1. The document discusses vector calculus concepts including the dot product, cross product, gradient, divergence, and curl of vectors. It provides examples and explanations of how to calculate these quantities. 2. The dot product yields a scalar quantity and is used to calculate the angle between two vectors. The cross product yields a vector that is perpendicular to the plane of the two input vectors. 3. The gradient is the vector of partial derivatives of a scalar function and points in the direction of maximum increase. The divergence and curl are used to describe vector fields and involve taking partial derivatives.

1. Calculus III
Summer 2020
Lesson 3
Problem Set 4
“Mathematics is the art of giving
the same name to different things.”
- Henri Poincaré -

2. Lehman College, Department of Mathematics
The Dot Product of Two Vectors (1 of 4)
Let 𝐫 be the vector given by:
The vector 𝐫 may also be the written in the form:
By the Pythagorean theorem, we know that:
Similar to the plane above, in space we have:
The Pythagorean theorem yields:
We will use the form above to construct the dot product
of two vectors 𝐮 and 𝐯.
𝐫 𝑥, 𝑦 = 𝑥 𝐢 + 𝑦 𝐣
𝑟2 = 𝑥2 + 𝑦2 + 𝑧2
𝐫 𝑥, 𝑦 = 𝑥, 𝑦
𝐫 𝑥, 𝑦, 𝑧 = 𝑥, 𝑦, 𝑧
𝑟2 = 𝑥2 + 𝑦2

3. Lehman College, Department of Mathematics
The Dot Product of Two Vectors (2 of 4)
Let 𝐮 and 𝐯 be two vectors in space given by:
The dot product of 𝐮 and 𝐯 is defined as the quantity:
The dot product or two vectors yields a scalar quantity.
Similar to the radius vector earlier, we define the
magnitude 𝐯 of a vector 𝐯 through the scalar quantity:
Example 1. Find the dot product of the vectors 𝐮 and 𝐯:
𝐮 = 𝑢1, 𝑢2, 𝑢3 𝐯 = 𝑣1, 𝑣2, 𝑣3and
𝐮 ⋅ 𝐯 = 𝑢1 𝑣1 + 𝑢2 𝑣2 + 𝑢3 𝑣3
𝐯 2
= 𝑣1
2
+ 𝑣2
2
+ 𝑣3
2
= 𝐯 ⋅ 𝐯
𝐮 = 1, 0, 3 𝐯 = −2, 1, 2and

4. Lehman College, Department of Mathematics
The Dot Product of Two Vectors (3 of 4)
Solution. The dot product of the vectors 𝐮 and 𝐯 for:
is given by:
Consider the following vector diagram:
𝐮 = 1, 0, 3 𝐯 = −2, 1, 2and
𝐮 ⋅ 𝐯 = 1 −2 + 0 1 + 3 2 = −2 + 0 + 6 = 4
Here 𝜃 is the angle between
the vectors 𝐮 and 𝐯.
By the Cosine Rule:
𝐯 − 𝐮 2
= 𝐮 2
+ 𝐯 2
− 2 𝐮 𝐯 cos 𝜃

5. Lehman College, Department of Mathematics
The Dot Product of Two Vectors (4 of 4)
However, we can also write 𝐯 − 𝐮 2
as a dot product:
Comparing this to the result obtained earlier:
We conclude that:
Two vectors 𝐮 and 𝐯 are orthogonal if and only if:
𝐯 − 𝐮 2 = 𝐯 − 𝐮 ⋅ 𝐯 − 𝐮
= 𝐯 ⋅ 𝐯 − 𝐯 ⋅ 𝐮 − 𝐮 ⋅ 𝐯 + 𝐮 ⋅ 𝐮
= 𝐯 2 − 2𝐮 ⋅ 𝐯 + 𝐮 2
𝐯 − 𝐮 2
= 𝐮 2
+ 𝐯 2
− 2 𝐮 𝐯 cos 𝜃
𝐮 ⋅ 𝐯 = 𝐮 𝐯 cos 𝜃
𝐮 ⋅ 𝐯 = 0

