Lehman Calculus III Summer 2020

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