Engineering Mathematics 233 Solutions: Double and triple integrals

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Engineering Mathematics 233. Solutions: Double and triple integrals. Double Integrals. 1. Sketch the region R in the xy-plane bounded by the curves y2 = 2x and ...
Engineering Mathematics 233 Solutions: Double and triple integrals Double Integrals 1. Sketch the region R in the xy-plane bounded by the curves y 2 = 2x and y = x, and find its area. Solution 1 The region R is bounded by the parabola x = y 2 and the straight line y = x. The points of intersection of the two 2 curves are given by y2 y= ⇐⇒ y 2 − 2y = 0 ⇐⇒ y(y − 2) = 0 ⇐⇒ y = 0, 2. 2 This gives the two points A = (0, 0) and B = (2, 2). y

2

B

A



x

y 2=2x 

The region R is a Type II region, and can be described by 0≤y≤2 y2 ≤ x ≤ y. 2

R:

Then, area(R)

Z Z

=

1 dA =

R

Z

2

y2 = y− dy = 2 0 8 2 = 2− = . 6 3 2. Evaluate the integral Z

3

y=0

Z



y



y2 y3 − 2 6

4−y

(x + y) dx dy

x=1

by interchanging the order of integration. Solution

1

1 dx dy

y 2 /2

0

2

Z

Z

 2 0

The region of integration is the Type II region R 0≤y≤3 p 1 ≤ x ≤ 4 − y.

R:

x=1 4 3 R x

2

1

x

x =

4−y

We have x=

p

4−y



x2 = 4 − y



y = 4 − x2 .

Then, from the drawing above, we can rewrite the region R as the Type I region 1≤x≤2 0 ≤ y ≤ 4 − x2 .

R:

Then, Z

3

y=0

Z



4−y

(x + y) dx dy

2

= =

x=1 y=0 Z 2

x=1

Z

4−x2

Z

xy +

1 2

(x + y) dy dx y2 2

 2 4−x 0

16 − 8x2 + x4 dx 2  1 4x3 x5 2 241 x4 = 2x2 − + 8x − + . = 4 3 10 1 60

=

Z

4x − x3 +

3

3. Find the volume of the region of R bounded by the paraboloid z = x2 + y 2 and by the planes z = 0, x = −a, x = a, y = −a and y = a. Solution

2

Let S be the 3D region bounded by the paraboloid and the planes. z a -a -a

a

region R

y

x

Then, volume(S) =

Z Z R

 x2 + y 2 − 0 dA,

where R is the projection of S in the xy-plane, i.e. −a ≤ x ≤ a −a ≤ y ≤ a.

R: Then, volume(S)

=

Z Z

2

2

x + y dA =

Z

−a

R

Z

a

x2 + y 2 dy dx

−a

 Z a y 3 a x2 + y 2 dy dx = 4 x2 y + dx 3 0 0 0 0  3  Z a 3 3 a a x a a x dx = 4 + = 4 x2 a + 3 3 3 0 0  4 2a 8a4 = 4 = . 3 3 =

4. Evaluate

4

Z

a

a

Z

a

Z Z p x2 + y 2 dA, where R is the region of the plane given by x2 + y 2 ≤ a2 . R

Solution p The region R and the integrand x2 + y 2 are best described with polar coordinates (r, θ). In those coordinates, the region R, which is the region inside the circle x2 + y 2 = a2 , becomes R:

0≤r≤a 0 ≤ θ ≤ 2π.

Then, Z Z p

x2 + y 2 dA

=

R

Z



0

=

Z

=

0

3

a

r r dr dθ

0 2π

0

Z

Z



Z

a

0 3

r2 dr dθ

a 2πa3 dθ = . 3 3

5. Evaluate

Z Z

2

e−(x

+y 2 )

dA, where R is the region of 4. above.

R

Solution Using polar coordinates again, we write Z Z 2 2 e−(x +y ) dA

=

Z

=

Z

=

Z

R



0 2π

0 2π

0

6. Evaluate the integral

R1R1 0

y2

2

yex dx dy.

a

Z

2

e−r r dr dθ

0 −r 2

a e dθ −2 0

2 2 1 (1 − e−a ) dθ = π(1 − e−a ). 2

Hint: First reverse the order of integration.

Solution If we try to evaluate the integral as written above, then the first step is to compute the indefinite integral Z 2 ex dx. 2

But ex does not have an indefinite integral that can be written in terms of elementary functions. Then, we will fist reverse the order of integration. The region of integration is the Type II region 0≤y≤1 y 2 ≤ x ≤ 1.