6. Lehman College, Department of Mathematics
The Cross Product of Two Vectors (1 of 5)
Let 𝐮 and 𝐯 be two vectors in space given by:
This is equivalent to:
The cross product of 𝐮 and 𝐯 is defined as the quantity:
Above, we use the determinant of a 3 × 3 matrix.
The cross product or two vectors yields a vector.
𝐮 = 𝑢1, 𝑢2, 𝑢3 𝐯 = 𝑣1, 𝑣2, 𝑣3and
𝐮 × 𝐯 =
𝐢 𝐣 𝐤
𝑢1 𝑢2 𝑢3
𝑣1 𝑣2 𝑣3
𝐮 = 𝑢1 𝐢 + 𝑢2 𝐣 + 𝑢3 𝐤 and 𝐯 = 𝑣1 𝐢 + 𝑣2 𝐣 + 𝑣3 𝐤

7. Lehman College, Department of Mathematics
The Cross Product of Two Vectors (2 of 5)
We can show that the magnitude of the cross product:
where 𝜃 is the angle between the vectors 𝐮 and 𝐯.
Two vectors 𝐮 and 𝐯 are parallel if and only if:
Example 2. Given the two vectors 𝐮 = 𝐢 − 2 𝐣 + 𝐤 and
𝐯 = 3𝐢 + 𝐣 − 2𝐤, find the following:
𝐮 = 𝑢1, 𝑢2, 𝑢3 𝐯 = 𝑣1, 𝑣2, 𝑣3and
𝐮 × 𝐯 = 𝐮 𝐯 sin 𝜃
𝐮 × 𝐯 = 0
𝐮 × 𝐯 𝐯 × 𝐮(a) (b)

8. Lehman College, Department of Mathematics
The Cross Product of Two Vectors (3 of 5)
We know that the magnitude of the cross product is:
𝐮 × 𝐯 = 𝐮 𝐯 sin 𝜃
The vector 𝐮 × 𝐯 is
perpendicular to the plane
determined by the vectors 𝐮
and 𝐯.
How about the direction of the
vector 𝐮 × 𝐯?
where 𝜃 is the angle between the vectors
𝐮 and 𝐯.
We use the right-hand rule.

9. Lehman College, Department of Mathematics
The Cross Product of Two Vectors (4 of 5)
Example 2. Given the two vectors 𝐮 = 𝐢 − 2 𝐣 + 𝐤 and
𝐯 = 3𝐢 + 𝐣 − 2𝐤, find the following:
Solution. By the definition of the cross product:
𝐮 × 𝐯 𝐯 × 𝐮(a) (b)
𝐮 × 𝐯 =
𝐢 𝐣 𝐤
1 −2 1
3 1 −2
1 1
3 −2
𝐣 +=
−2 1
1 −2
𝐢 −
1 −2
3 1
𝐤
= 3𝐢 − (−5) 𝐣 + 7 𝐤
= 3𝐢 + 5 𝐣 + 7 𝐤

10. Lehman College, Department of Mathematics
The Cross Product of Two Vectors (5 of 5)
Solution (cont’d). From the previous slide:
This is an important property of the cross product. For
any two vectors 𝐮 and 𝐯:
𝐯 × 𝐮 =
𝐢 𝐣 𝐤
3 1 −2
1 −2 1
3 −2
1 1
𝐣 +=
1 −2
−2 1
𝐢 −
3 1
1 −2
𝐤
= −3𝐢 − 5 𝐣 + (−7) 𝐤
= −3𝐢 − 5 𝐣 − 7 𝐤
= −(𝐮 × 𝐯)
𝐯 × 𝐮 = −(𝐮 × 𝐯)

