region R is shown below

-2 0 2 4 6

0

1

2

3

It is found by noting that

2−√

7 + 12y − 4y2 ≤ x ≤ 2 +√

7 + 12y − 4y2 =⇒ (x− 2)2 ≤ 7 + 12y − 4y2

=⇒ (x− 2)2 + 4(y − 3

2)2 ≤ 16 =⇒ (x− 2)2

16+

(y − 32)2

4≤ 1 .

(3)Thus it is a closed region bounded by an ellipse centred at (2, 3/2) with semi-axis 4and 2. Reversing the order of integration one gets

∫ 7/2

−1/2

∫ 2+√

7+12y−4y2

2−√

7+12y−4y2f(x, y)dxdy =

∫ 6

−2

∫ 32+

√3+x−x2

4

32−√

3+x−x2

4

f(x, y)dydx . (4)

(b) ∫ 1

0

∫ √3−x2x2/2

f(x, y)dydx (5)

Solution: The region R is shown below

0.0 0.2 0.4 0.6 0.8 1.0

0.0

0.5

1.0

1.5

2

Reversing the order of integration one gets the sum of three repeated integrals∫ 1

0

∫ √3−x2x2/2

f(x, y)dydx

=

∫ 1/2

0

∫ √2y0

f(x, y)dxdy +

∫ √21/2

∫ 1

0

f(x, y)dxdy +

∫ √3√2

∫ √3−y2

0

f(x, y)dxdy .

(6)

3. Sketch the integration region R and write the expression as one repeated integral byreversing the order of integration∫ 6

2

∫ 3

9/(x+1)

f(x, y)dydx+

∫ 8

6

∫ 9−x

9/(x+1)

f(x, y)dydx (7)

Solution: The region R is shown below

2 3 4 5 6 7 81.0

1.5

2.0

2.5

3.0

Reversing the order of integration one gets∫ 6

2

∫ 3

9/(x+1)

f(x, y)dydx+

∫ 8

6

∫ 9−x

9/(x+1)

f(x, y)dydx =

∫ 3

1

∫ 9−y

−1+9/y

f(x, y)dxdy . (8)

4. Prove the Dirichlet formula∫ b

a

∫ x

a

f(x, y)dydx =

∫ b

a

∫ b

y

f(x, y)dxdy , (9)

and use it to prove that∫ x

a

∫ t1

a

(t1 − t)n−1f(t)dtdt1 =1

n

∫ x

a

(x− t)nf(t)dt . (10)

Solution: The Dirichlet formula follows from the fact that the repeated integrals are equalto the double integral over the triangle enclosed by the lines y = a, x = b, y = x.

Thus, we get∫ x

a

∫ t1

a

(t1 − t)n−1f(t)dtdt1 =

∫ x

a

∫ x

t

(t1 − t)n−1f(t)dt1dt

=

∫ x

a

1

n(t1 − t)n|t1=xt1=t f(t)dt =

1

n

∫ x

a

(x− t)nf(t)dt .

(11)

3

5. Find the volume V of the solid bounded by

(a) the planes x = 0, y = 0, z = 0, 2x+ y = 1, and the surface z = x+ y2 + 1.

Solution: The solid, and its projection R onto the xy-plane are shown below

0.0 0.1 0.2 0.3 0.4 0.5

0.0

0.2

0.4

0.6

0.8

1.0

Thus, the volume is

V =

∫∫R

(x+ y2 + 1)dxdy =

∫ 1

0

∫ 1−y2

0

(x+ y2 + 1)dxdy

=

∫ 1

0

(1

8(1− y)2 +

1

2(y2 + 1)(1− y))dy =

1

8

∫ 1

0

(−4y3 + 5y2 − 6y + 5)dy

=1

8(−1 +

5

3− 3 + 5) =

1

3.

(12)

(b) the planes x = 1, z = 0, the parabolic cylinder x − y2 = 0, and the paraboloidz = x2 + y2.