R:

y x=y2

1

x 1

Then, R can also be described as the Type I region R:

0≤x≤1 √ 0 ≤ y ≤ x.

This gives Z 0

1

Z

1

x2

ye

dx dy

=

Z

=

Z

y2

1

1



Z

1

0

=

1 2

x

2

yex dy dx

0

0

=



Z

y 2 x2 e 2 2

 √ x dx 0

xex dx 0 ! 2 1 1 ex 1 = (e − 1). 2 2 4 0 4

7. Find the volume of the tetrahedron bounded by the coordinate axes and the plane 3x + 6y + 4z = 12. Solution We have to find the volume of the tetrahedron S bounded by the plane 3x + 6y + 4z = 12 ⇐⇒ z =

12 − 3x − 6y 4

and the coordinate axes. This is the portion of the plane in the first octant, as one can see from Picture (1).

z 3

line 3x + 6y = 12

Region R

plane 3x + 6y + 4z = 12

3 2

4

y

4

y

R

x

x

(1)

(2)

Then, we have volume(S) =

Z Z R

12 − 3x − 6y dA, 4

where R is the projection of the tetrahedron in the xy-plane. Then, R is the Type I region (see Picture (2)) 0≤x≤4 12 − 3x 0≤y≤ . 6

R:

Finally, this gives volume(S)

= = = = =

1 4

Z

1 4

Z

1 4

Z

4

0

Z

12−3x/6

(12 − 3x − 6y) dy dx

0 4

 2− x2 (12 − 3x)y − 3y 2 dx 0

0 4



(12 − 3x) 2 −

0

 x 2 −3 2− dx 2 2

x

Z 1 4 3 2 x − 6x + 12 dx 4 0 4  3  1 x 4 2 − 3x + 12x = 4. 4 4 0

5

8. Evaluate the integral 1

Z 0

Z √4−y2 p x2 + y 2 dx dy. √ 3y

Hint: Use polar coordinates. Solution The region R of integration is the Type II region 0≤y≤1 p √ 3 y ≤ x ≤ 4 − y2

R:

We have x=

p 4 − y2

Then, x varies between the straight line x =





x2 = 4 − y 2

x2 + y 2 = 4.



3 y and the circle x2 + y 2 = 4.

The region R is

P = ( 3 , 1)

1

2

In polar coordinates, the region R is R:

0≤r≤2 0 ≤ θ ≤ α,

√ where α is the angle made by the straight line x = 3 y. The straight line and the cicle meet at the points  √ 2 3y + y 2 = 4 ⇐⇒ 4y 2 = 4 ⇐⇒ y 2 = 1 ⇐⇒ y = ±1. √ The intersection point in the first quadrant is then P = ( 3, 1) = (2 cos α, 2 sin α). Then,   1 π α = arcsin = . 2 6 p Finally, the integrand x2 + y 2 is r in polar coordinates. This gives Z 1 Z √4−y2 p Z π/6 Z 2 2 + y 2 dx dy = x r r dr dθ √ 0

3y

0

= =

0

π/6

r3 2 dθ 3 0 0 π 8 4π = . 6 3 9

Z

9. Find the volume below the surface z = x2 + y 2 , above the plane z = 0, and inside the cylinder x2 + y 2 = 2y.

6

Solution Completing the squares, we rewrite the equation of the cylinder as x2 + y 2 = 2y ⇐⇒ x2 + (y 2 − 2y) = 0 ⇐⇒ x2 + (y − 1)2 = 1. The base of the cylinder is then the circle of radius 1 centered at (0, 1). Then, we have to find the volume of the 3D region: z

z = x2 + y2 volume

y region R

x

From the picture above, we write V =

Z Z

x2 + y 2 dA,

R

where R is the projection of the 3D region in the plane, i.e. the circle x2 + y 2 = 2y. Using polar coordinates, this gives Z Z V = r2 r dr dθ. R

In polar coordinates x = r cos θ and y = r sin θ, the circle writes as x2 + y 2 = 2y ⇐⇒ r2 = 2r sin θ ⇐⇒ r = 2 sin θ, and R is the region 0≤θ≤π 0 ≤ r ≤ 2 sin θ.