11. Lehman College, Department of Mathematics
Divergence, Gradient and Curl (1 of 5)
Let 𝑓 be a scalar function of 𝑥, 𝑦, 𝑧 with continuous
partial derivatives:
Then the gradient of 𝑓 is the vector-valued function:
Example 3. Given 𝑓 𝑥, 𝑦, 𝑧 = 𝑥3
+ 2𝑥𝑦2
+ 3𝑦𝑧2
, find ∇𝑓.
Solution. Determine the partial derivatives:
𝑓𝑥(𝑥, 𝑦, 𝑧) and𝑓𝑦(𝑥, 𝑦, 𝑧) 𝑓𝑧(𝑥, 𝑦, 𝑧)
∇𝑓 = 𝑓𝑥 𝑥, 𝑦, 𝑧 𝐢 + 𝑓𝑦 𝑥, 𝑦, 𝑧 𝐣 + 𝑓𝑧 𝑥, 𝑦, 𝑧 𝐤
𝑓𝑥 𝑥, 𝑦, 𝑧 = 3𝑥2
+ 2𝑦2 𝑓𝑦 𝑥, 𝑦, 𝑧 = 4𝑥𝑦 + 3𝑧2
𝑓𝑧 𝑥, 𝑦, 𝑧 = 6𝑦𝑧
∇𝑓 = 3𝑥2 + 2𝑦2 𝐢 + 4𝑥𝑦 + 3𝑧2 𝐣 + 6𝑦𝑧 𝐤

12. Lehman College, Department of Mathematics
Divergence, Gradient and Curl (2 of 5)
Let 𝐅(𝑥, 𝑦, 𝑧) be a vector field given by:
for functions 𝑀, 𝑁 and 𝑃 of 𝑥, 𝑦, 𝑧 with continuous
partial derivatives in an open sphere.
Then the divergence of 𝐅 is the scalar function:
Example 3. Given 𝐅 𝑥, 𝑦, 𝑧 = 𝑥3
𝑦2
𝑧 𝐢 + 𝑥2
𝑧 𝐣 + 𝑥2
𝑦 𝐤,
find the divergence of 𝐅 at the point 2, 1, −1 .
Solution. First determine the functions 𝑀, 𝑁 and 𝑃.
𝐅 𝑥, 𝑦, 𝑧 = 𝑀 𝐢 + 𝑁 𝐣 + 𝑃 𝐤
div 𝐅 𝑥, 𝑦, 𝑧 =
𝜕𝑀
𝜕𝑥
+
𝜕𝑁
𝜕𝑦
+
𝜕𝑃
𝜕𝑧

13. Lehman College, Department of Mathematics
Divergence, Gradient and Curl (3 of 5)
Solution. Since 𝐅 𝑥, 𝑦, 𝑧 = 𝑥3
𝑦2
𝑧 𝐢 + 𝑥2
𝑦 𝐣 + 𝑥𝑧2
𝐤, then
From the definition of divergence:
Therefore, we find the partial derivatives:
At the point 2, 1, −1 :
div 𝐅 𝑥, 𝑦, 𝑧 =
𝜕𝑀
𝜕𝑥
+
𝜕𝑁
𝜕𝑦
+
𝜕𝑃
𝜕𝑧
𝑀 𝑥, 𝑦, 𝑧 = 𝑥3 𝑦2 𝑧, 𝑁 𝑥, 𝑦, 𝑧 = 𝑥2 𝑦, 𝑃 𝑥, 𝑦, 𝑧 = 𝑥𝑧2
div 𝐅 𝑥, 𝑦, 𝑧 =
𝜕
𝜕𝑥
𝑥3
𝑦2
𝑧 +
𝜕
𝜕𝑦
𝑥2
𝑦 +
𝜕
𝜕𝑧
𝑥𝑧2
= 3𝑥2
𝑦2
𝑧 + 𝑥2
+ 2𝑥𝑧
div 𝐅 2, 1, −1 = 3 2 2
1 2
−1 + 2 2
+ 2 2 −1 =

14. Lehman College, Department of Mathematics
Divergence, Gradient and Curl (4 of 5)
We see that if 𝐅 𝑥, 𝑦, 𝑧 = 𝑀 𝐢 + 𝑁 𝐣 + 𝑃 𝐤, then
So we can write divergence as a symbolic dot product:
Where ∇ is the vector operator:
div 𝐅 𝑥, 𝑦, 𝑧 =
𝜕𝑀
𝜕𝑥
+
𝜕𝑁
𝜕𝑦
+
𝜕𝑃
𝜕𝑧
div 𝐅 𝑥, 𝑦, 𝑧 = ∇ ⋅ 𝐅
∇ =
𝜕
𝜕𝑥
,
𝜕
𝜕𝑦
,
𝜕
𝜕𝑧