Solution: The solid, and its projection R onto the xy-plane are shown below

0.0 0.2 0.4 0.6 0.8 1.0

-1.0

-0.5

0.0

0.5

1.0

Thus, the volume is

V =

∫∫R

(x2 + y2)dxdy =

∫ 1

−1

∫ 1

y2(x2 + y2)dxdy =

∫ 1

−1(y2(1− y2) +

1

3(1− y6))dy

= 2

∫ 1

0

(y2 − y4 +1

3− 1

3y6)dy = 2(

1

3− 1

5+

1

3− 1

21) =

88

105.

(13)

4

(c) the planes x = 0, y = 0, z = 0, the cylinders az = x2, a > 0, x2 + y2 = b2, andlocated in the first octant x ≥ 0, y ≥ 0, z ≥ 0.

Solution: The solid, and its projection R onto the xy-plane are shown below (a = 2, b = 1)

Thus, the volume is

V =

∫∫R

1

ax2dxdy =

1

a

∫ b

0

∫ √b2−x20

x2dydx =1

a

∫ b

0

x2√b2 − x2dx

=b4

a

∫ 1

0

x2√

1− x2dx =b4

a

∫ π/2

0

sin2 t cos2 tdt =b4

a

∫ π/2

0

1

4sin2 2tdt

=b4

4a

∫ π/2

0

1

2(1− cos 4t)dt =

πb4

16a.

(14)

(d) the planes z = zmin, z = λx+ µy + h (λ > 0, µ > 0, h > 0), and the elliptic cylinderx2

a2+ y2

b2= 1, where zmin is the z-coordinate of the lowest point on the intersection of

z = λx+ µy + h (λ > 0, µ > 0, h > 0) and x2

a2+ y2

b2= 1.

Solution: The intersection, the solid, and its projection R onto the plane z = zmin are shownbelow (a = 1, b = 2, λ = 1, µ = 2, h = 5)

-1.0 -0.5 0.0 0.5 1.0

-2

-1

0

1

2

5

To find zmin we use the Lagrange multiplier method, and get the equations

λ = 2Λx

a2, µ = 2Λ

y

b2,

x2

a2+y2

b2= 1 . (15)

Multiplying the first equation by x, the second equation by y and summing up theresulting equations, one gets

λx+ µy = 2Λ . (16)

Thus, one derives the equations for x and y

λ = (λx+ µy)x

a2, µ = (λx+ µy)

y

b2, (17)

or by using the constraint x2

a2+ y2

b2= 1

y2

b2=

µ

λa2xy ,

x2

a2=

λ

µb2xy . (18)

Summing up these equations one finds

xy =λµa2b2

λ2a2 + µ2b2, x2 =

λ2a4

λ2a2 + µ2b2, y2 =

µ2b4

λ2a2 + µ2b2. (19)

Thus there are two solutions

xmin = − λa2√λ2a2 + µ2b2

, ymin = − µb2√λ2a2 + µ2b2

,

xmax = +λa2√

λ2a2 + µ2b2, ymax = +

µb2√λ2a2 + µ2b2

,

(20)

corresponding to the lowest and highest points on the intersection, respectively. Thez-coordinates of these points are

zmin = h−√λ2a2 + µ2b2 , zmax = h+

√λ2a2 + µ2b2 . (21)

To find the volume we first shift the z-coordinate z → z + zmin so that the planez = zmin becomes the plane z = 0, and the plane z = λx+ µy + h becomes

z = λx+ µy + h− zmin. Then, the volume is

V =

∫∫R

(λx+ µy + h− zmin)dxdy =

∫ a

−a

∫ b√

1−x2/a2

−b√

1−x2/a2(λx+ µy + h− zmin)dydx

= 2b

∫ a

−a(λx+ h− zmin)b

√1− x2

a2dx = 4b(h− zmin)

∫ a

0

√1− x2

a2dx

= 4ab(h− zmin)

∫ 1

0

√1− x2dx = 4ab(h− zmin)

∫ π/2

0

cos2 tdt = πab√λ2a2 + µ2b2 .

(22)

6