In some cases it is possible to use Definition 15.5.1 along with symmetry considerations to evaluate a surface integral without reference to a parametrization of the surface. In these exercises, σ denotes the unit sphere centered at the origin. Use the argument in Exercise 15 to prove that if f x is a continuous odd function of x , and if g y , z is a continuous function, then ∬ σ f x g y , z d S = 0
In some cases it is possible to use Definition 15.5.1 along with symmetry considerations to evaluate a surface integral without reference to a parametrization of the surface. In these exercises, σ denotes the unit sphere centered at the origin. Use the argument in Exercise 15 to prove that if f x is a continuous odd function of x , and if g y , z is a continuous function, then ∬ σ f x g y , z d S = 0
In some cases it is possible to use Definition 15.5.1 along with symmetry considerations to evaluate a surface integral without reference to a parametrization of the surface. In these exercises,
σ
denotes the unit sphere centered at the origin.
Use the argument in Exercise 15 to prove that if
f
x
is a continuous odd function of
x
,
and if
g
y
,
z
is a continuous function, then
∬
σ
f
x
g
y
,
z
d
S
=
0
With differentiation, one of the major concepts of calculus. Integration involves the calculation of an integral, which is useful to find many quantities such as areas, volumes, and displacement.
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, calculus and related others by exploring similar questions and additional content below.
RELATIONS-DOMAIN, RANGE AND CO-DOMAIN (RELATIONS AND FUNCTIONS CBSE/ ISC MATHS); Author: Neha Agrawal Mathematically Inclined;https://www.youtube.com/watch?v=u4IQh46VoU4;License: Standard YouTube License, CC-BY