15. Lehman College, Department of Mathematics
Divergence, Gradient and Curl (5 of 5)
Let 𝐅(𝑥, 𝑦, 𝑧) be a vector field given by:
for functions 𝑀, 𝑁 and 𝑃 of 𝑥, 𝑦, 𝑧 with continuous
partial derivatives in an open sphere.
Then the curl of 𝐅 is the vector-valued function:
From the definition of ∇, we see that (symbolically):
𝐅 𝑥, 𝑦, 𝑧 = 𝑀 𝐢 + 𝑁 𝐣 + 𝑃 𝐤
curl 𝐅 =
𝐢 𝐣 𝐤
𝜕
𝜕𝑥
𝜕
𝜕𝑦
𝜕
𝜕𝑧
𝑀 𝑁 𝑃
curl 𝐅 = ∇ × 𝐅

16. Lehman College, Department of Mathematics
Question 1 (1 of 7)
Consider the vector field
(b) Prove that 𝐅 is conservative and find a potential
function for 𝐅.
(c) Using the Fundamental Theorem of Line Integrals,
evaluate the line integral:
where 𝐶 is any smooth curve from (1, 2) to (3, 4).
Solution. A vector field 𝐅 is conservative if and only if:
𝐅(𝑥, 𝑦) =
1
1 + 𝑥𝑦
𝑦 𝐢 + 𝑥 𝐣
𝐶
𝑦 𝑑𝑥 + 𝑥 𝑑𝑦
1 + 𝑥𝑦
curl 𝐅 = 0

17. Lehman College, Department of Mathematics
Question 1 (2 of 7)
Solution (cont’d). To find curl 𝐅. If 𝐅 is of the form:
for functions 𝑀, 𝑁 and 𝑃 of 𝑥, 𝑦, 𝑧 with continuous
partial derivatives in an open sphere. Then:
If 𝑀 and 𝑁 are not dependent on 𝑧 and 𝑃 = 0, then the
above condition is equivalent to:
curl 𝐅 =
𝐢 𝐣 𝐤
𝜕
𝜕𝑥
𝜕
𝜕𝑦
𝜕
𝜕𝑧
𝑀 𝑁 𝑃
𝐅 𝑥, 𝑦, 𝑧 = 𝑀 𝐢 + 𝑁 𝐣 + 𝑃 𝐤
𝜕𝑁
𝜕𝑥
=
𝜕𝑀
𝜕𝑦

18. Lehman College, Department of Mathematics
Question 1 (3 of 7)
Solution (cont’d). For:
the functions 𝑀, 𝑁 are given by:
Therefore:
𝑀 =
𝑦
1 + 𝑥𝑦
𝐅(𝑥, 𝑦) =
1
1 + 𝑥𝑦
𝑦 𝐢 + 𝑥 𝐣
𝜕𝑁
𝜕𝑥
=
𝑁 =
𝑥
1 + 𝑥𝑦
and
𝜕
𝜕𝑥
𝑥 1 + 𝑥𝑦 −
1
2
= 1 ⋅ 1 + 𝑥𝑦 −
1
2 −
1
2
𝑥𝑦 1 + 𝑥𝑦 −
3
2
2 + 𝑥𝑦
2 1 + 𝑥𝑦
3
2
=
2 1 + 𝑥𝑦 − 𝑥𝑦
2 1 + 𝑥𝑦
3
2
=

19. Lehman College, Department of Mathematics
Question 1 (4 of 7)
Solution (cont’d). Similarly:
It follows that 𝐅 is a conservative vector field.
(c) To find the potential function 𝑓, we know that:
Therefore:
and:
𝜕𝑀
𝜕𝑦
=
2 + 𝑥𝑦
2 1 + 𝑥𝑦
3
2
𝐅 = ∇𝑓 = 𝑓𝑥 𝐢 + 𝑓𝑦 𝐣
𝑓𝑥 𝑥, 𝑦 = 𝑀 𝑥, 𝑦 =
𝑦
1 + 𝑥𝑦
𝑓𝑦 𝑥, 𝑦 = 𝑁 𝑥, 𝑦 =
𝑥
1 + 𝑥𝑦

