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A step-by-step guide to understanding and applying stokes' theorem in multivariable calculus. It includes detailed explanations, solved examples, and various problem types to help students grasp the concepts and practice their skills. Topics such as finding bounding curves, evaluating surface integrals, and applying stokes' theorem to different surfaces, including hemispheres, cubes, and triangles. It is a valuable resource for students studying vector calculus and related fields.
Typology: Exercises
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Dr. Z’s Math251 Handout #16.8 [Stokes’ Theorem]
By Doron Zeilberger
Problem Type 16.8a: Use Stokes’ Theorem to evaluate
S curl^ F^ ·^ dS, where
F(x, y, z) = P (x, y, z) i + Q(x, y, z) j + R(x, y, z) k ,
and S is some surface with a given orientation (that should boil down to either outwards or inwards).
Example Problem 16.8a: Use Stokes’ Theorem to evaluate
S curl^ F^ ·^ dS, where
F(x, y, z) = x^2 e^2 yz^ i + y^2 e^3 xz^ j + z^2 e^4 xy^ k ,
and S is the hemisphere x^2 + y^2 + z^2 = 9, z ≥ 0, oriented upwards.
Steps Example
C
F · d r.
The challenge is to find the bounding curve C.
For hemispheres x^2 + y^2 + z^2 = R^2 , z ≥ 0, like in this problem, the bounding curve is simply the circle x^2 + y^2 = R^2 , on the xy-plane z = 0 and the parametric repre- sentation is
x = R cos t , y = R sin t , z = 0.
If it is the hemisphere z ≤ 0 then it is the same but in the negative direction. If it is the hemisphere x^2 + y^2 + z^2 = R^2 , y ≤ 0, then the parametric representation is
x = R cos t , z = R sin t , y = 0 ,
etc.
If it is the part of a surface z = g(x, y) that lies above a plane z = a, oriented outwards, then C is obtained by solving g(x, y) = a and representing g(x, y) = a in parametric notation and adding to it z = a. The orientation of C is such as to obey the right-hand rule.
At the end you need to represent C in parametric form
r(t) = 〈x(t), y(t), z(t)〉 , a ≤ t ≤ b ,
for some expressions x(t), y(t), z(t) of t and some numbers a and b.
x = 3 cos t , y = 3 sin t , z = 0.
So
r(t) = 3 cos t i + 3 sin t j + 0 k ,
0 ≤ t ≤ 2 π.
− 2 ≤ x, y ≤ 2 , z = − 2.
curl F = (z^2 − y^2 ) i + 0 j − 2 x^2 y k
(You do it!)
{(x, y) | − A ≤ x ≤ A, −A ≤ y ≤ A }.
Do the integration
curl F = (4 − y^2 ) i + 0 j − 2 x^2 y k.
So
curl F·dS = ((4−y^2 ) i+0 j− 2 x^2 y k)·k = − 2 x^2 y.
Finally, ∫ ∫
S
curl F · dS =
− 2
− 2
− 2 x^2 y dx dy
− 2
− 2
− 2 x^2 y dx
dy =
− 2
(− 2 y)
x^3 3
2 − 2
dy
− 2
(− 2 y)^163 dy
− 2
y dy =
y^2 2
2 − 2 =^
Ans.: 0.
Problem Type 16.8c: Use Stokes’ Theorem to evaluate
C F^ ·^ d^ r, where
F(x, y, z) = P (x, y, z) i + Q(x, y, z) j + R(x, y, z) k ,
and C is a curve that bounds some surface (that you have to figure out!), z = g(x, y), above the region {(x, y)|(x, y) ∈ D } (that you have to find!). C is oriented counterclockwise as viewed from above.
Example Problem 16.8c: Use Stokes’ Theorem to evaluate
C F^ ·^ d^ r, where
F(x, y, z) = (2x + y^2 ) i + (2y + z^2 ) j + (3z + x^2 ) k ,
and C is the triangle with vertices (2, 0 , 0),(0, 2 , 0), (0, 0 , 2), and is oriented counterclockwise as viewed from above.
Steps Example
D = {(x, y) | 0 ≤ x ≤ 2 , 0 ≤ y ≤ 2 −x }.
curl F = − 2 z i − 2 x j − 2 y k.
(You do it!)