R:

Then, V

Z π 4 2 sin θ r r3 dr dθ = dθ 4 0 0 0 0  Z π π  3θ 1 1 4 = 4 sin θ dθ = 4 − sin 2θ + sin 4θ 8 4 32 0 0   3π 3π = 4 = . 8 2

=

Z

π

Z

2 sin θ

Triple Integrals 10. Evaluate Z

1

x=0

Z

1

y=0

Z

2



z=

xyz dz dy dx. x2 +y 2

7

Solution

Z 0

1

Z

1

0

Z

2



xyz dz dy dx =

Z

=

Z

=

Z

x2 +y 2

1

Z

1

Z

0

1

0

0

1

xyz 2 2 dy dx √ 2 z= x2 +y2 2xy −

0 1

xy(x2 + y 2 ) dy dx 2

1

x3 y y 3 x − dy dx 2 2 0 0   Z 1 y 4 x 1 x3 y 2 xy 2 − − dx 4 8 0 0 Z 1 3 x x x− − dx 4 8 0 Z 1 3 7x x − dx 8 4 0   2 x4 1 7x − 16 16 0 7 1 3 − = . 16 16 8

= = = = =

Z

2xy −

11. Find the mass of the 3D region B given by x2 + y 2 + z 2 ≤ 4, x ≥ 0, y ≥ 0, z ≥ 0, if the density is equal to xyz. Solution We have mass(B) =

Z Z Z

xyz dV,

B

and the region B is the portion of the sphere of radius 2 in the first octant. z sphere x 2 + y2 + z2 = 4

y

y

x 2 + y2 ≤ 4

x x x

8

3D region B

2D region R

Then, B can be described as p

4 − x2 − y 2 , for all (x, y) ∈ R,

0≤z≤

where R is the projection of B in the xy-plane. Describing R as a Type I region, this gives 0≤x≤2 p 0 ≤ y ≤ 4 − x2 p 0 ≤ z ≤ 4 − x2 − y 2 .

B:

Then, mass(B)

=



2

Z

Z

0

=

0 2

Z

1 2

Z

2

1 2

Z

Z 0

= = = = = = =



4−x2

0 4−x2

0

0



Z 0

2

Z √4−x2 −y2

Z



z2 xy 2

4−x2

xyz dz dy dx

 √ 2 2 4−x −y dy dx 0

xy(4 − x2 − y 2 ) dy dx √

4−x2

4xy − x3 y − y 3 x dy dx  √ Z  1 2 x3 y 2 y 4 x 4−x2 2xy 2 − − dx 2 0 2 4 0 Z 1 2 x3 (4 − x2 ) (4 − x2 )2 x 2x(4 − x2 ) − − dx 2 0 2 4 Z 2 5 1 x − 2x3 + 4x dx 2 0 4   x4 4x2 2 1 x6 − + 2 24 2 2 0   1 64 4 −8+8 = . 2 24 3 0

0

12. Find the volume of the region B bounded by the paraboloid z = 4 − x2 − y 2 and the xy-plane. Solution

9

We have volume(B) =

Z Z Z

1 dV.

B

z (0,0,4) z = 4 - x2 - y2

y x

R

Then, B can be described as 0 ≤ z ≤ 4 − x2 − y 2 , for all (x, y) ∈ R, where R is the projection of B in the xy-plane. Then, R is the interior of the circle x2 + y 2 = 4. In polar coordinates, the region R is 0 ≤ θ ≤ 2π 0 ≤ r ≤ 2,

R:

and in cylindrical coordinates, the region B is 0 ≤ z ≤ (4 − r2 ) 0 ≤ θ ≤ 2π 0 ≤ r ≤ 2.

B:

Then, Z

volume(B) =



0

Z

=

2

Z 0



0

Z

4−r 2

Z

r dz dr dθ

0 2

Z

(4 − r2 )r dr dθ

0 2

4r − r3 dr 0  r4 2 = 2π 2r2 − = 2π(4) = 8π. 4 0 = 2π

13. Find the center of gravity of the region in 12., assuming constant density σ. Solution By symmetry, x = y = 0. Also, as the density is d(x, y, z) = σ, Z Z Z Z Z Z d(x, y, z)z dV σ z dV B B Z Z Z Z Z Z z = = σ 1 dV d(x, y, z) dV B

B

Z Z Z =

z dV

B

volume(B)

1 = 8π 10

Z Z Z B

z dV .

Using the description of the region B in cylindrical coordinates of 12., we get Z Z Z

z dV



Z

=

B

Z

0

0 2π

Z

=

2

0

Z

Z

4−r 2

Z

z r dz dr dθ

0 2

0 2

(4 − r2 )2 r dr dθ 2

2π 16r − 8r3 + r5 dr 2 0   r6 2 32 = π 8r2 − 2r4 + = π. 6 0 3

=

Then, z=

14. Evaluate



1 8π

32 π 3



=

4 . 3

Z Z Z p

x2 + y 2 + z 2 dV,

B

where B is the region bounded by the plane z = 3 and the cone z =

p x2 + y 2 .