20. Lehman College, Department of Mathematics
Question 1 (5 of 7)
Solution (cont’d). It follows that:
Let 𝑢 𝑥, 𝑦 = 1 + 𝑥𝑦, so
𝜕𝑢
𝜕𝑥
= 𝑦, and:
where 𝐾 is a constant.
𝑓(𝑥, 𝑦) =
2 1 + 𝑥𝑦 + 𝐾
𝑦
1 + 𝑥𝑦
𝑑𝑥
= 𝑦 1 + 𝑥𝑦 −
1
2 𝑑𝑥
𝑓(𝑥, 𝑦) =
𝑓𝑥 𝑥, 𝑦 𝑑𝑥 =

21. Lehman College, Department of Mathematics
Question 1 (6 of 7)
Fundamental Theorem of Line Integrals:
Let 𝐶 be a piecewise smooth curve lying in an open
region 𝑅 and given by:
If 𝐅 𝑥, 𝑦 = 𝑀 𝐢 + 𝑁 𝐣 is conservative in 𝑅, with 𝑀 and 𝑁
continuous in 𝑅, then:
where 𝑓 is a potential function of 𝐅 in 𝑅:
𝐫 𝑡 = 𝑥 𝑡 𝐢 + 𝑦 𝑡 𝐣 for 𝑎 ≤ 𝑡 ≤ 𝑏
𝐶
𝐅 ⋅ 𝑑𝐫 =
𝐶
𝛻𝑓 ⋅ 𝑑𝐫
= 𝑓 𝑥 𝑏 , 𝑦 𝑏 − 𝑓 𝑥 𝑎 , 𝑦 𝑎
𝐅 = 𝛻𝑓

22. Lehman College, Department of Mathematics
Question 1 (7 of 7)
(c) Using the Fundamental Theorem of Line Integrals,
evaluate the line integral:
where 𝐶 is any smooth curve from (1, 2) to (3, 4).
We see that the integrand represents 𝛻𝑓 ⋅ 𝑑𝐫 for:
By the Fundamental Theorem of Line Integrals:
𝐶
𝑦 𝑑𝑥 + 𝑥 𝑑𝑦
1 + 𝑥𝑦
2 1 + 𝑥𝑦𝑓(𝑥, 𝑦) =
𝐶
𝑦 𝑑𝑥 + 𝑥 𝑑𝑦
1 + 𝑥𝑦
= 𝑓 3, 4 − 𝑓 1, 2 = 2 13 − 3

23. Lehman College, Department of Mathematics
Fund. Theorem of Line Integrals (1 of 6)
Example 3 (Lar., p. 1085). Using the Fundamental
Theorem of Line Integrals, evaluate the line integral:
where 𝐶 is a smooth curve from (1, 1, 0) to (0, 2, 3).
Solution. Note that the line integral represents:
for the vector field 𝐅(𝑥, 𝑦, 𝑧) given by:
We will first establish that 𝐅 is conservative and find a
potential function 𝑓 for 𝐅.
𝐶
2𝑥𝑦 𝑑𝑥 + 𝑥2 + 𝑧2 𝑑𝑦 + 2𝑦𝑧 𝑑𝑧
𝐶
𝐅 ⋅ 𝑑𝐫
𝐅 𝑥, 𝑦, 𝑧 = 2𝑥𝑦 𝐢 + 𝑥2 + 𝑧2 𝐣 + 2𝑦z 𝐤

24. Lehman College, Department of Mathematics
Fund. Theorem of Line Integrals (2 of 6)
Solution. 𝐅 is conservative if and only if:
where:
with 𝑀 = 2𝑥𝑦, 𝑁 = 𝑥2
+ 𝑧2
, and 𝑃 = 2𝑦𝑧, since:
Now, curl 𝐅 is given by:
so, we need to evaluate the partial derivatives.
𝐅 𝑥, 𝑦, 𝑧 = 2𝑥𝑦 𝐢 + 𝑥2 + 𝑧2 𝐣 + 2𝑦z 𝐤
curl 𝐅 = 0
curl 𝐅 =
𝐢 𝐣 𝐤
𝜕
𝜕𝑥
𝜕
𝜕𝑦
𝜕
𝜕𝑧
𝑀 𝑁 𝑃
curl 𝐅 =
𝜕𝑃
𝜕𝑦
−
𝜕𝑁
𝜕𝑧
𝐢 −
𝜕𝑃
𝜕𝑥
−
𝜕𝑀
𝜕𝑧
𝐣 +
𝜕𝑁
𝜕𝑥
−
𝜕𝑀
𝜕𝑦
𝐤

