Homework 8: Solutions

5.4.2: Change order and evaluate:

1 1
2
∫0 ∫y sin(x ) dx dy

The region we’re integrating over is the triangle with end-points (0, 0), (1, 0)
and (1, 1). Thus, changing the order yields:
1 x
2
∫0 ∫0 sin(x ) dy dx

Which can be evaluated to:

1 x 1
2 2 x
∫0 ∫0 sin(x ) dy dx = ∫0 [y sin(x )]0 dx
1
= ∫ x sin(x2 ) dx
0
1 1
= ∫ sin(x2 ) 2xdx
2 0
1 1
= ∫ sin(u) du
2 0
1 1 − cos(1)
= (− cos(1) + cos(0)) =
2 2

5.4.5: Change order and evaluate:

1 1 3
x
∫0 ∫√y e dx dy
√ √
x ranges from x = y to x = 1. Notice x = y is the same as saying y = x2
and x ≥ 0. Thus the region we’re integrating over is bounded by the x-axis,
the parabola y = x2 and the line x = 1. Thus integrating in the y direction
first, we would have to go from 0 to x2 .

1
 1 1 1 x2
x3 x 3
∫0 ∫√y e dx dy = ∫0 ∫0 e dy dx
1 3
= ∫ ex x2 dx
0
1 1 3
= ∫ ex 3x2 dx
3 0
1 1
= ∫ eu du
3 0
1
= (e − 1)
3

5.4.8: Show:
1 1 1 sin(x)
(1 − cos(1)) ≤ ∫ ∫ dx dy ≤ 1.
2 0 0 1 + (xy)4

Since x and y are both between 0 and 1, the denominator, 1 + (xy)4 , is at

least 1 and at most 2. i.e.
1 1 1
≤ 4
≤ .
2 1 + (xy) 1

Also, since sin(x) is positive for x in the interval [0, 1], we can multiple all
sides by sin(x). Then notice that sin(x) ≤ 1 to obtain:

sin(x) sin(x)
≤ ≤ sin(x) ≤ 1.
2 1 + (xy)4

Finally, integrating over the square [0, 1] × [0, 1] yields:

1 1 1 1 1 sin(x) 1 1
∫ ∫ sin(x) dx dy ≤ ∫ ∫ dx dy ≤ ∫ ∫ dx dy.
2 0 0 0 0 1 + (xy)4 0 0

The left hand side evaluates to

1 1 1 1 1 1
∫ ∫ sin(x) dx dy = ∫ (− cos(1)) − (− cos(0)) dy = (1 − cos(1)),
2 0 0 2 0 2

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giving us:

1 1 1 sin(x)
(1 − cos(1)) ≤ ∫ ∫ dx dy ≤ 1.
2 0 0 1 + (xy)4

5.5.12: Find the volume of the solid bounded by

x2 + 2y 2 = 2, z = 0, and x + y + 2z = 2

x2 + 2y 2 = 2 defines a vertical cylinder that crosses the xy-plane in an ellipse.

z = 0 is the xy-plane. Since the ellipse in the xy-plane given by x2 + 2y 2 = 2
does not intersect the line x+y = 2, the plane x+y +2z = 2 crosses the cylinder
above the xy-plane. Thus the region looks like:

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For fixed (x, y), the values of z range between the planes z = 0 and x+y +2z =
2, (so 0 to (2 2 2
√− x − y)/2).√For a fixed y, since x + 2y = 2, we get that x ranges
between, − 2 − 2y 2 to 2 − 2y 2 . Thus we get the volume V is given by:
√
1 2−2y 2 1− x+y
2
V =∫ ∫ √ ∫ dz dx dy
−1 − 2−2y 2 0
√
12−2y 2 y x
=∫ ∫ √ [(1 − ) − ] dx dy
−1 − 2−2y 2 2 2
1 y √
=∫ 2(1 − ) 2 − 2y 2 − 0 dy
−1 2
√ 1 √ √
= 2 ∫ 2 1 − y 2 − y 1 − y 2 dy
−1

√ 1√ √ 1 √
= 2 2∫ 1 − y 2 dy − 2∫ y 1 − y 2 dy
−1 −1
√ √
= 2π − 0 = 2π.

