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. (a) Explain why ∬ σ x 2 d S = ∬ σ y 2 d S = ∬ σ z 2 d S (b) Conclude from part (a) that ∬ σ x 2 d S = 1 3 ∬ σ x 2 d S + ∬ σ y 2 d S + ∬ σ y 2 d S (c) Use part (b) to evaluate ∬ σ x 2 d S without performing an integration .
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. (a) Explain why ∬ σ x 2 d S = ∬ σ y 2 d S = ∬ σ z 2 d S (b) Conclude from part (a) that ∬ σ x 2 d S = 1 3 ∬ σ x 2 d S + ∬ σ y 2 d S + ∬ σ y 2 d S (c) Use part (b) to evaluate ∬ σ x 2 d S without performing an integration .
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.
(a) Explain why
∬
σ
x
2
d
S
=
∬
σ
y
2
d
S
=
∬
σ
z
2
d
S
(b) Conclude from part (a) that
∬
σ
x
2
d
S
=
1
3
∬
σ
x
2
d
S
+
∬
σ
y
2
d
S
+
∬
σ
y
2
d
S
(c) Use part (b) to evaluate
∬
σ
x
2
d
S
without performing an integration.
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.
Find the "Area" of the "Surface"
The part of the paraboloid (x
=
y² + z²)
that lies within the cylinder (y² + z²
=
Hint: Consider parametric representations of the surface
25)
Can someone explain why I am getting this question wrong?
Sketch the graph of the parametric surface. x= sinh u, y=v, z=2cosh u.
Chapter 15 Solutions
Calculus Early Transcendentals, Binder Ready Version
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