25. Lehman College, Department of Mathematics
Fund. Theorem of Line Integrals (3 of 6)
Solution (cont’d). curl 𝐅 is given by:
where 𝑀 = 2𝑥𝑦, 𝑁 = 𝑥2
+ 𝑧2
, and 𝑃 = 2𝑦𝑧.
curl 𝐅 =
𝜕𝑃
𝜕𝑦
−
𝜕𝑁
𝜕𝑧
𝐢 −
𝜕𝑃
𝜕𝑥
−
𝜕𝑀
𝜕𝑧
𝐣 +
𝜕𝑁
𝜕𝑥
−
𝜕𝑀
𝜕𝑦
𝐤
𝜕𝑃
𝜕𝑦
=
𝜕
𝜕𝑦
2𝑦𝑧 = 2𝑧
𝜕𝑁
𝜕𝑧
=
𝜕
𝜕𝑧
𝑥2 + 𝑧2 = 2𝑧
𝜕𝑃
𝜕𝑥
=
𝜕
𝜕𝑥
2𝑦𝑧 = 0
𝜕𝑀
𝜕𝑧
=
𝜕
𝜕𝑧
2𝑥𝑦 = 0
𝜕𝑁
𝜕𝑥
=
𝜕
𝜕𝑥
𝑥2 + 𝑧2 = 2𝑥
𝜕𝑀
𝜕𝑦
=
𝜕
𝜕𝑦
2𝑥𝑦 = 2𝑥

26. Lehman College, Department of Mathematics
Fund. Theorem of Line Integrals (4 of 6)
Solution (cont’d). From the previous slide, we note:
It follows that:
Since curl 𝐅 is the zero vector, then 𝐅 is a conservative
vector field.
Hence, 𝐅 has a potential function 𝑓, such that:
It follows that:
curl 𝐅 = 0
𝜕𝑃
𝜕𝑦
=
𝜕𝑁
𝜕𝑧
= 2𝑧
𝜕𝑃
𝜕𝑥
=
𝜕𝑀
𝜕𝑧
= 0
𝜕𝑁
𝜕𝑥
=
𝜕𝑀
𝜕𝑦
= 2𝑥
𝐅 = ∇𝑓
𝑓𝑥 𝑥, 𝑦, 𝑧 = 𝑀 𝑥, 𝑦, 𝑧 = 2𝑥𝑦
𝑓𝑦 𝑥, 𝑦, 𝑧 = 𝑁 𝑥, 𝑦, 𝑧 = 𝑥2
+ 𝑧2
𝑓𝑧 𝑥, 𝑦, 𝑧 = 𝑃 𝑥, 𝑦, 𝑧 = 2𝑦𝑧

27. Lehman College, Department of Mathematics
Fund. Theorem of Line Integrals (5 of 6)
Solution (cont’d). Therefore:
It follows that:
Back to evaluating the line integral:
𝑓(𝑥, 𝑦, 𝑧) = 𝑓𝑥 𝑥, 𝑦, 𝑧 𝑑𝑥 = 2𝑥𝑦 𝑑𝑥 = 𝑥2 𝑦 + 𝐾1(𝑦, 𝑧)
𝑓(𝑥, 𝑦, 𝑧) = 𝑓𝑦 𝑥, 𝑦, 𝑧 𝑑𝑦 = 𝑥2 + 𝑧2 𝑑𝑦
= 𝑥2 𝑦 + 𝑦𝑧2 + 𝐾2(𝑥, 𝑧)
𝑓(𝑥, 𝑦, 𝑧) = 𝑓𝑧 𝑥, 𝑦, 𝑧 𝑑𝑧 = 2𝑦𝑧 𝑑𝑧 = 𝑦𝑧2 + 𝐾3(𝑥, 𝑦)
𝑓(𝑥, 𝑦, 𝑧) = 𝑥2 𝑦 + 𝑦𝑧2 + 𝐾
𝐶
2𝑥𝑦 𝑑𝑥 + 𝑥2
+ 𝑧2
𝑑𝑦 + 2𝑦𝑧 𝑑𝑧