Solution We will use the spherical coordinates z

ρ φ

y θ

x

to describe the region B. In those coordinates, x = ρ sin φ cos θ y = ρ sin φ sin θ z = ρ cos φ.

Then, the cone z =

p x2 + y 2 writes as ρ cos φ = ρ sin φ ⇐⇒

sin φ π = tan φ = 1 ⇐⇒ φ = , cos φ 4

the plane z = 3 as 3 = ρ cos φ ⇐⇒ ρ =

11

3 , cos φ

and the region B can be described as π 4 0 ≤ φ ≤ 2π 3 0≤ρ≤ . cos φ 0≤φ≤

B:

z=3

x 2 + y2

z =

y

x

Finally, in spherical coordinates, Then,

p

x2 + y 2 + z 2 = ρ.

Z Z Z p x2 + y 2 + z 2 dV

Z

=

B



Z

0

= 2π

Z

=

Z

π/4

0 π/4

Z

3/ cos φ

sin φ

Z

sin φ



0



ρ3 dρ dφ ρ4 3/ cos φ 4 0





Z π/4 3 sin φ dφ 4 0 cos4 φ   81π (cos φ)−3 π/4 2 3 0   √ !−3  81π  2 27π  √ − 1 = 2 2−1 . 6 2 2

=



=

=

Z Z Z

3/ cos φ

0 π/4

0 4

15. Evaluate

ρ ρ2 sin φ dρ dφ dθ

0

x2 + y 2 + z 2

B

−3/2

dV,

where B is the region bounded by the spheres x2 + y 2 + z 2 = a2 and x2 + y 2 + z 2 = b2 , where a > b > 0. Solution Using spherical coordinates, the region B between the 2 spheres can be described as B:

0 ≤ θ ≤ 2π 0≤φ≤π a ≤ ρ ≤ b.

12

z

B

y

x

Then, Z Z Z

2

2

2 −3/2

(x + y + z )

dV

Z

=

B



0

Z

Z

π

Z

b

ρ−3 ρ2 sin φ dρ dφ dθ

0 a πZ b

1 sin φ dρ dφ 0 a ρ  Z π b  = 2π ln ρ sin φ dφ a 0  Z π b = 2π ln sin φ dφ a     0 π  b b = 2π ln − cos φ = 4π ln . a a 0

= 2π

16. Find the volume of the region B bounded above by the sphere x2 + y 2 + z 2 = a2 and below by the plane z = b, where a > b > 0. Solution 3

The region B can be described as the set of (x, y, z) ∈ R such that p b ≤ z ≤ a2 − x2 − y 2

13

for all (x, y) in the plane region bounded by the circle x2 + y 2 + b2 = a2 ⇐⇒ x2 + y 2 = a2 − b2 . (0, 0, a) B z=b

x2 + y2 + z2 = a2

sphere

The region B is best described in cylindrical coordinates, and this gives p B: b ≤ z ≤ a2 − r2 p 0 ≤ r ≤ a2 − b2 0 ≤ θ ≤ 2π. Then, volume(B)

Z Z Z

=

Z

=



0

Z

=

Z 0



0

Z

1 dV

B √

Z



a2 −b2



Z

a2 −r 2

r dz dr dθ

b a2 −b2

0 √ a2 −b2

p ( a2 − r2 − b)r dr dθ

p a2 − r2 r − br dr 0 √a2 −b2 ! (a2 − r2 )3/2 r2 = 2π −b (3/2)(−2) 2 0     b 1 2 3/2 2 3/2 2 2 = 2π − (b ) − (a ) − (a − b ) 3 2  3   3  a − b3 a2 b b3 a a2 b b3 = 2π − + = 2π − + . 3 2 2 3 2 6

=



17. Sketch the region B whose volume is given by the triple integral Z 4 Z (4−x)/2 Z (12−3x−6y)/4 1 dz dy dx. 0

0

0

Rewrite the triple integral using the order of integration dV = dy dx dz. Solution From the triple integral, the region B is described by 0≤z≤

12 − 3x − 6y , 4 14

i.e. z varies between the planes z = 0 and z = the region R described by

12 − 3x − 6y ⇐⇒ 3x + 6y + 4z = 12. Furthermore, (x, y) are in 4 4−x 2 0 ≤ x ≤ 4, 0≤y≤

R:

which is the projection of the plane 3x + 6y + 4z = 12 in the xy-plane. z

(0,0,3)

y (0,2,0) (4,0,0)

x

region R

We now use the ordre of integration dV = dy dx dz. The region B be can be described as 0≤y≤

12 − 3x − 4x 6

for all (x, z) in the region R which is the projection of B in the xz-plane. Then, R can be described as 0≤z≤3 (12 − 4z) 0≤x≤ . 3

R:

This gives Z 0

4

Z 0

(4−x)/2

Z

(12−3x−6y)/4

1 dz dy dx =

Z

0

0

18. Evaluate

3

Z

(12−4z)/3

0

Z

(12−3x−4z)/6

0

Z Z Z p x2 + y 2 dV, B

where B is the region lying above the xy-plane, and below cone z = 4 −

p x2 + y 2 .