Where in the second last line, the first integral we identify as the area of a
semi-circle of radius 1, and the second integral is 0, since it is the integral of
an odd function over a symmetric interval.

5.5.22: Evaluate
2 2
∭W (x + y ) dx dy dz
where W is the pyramid with top vertex (0, 0, 1) and base vertices (0, 0, 0), (1, 0, 0), (0, 1, 0),
and (1, 1, 0).

The horizontal cross-section of this pyramid at height z is a square with

one corner at (0, 0, z) and side length 1 − z. Thus both x and y range from
0 to 1 − z, and the integral becomes:

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 1 1−z 1−z
2 2
∭W (x + y ) dx dy dz = ∫0 ∫0 ∫0 (x2 + y 2 ) dx dy dz

1 1−z (1 − z)3
=∫ ∫ ( + (1 − z)y 2 ) dy dz
0 0 3
1 (1 − z)4 (1 − z)4
=∫ ( + ) dz
0 3 3
2 1
= (1 − z)4 dz
3 ∫0
1
2 (1 − z)5
= [ ]
3 −5 0

2 1 2
= (0 − ) = .
3 −5 15

5.5.24: (a) Sketch the region for

1 x y
∫0 ∫0 ∫0 f (x, y, z) dz dy dx.

The region is a tetrahedron bounded by the 4 planes z = 0, z = y, y = x and

x = 1. In other words, it is the tetrahedron with its four vertices the points
(0, 0, 0), (1, 0, 0), (1, 1, 0) and (1, 1, 1):

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(b) Change order to dx dy dz.

x ranges between the planes x = y and x = 1. After x is integrated, we’re

left with an integral dy dz. We want the region in the yz-plane that we’re
integrating over. This is the projection of the tetrahedron to the yz-plane,
and we see it is a triangle with vertices (0, 0, 0), (0, 1, 0) and (0, 1, 1). Thus
we need to integrate y from y = z to y = 1. z then ranges from 0 to 1. Thus
we get that the integral is:
1 1 1
∫0 ∫z ∫y f (x, y, z) dx dy dz.

6.1.2: Determine if the functions T ∶ R3 → R3 are one-to-one and/or onto:

(a) T (x, y, z) = (2x + y + 3z, 3y − 4z, 5x)

First note that if we let A be the matrix

⎛ 2 1 3 ⎞
A = ⎜ 0 3 −4 ⎟ ,
⎝ 5 0 0 ⎠

we have (if we think of (x, y, z) as a column vector)

T (x, y, z) = A(x, y, z).

The matrix A is invertible, [since its determinant is 5 × (−4 − 9) ≠ 0], so for

any vector (a, b, c), we can solve T (x, y, z) = (a, b, c) by simply multiplying
both sides by A−1 to get (x, y, z) = A−1 (a, b, c)

Thus for every (a, b, c) we can find an (x, y, z) that is sent by T to it (hence
onto), and we can only find one such (x, y, z) (hence one-to-one). So the
function is both onto and one-to-one.

(b) T (x, y, z) = (y sin(x), z cos(y), xy)

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Both (0, 0, 0) and (1, 0, 0) are sent to the point (0, 0, 0), so the function is
not one-to-one.

Also, the point (0, 0, 1) is not in the image, since for the last coordinate
to be 1, we must have xy = 1 so both x and y are ±1. But this forces the
first coordinate to be sin(1) which is not equal to 0. Thus the function is not
onto either.

(c) T (x, y, z) = (xy, yz, xz)

Both (1, 1, 1) and (−1, −1, −1) are sent to the point (1, 1, 1), so the function
is not one-to-one.

As for onto, nothing is sent to (1, 1, 0), since xz = 0 implies either x = 0

or z = 0. Either way, one of the first two coordinates is 0. So the function is
not onto either.