28. Lehman College, Department of Mathematics
Fund. Theorem of Line Integrals (6 of 6)
Solution (cont’d). Recall that 𝐶 is a smooth curve from
(1, 1, 0) to (0, 2, 3), and that the potential function is:
By the Fundamental Theorem of Line Integrals:
= 𝑓 0, 2, 3 − 𝑓 1, 1, 0
𝐶
2𝑥𝑦 𝑑𝑥 + 𝑥2 + 𝑧2 𝑑𝑦 + 2𝑦𝑧 𝑑𝑧
𝑓(𝑥, 𝑦, 𝑧) = 𝑥2 𝑦 + 𝑦𝑧2 + 𝐾
= 18 − 1 = 17

29. Lehman College, Department of Mathematics
Question 2 (1 of 3)
Consider the vector field
(a) Find curl 𝐅 and div 𝐅.
(b) Prove that 𝐅 is not conservative.
Solution. (a) curl 𝐅 is given by:
where 𝑀 = 𝑥2
, 𝑁 = 𝑥𝑦2
, 𝑃 = 𝑥2
𝑧 are functions of 𝑥, 𝑦, 𝑧.
𝐅 𝑥, 𝑦, 𝑧 = 𝑥2 𝐢 + 𝑥𝑦2 𝐣 + 𝑥2 𝑧 𝐤
curl 𝐅 =
𝐢 𝐣 𝐤
𝜕
𝜕𝑥
𝜕
𝜕𝑦
𝜕
𝜕𝑧
𝑀 𝑁 𝑃

30. Lehman College, Department of Mathematics
Question 2 (2 of 3)
Solution (cont’d). curl 𝐅 is given by:
where 𝑀 = 𝑥2
, 𝑁 = 𝑥𝑦2
, 𝑃 = 𝑥2
𝑧 are functions of 𝑥, 𝑦, 𝑧.
curl 𝐅 =
𝜕𝑃
𝜕𝑦
−
𝜕𝑁
𝜕𝑧
𝐢 −
𝜕𝑃
𝜕𝑥
−
𝜕𝑀
𝜕𝑧
𝐣 +
𝜕𝑁
𝜕𝑥
−
𝜕𝑀
𝜕𝑦
𝐤
𝜕𝑃
𝜕𝑦
=
𝜕
𝜕𝑦
𝑥2
𝑧 = 0
𝜕𝑁
𝜕𝑧
=
𝜕
𝜕𝑧
𝑥𝑦2
= 0
𝜕𝑃
𝜕𝑥
=
𝜕
𝜕𝑥
𝑥2
𝑧 = 2𝑥𝑧
𝜕𝑀
𝜕𝑧
=
𝜕
𝜕𝑧
𝑥2
= 0
𝜕𝑁
𝜕𝑥
=
𝜕
𝜕𝑥
𝑥𝑦2 = 𝑦2 𝜕𝑀
𝜕𝑦
=
𝜕
𝜕𝑦
𝑥2
= 0

31. Lehman College, Department of Mathematics
Question 2 (3 of 3)
Solution (cont’d). From the previous slide, curl 𝐅 is:
To find div 𝐅, we know that it is defined as:
(b) Since curl 𝐅 is not the zero vector, then 𝐅 is not a
conservative vector field.
curl 𝐅 = −2𝑥𝑧 𝐣 + 𝑦2 𝐤
div 𝐅 = ∇ ⋅ 𝐅 =
𝜕
𝜕𝑥
,
𝜕
𝜕𝑦
,
𝜕
𝜕𝑧
⋅ 𝑀, 𝑁, 𝑃
=
𝜕𝑀
𝜕𝑥
+
𝜕𝑁
𝜕𝑦
+
𝜕𝑃
𝜕𝑧
=
𝜕
𝜕𝑥
𝑥2 +
𝜕
𝜕𝑦
𝑥𝑦2 +
𝜕
𝜕𝑧
𝑥2 𝑧
= 2𝑥 + 2𝑥𝑦 + 𝑥2
= 𝑥 1 + 𝑥 + 2𝑦