Solution The region B be can be described as 0≤z ≤4−

p

x2 − y 2 ,

for all (x, y) ∈ R,

where R is the projection of B in the xy-plane. Then, R is the region inside the curve p 0 = 4 − x2 + y 2 ⇐⇒ x2 + y 2 = 16,

15

1 dy dx dz.

which is a circle of radius 4. z

z = 4−

x

2

+y

2

y x R

We then use cylindrical coordinates to describe the region B. This gives 0≤z ≤4−r 0≤r≤4 0 ≤ θ ≤ 2π.

B:

Then, Z Z Z p x2 + y 2 dV

=

Z

B



Z

0

=

Z

0 2π

Z

0

=

Z Z



Z

19. Evaluate the integral Z 0

4



Z

16−x2

0

Z



64 3

(16−x2 −y 2 )

0

4−r

r r dz dr dθ

0 4

r2 (4 − r) dr dθ 4

4r2 − r3 dr

0

0

=

Z

0

0

=

4

Z

4 r4 4r3 − dθ 3 4 0

0



dθ =

64 (2π). 3

p x2 + y 2 dz dy dx.

Hint: First convert to cylindrical coordinates. Solution The region B described by the integral is the region given by 0 ≤ z ≤ (16 − x2 − y 2 ), i.e. bounded below by √the plane z = 0 and above by the paraboloid z = 16−x2 −y 2 , for all (x, y) ∈ R. For the region R, we have 0 ≤ y ≤ 16 − x2 , i.e. y varies between the straight line y = 0 and the top part of the circle x2 + y 2 = 16. Similarly, x varies between

16

the straight lines x = 0 and x = 4. Then, R is the portion of the circle x2 + y 2 = 16 in the first quadrant. z

(0,0,16)

y =

z = 16 - (x2 + y2)

16 − x

2

y 0

R

x

3D region B

4

2D region R

The region B is best described in cylindrical coordinates as 0 ≤ z ≤ 16 − r2 0 ≤ θ ≤ π/2 0 ≤ r ≤ 4.

B:

Then, Z 0

4



Z 0

16−x2

Z 0

16−x2 −y 2

Z p 2 2 x + y dz dy dx =

0

= = = = 20. Evaluate

π/2

π 2

Z 0

Z

4

Z

16−r 2

r r dz dr dθ

0

4

r2 (16 − r2 ) dr

0

Z π 4 16r2 − r4 dr 2 0   π 16r3 r5 4 − 2 3 5 0   π 2048 1024π = . 2 15 15

Z Z Z p x2 + y 2 + z 2 dV, B

where B is the region above the xy-plane bounded by the cone z 2 = 3(x2 + y 2 ) and by the sphere x2 + y 2 + z 2 = 1. Solution In spherical coordinates (ρ, θ, φ), the sphere is x2 + y 2 + z 2 = 1 ⇐⇒ ρ2 = 1 ⇐⇒ ρ = 1, and the cone is z 2 = 3(x2 + y 2 ) ⇐⇒ ρ2 cos2 φ = 3ρ2 sin2 φ ⇐⇒ tan2 φ =

17

1 π 4π 1 . ⇐⇒ tan φ = ± √ ⇐⇒ φ = or 6 6 3 3

4π . 6

The part of the cone above the xy-plane corresponds to φ =

sphere x2 + y2 + z2 = 1

z

top part of cone

B

z2 = 3(x2 + y2 )

y

x

Then, in spherical coordinates, the region B is 0≤ρ≤1 π 0≤φ≤ 6 0 ≤ θ ≤ 2π.

B:

Then, Z Z Z p x2 + y 2 + z 2 dV

=

Z

= = = =

Z

π/6

Z

1

ρ ρ2 sin φ dρ dφ dθ 1 ! Z 2π Z π/6 ρ4 sin φ dφ dθ 4 0 0 0 Z Z 1 2π π/6 sin φ dφ dθ 4 0 0 π/6 ! Z 1 2π − cos φ dθ 4 0 0 √ Z 1 2π 3 1− dθ 4 0 2 √ ! 3 π 1− . 2 2 0

=



18

0

0