(d) T (x, y, z) = (ex , ey , ez )

This function cannot be onto since e is positive, and the power of a positive
number is always (strictly) positive. Thus, for instance, nothing is sent to
(0, 0, 0) or to (−1, −3, −11).

The function is one-to-one though, as can be seen as follows. Suppose, two

′
points get sent to the same value, i.e. T (x, y, z) = T (x′ , y ′ , z ′ ). Then ex = ex ,
and taking logarithms on both sides yields, x = x′ as the only real solution.
Similarly y = y ′ and z = z ′ . Thus the points (x, y, z) and (x′ , y ′ , z ′ ) are ac-
tually the same. Since this works in general, it proves the claim. Thus T is
one-to-one but not onto.

6.1.10: Find a T that sends the parallelogram D∗ with vertices (−1, 3), (0, 0), (2, −1), (1, 2)
to the square D = [0, 1] × [0, 1].

Linear maps (those given by a matrix) send squares to parallelograms, so

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 a b
we can try to find a linear map. The matrix ( ) sends (1, 0) to its first
c d
column, and (0, 1) to its second column (as you may easily check).

2 −1
Thus the matrix A = ( ) sends (1, 0) to (2, −1) and sends (0, 1) to
−1 3
(−1, 3) (and hence sends D to D∗ ). Since we want a map from D∗ to D, we
are looking for the inverse matrix,

1 3 1
A−1 = ( ).
5 1 2

Thus T is given by,

x 1 3 1 x 1
T (x, y) = A−1 ( )= ( ) ( ) = (3x + y, x + 2y) .
y 5 1 2 y 5

6.2.1: Suggest substitution and find its Jacobian.

(a) ∬R (3x + 2y) sin(x − y) dA

Letting u = 3x + 2y and v = x − y would greatly simplify the integral. The

resulting integral would become

∂(x, y)
∬R′ u sin(v) ∣ ∂(u, v) ∣ dA

∂(x,y)
where ∣ ∂(u,v) ∣ is the absolute value of the determinant of the matrix

∂x/∂u ∂x/∂v
( )
∂y/∂u ∂y/∂v

is just a number in this case.

Moreover, the transformation (u, v) = T (x, y) = (3x + 2y, x − y) is one-to-

one and onto, as it must be in order to be a valid change of coordinates. So

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it has an inverse T −1 such that T ○ T −1 = Identity. Using chain rule, we thus
get,
∂(x, y) ∂(u, v)
= 1.
∂(u, v) ∂(x, y)

We do this since, ∣ ∂(u,v)

∂(x,y) ∣ is much easier to compute, (you may also solve for
(x, y) in terms of (u, v) and compute the Jacobian directly).

∂(u, v) 3 2
=∣ ∣ = −5.
∂(x, y) 1 −1

Thus
∂(x, y) 1
=
∂(u, v) −5

and the Jacobian is its absolute value 51 .

∬R e cos(7x − 2y) dA
(−4x+7y)
(b)

Let u = −4x + 7y and v = 7x − 2y. Then

∂(u, v) −4 7
=∣ ∣ = 8 − 49 = −41.
∂(x, y) 7 −2

Thus
∂(x, y) 1
=
∂(u, v) −41

and the Jacobian is its absolute value 1/41.

6.2.2: Suggest substitution and find its Jacobian.

3 4
(a) ∬R (5x + y) (x + 9y) dA

Let u = 5x + y and v = x + 9y. Then

∂(u, v) 5 1
=∣ ∣ = 45 − 1 = 44.
∂(x, y) 1 9

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Thus
∂(x, y) 1
=
∂(u, v) 44

is the Jacobian.

(b) ∬R x sin(6x + 7y) − 3y sin(6x + 7y) dA

Note that the integral may be written as ∬R (x − 3y) sin(6x + 7y) dA, so let
u = x − 3y and v = 6x + 7y. Then

∂(u, v) 1 −3
=∣ ∣ = 7 + 18 = 25.
∂(x, y) 6 7

Thus
∂(x, y) 1
=
∂(u, v) 25

is the Jacobian.

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