32. Lehman College, Department of Mathematics
Question 3 (1 of 2)
Find the work done by the force field:
on a particle moving along the path 𝐶 given by:
Solution. The work 𝑊 done by the force is given by:
where:
𝑊 =
𝐶
𝐅 ⋅ 𝑑𝐫 =
𝐅 𝑥, 𝑦 = 𝑥 𝐢 + 𝑦 𝐣 − 5z 𝐤
𝐫 𝑡 = 2 cos 𝑡 𝐢 + 2 sin 𝑡 𝐣 + 𝑡 𝐤
0
2𝜋
𝐅(𝑟 𝑡 ) ⋅ 𝐫′ t 𝑑𝑡
𝐫 𝑡 = 2 cos 𝑡 𝐢 + 2 sin 𝑡 𝐣 + 𝑡 𝐤
𝐫′ 𝑡 = −2 sin 𝑡 𝐢 + 2 cos 𝑡 𝐣 + 𝐤
𝐅 𝑟(𝑡) = 2 cos 𝑡 𝐢 + 2 sin 𝑡 𝐣 − 5t 𝐤

33. Lehman College, Department of Mathematics
Question 3 (2 of 2)
Solution. The work 𝑊 done by the force is given by:
where:
So:
𝑊 =
0
2𝜋
𝐅(𝑟 𝑡 ) ⋅ 𝐫′
t 𝑑𝑡
𝐫 𝑡 = 2 cos 𝑡 𝐢 + 2 sin 𝑡 𝐣 + 𝑡 𝐤
𝐫′ 𝑡 = −2 sin 𝑡 𝐢 + 2 cos 𝑡 𝐣 + 𝐤
𝐅 𝑟(𝑡) = 2 cos 𝑡 𝐢 + 2 sin 𝑡 𝐣 − 5t 𝐤
𝑊 =
0
2𝜋
−4 sin 𝑡 cos 𝑡 + 4 sin 𝑡 cos 𝑡 − 5𝑡 𝑑𝑡
=
0
2𝜋
−5𝑡 𝑑𝑡 = −5
𝑡2
2 0
2𝜋
= −10𝜋2

34. Lehman College, Department of Mathematics
Question 4 (1 of 3)
Use Green’s theorem to evaluate the integral:
where 𝐶 is the boundary, oriented clockwise of the
region lying inside the semicircle 𝑦 = 25 − 𝑥2 and
outside the semicircle 𝑦 = 9 − 𝑥2.
Solution. By Green’s Theorem:
Here 𝑀(𝑥, 𝑦) = 𝑦 − 𝑥 and 𝑁(𝑥, 𝑦) = 2𝑥 − 𝑦. The region
𝐷 is bounded by the piecewise-smooth curve 𝐶
𝐶
𝑦 − 𝑥 𝑑𝑥 + 2𝑥 − 𝑦 𝑑𝑦
𝐶
𝑀𝑑𝑥 + 𝑁𝑑𝑦 =
𝐷
𝜕𝑁
𝜕𝑥
−
𝜕𝑀
𝜕𝑦
𝑑𝐴

35. Lehman College, Department of Mathematics
Question 4 (2 of 3)

36. Lehman College, Department of Mathematics
Question 4 (3 of 3)
Solution (cont’d). For the functions 𝑀(𝑥, 𝑦) = 𝑦 − 𝑥,
and 𝑁(𝑥, 𝑦) = 2𝑥 − 𝑦, we have:
It follows that:
The latter is the area of the region:
𝐷
2 − 1 𝑑𝐴
𝜕𝑁
𝜕𝑥
= 2 and
𝜕𝑀
𝜕𝑦
= 1
𝐶
𝑦 − 𝑥 𝑑𝑥 + 2𝑥 − 𝑦 𝑑𝑦 =
=
𝐷
𝑑𝐴
1
2
𝜋 25 − 9 = 8𝜋