Math, asked by klohithkumar26, 7 months ago

Q 25
So'S
4-4(x + y) dxdy is
(A) 35
(B) 45
(C)241/60
(D) 60​

Answers

Answered by sudershanswain120995
0

Answer:

Problem 1 (16.1.15). Use symmetry to evaluate Z Z

R

x

3

dA for R = [−4, 4] × [0, 5].

Solution. For each y, the x-integral goes from −4 to 4, so is symmetric about the x-axis, while the

function x

3

is odd, meaning that (−x)

3 = −x

3

. This means that the negative and positive portions

of the integral cancel, so each x-integral is 0, hence the entire area integral is 0.

Problem 2 (16.1.17). Use symmetry to evaluate Z Z

R

sin x dA for R = [0, 2π] × [0, 2π].

Solution. For each y, the x-integral is the integral of sin x over a whole period, so is 0. Thus the

entire area integral is 0.

Problem 3 (16.1.24). Evaluate Z 1

−1

Z π

0

x

2

sin y dx dy.

Solution.

Z y=1

y=−1

Z x=π

x=0

x

2

sin y dx dy =

Z y=1

y=−1

x

3

3

sin y

x=π

x=0

dy

=

Z y=1

y=−1

π

3

3

sin y dy

=

π

3

3

[− cos y]

y=1

y=−1

=

π

3

3

(cos(−1) − cos 1)

= 0.

Problem 4 (16.1.31). Evaluate Z 2

1

Z 4

0

dy dx

x + y

.

Solution.

Z x=2

x=1

Z y=4

y=0

dy dx

x + y

=

Z x=2

x=1

[ln(x + y)]y=4

y=0 dx

=

Z x=2

x=1

(ln(x + 4) − ln(x)) dx

= [(x + 4) ln(x + 4) − (x + 4)]x=2

x=1 − [x ln x − x]

x=2

x=1

= [(6 ln 6 − 6) − (5 ln 5 − 5)] − [(2 ln 2 − 2) − (1 ln 1 − 1)]

= 6 ln 6 − 5 ln 5 − 2 ln 2.

3

Homework Solutions MATH 32B-2 (18W)

Problem 5 (16.1.34). Evaluate Z 8

0

Z 2

1

x dx dy

p

x

2 + y

.

Solution.

Z y=8

y=0

Z x=2

x=1

x dx dy

p

x

2 + y

=

Z y=8

y=0

Z u=4+y

u=1+y

(1/2) du dy

u

(u = x

2 + y)

=

Z y=8

y=0

u

u=4+y

u=1+y

dy =

Z y=8

y=0

p

4 + y −

p

1 + y

dy

=

2

3

(4 + y)

3/2 −

2

3

(1 + y)

3/2

y=8

y=0

=

2

3

h123/2 − 9

3/2

4

3/2 − 1

3/2

i

=

2

3

h

24√

3 − 34i

=

−68 + 48√

3

3

.

Problem 6 (16.1.40). Evaluate Z Z

R

y

x + 1

dA for R = [0, 2] × [0, 4].

Solution.

Z Z

R

y

x + 1

dA =

Z x=2

x=0

Z y=4

y=0

y

x + 1

dy dx (Fubini)

=

Z x=2

x=0

1

2

y

2

x + 1y=4

y=0

dx =

Z x=2

x=0

8

x + 1

dx

= 8 [ln(x + 1)]x=2

x=0 = 8 ln 3.

Problem 7 (16.1.42). Evaluate Z Z

R

e

3x+4y

dA for R = [0, 1] × [1, 2].

Solution.

Z Z

R

e

3x+4y

dA =

Z y=2

y=1

Z x=1

x=0

e

3x+4y

dx dy (Fubini)

=

Z y=2

y=1

1

3

e

3x+4y

x=1

x=0

dy =

1

3

Z y=2

y=1

e

4y+3 − e

4y

dy

=

e

3 − 1

3

Z y=2

y=1

e

4y

dy =

e

3 − 1

3

1

4

e

4y

y=2

y=1

=

(e

3 − 1)(e

8 − e

4

)

12

.

4

MATH 32B-2 (18W) (L) G. Liu / (TA) A. Zhou Calculus of Several Variables

Problem 8 (16.1.47). Evaluate Z 1

0

Z 1

0

y

1 + xy

dy dx.

Solution.

Z x=1

x=0

Z y=1

y=0

y

1 + xy

dy dx =

Z y=1

y=0

Z x=1

x=0

y

1 + xy

dx dy (Fubini to change order)

=

Z y=1

y=0

Z u=1+y

u=1

1

u

du dy =

Z y=1

y=0

[ln u]

u=1+y

u=1 dy (u = 1 + xy)

=

Z y=1

y=0

ln(1 + y) dy = [(1 + y) ln(1 + y) − (1 + y)]y=1

y=0

= [2 ln 2 − 2] − [1 ln 1 − 1] = 2 ln 2 − 1.

Problem 9 (16.1.49). Using Fubini’s theorem, argue that the solid in Figure 1 has volume AL,

where A is the area of the front face of the solid.

Figure 1

Solution. The surface bounding the solid from above is the graph of a positive function z = f(y)

that does not depend on x. (Here a is the largest value that y can take, which is not labeled in the

diagram.) The volume of the solid is

Z Z

[0,L]×[0,a]

f(y) dA =

Z y=a

y=0

Z x=L

x=0

f(y) dx dy (Fubini)

=

Z y=a

y=0

[xf(y)]x=L

x=0 dy =

Z y=a

y=0

Lf(y) dy

= L

Z y=a

y=0

f(y) dy = LA,

where in the last step, we have used the fact that A is the area under the graph of z = f(y) in the

(y, z)-plane (which is equal to the area of the front face of the solid), which is then given by the

integral of f over the interval [0, a].

5

Homework Solutions MATH 32B-2 (18W)

Problem 10 (16.2.10). Sketch the region D between y = x

2 and y = x(1 − x). Express D as a

simple region and calculate the integral of f(x, y) = 2y over D.

Solution.

y = x

2

y = x(1 − x)

(1/2, 1/4)

The region D is both vertically simple and horizontally simple, but the bounds for y in terms of x

are simpler than the bounds for x in terms of y, so when we use Fubini’s theorem to evaluate the

integral, we take the y-integral on the inside and the x-integral on the outside. This gives us

Z Z

D

f(x, y) dA =

Z x=1/2

x=0

Z y=x(1−x)

y=x2

2y dy dx (Fubini)

=

Z x=1/2

x=0

[y

2

]

y=x(1−x)

y=x2 dx

=

Z x=1/2

x=0

x

2

(1 − x)

2 − x

4

dx

=

Z x=1/2

x=0

x

2

(1 − 2x) dx

=

1

3

x

3 −

1

2

x

4

x=1/2

x=0

=

1

24

1

32

=

1

96

.

Step-by-step explanation:

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