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.
Use Euler's method to numerically integrate
dy
dx
-2x+12x² - 20x +8.5
from x=0 to x=4 with a step size of 0.5. The initial condition at x=0 is y=1. Recall
that the exact solution is given by y = -0.5x+4x³- 10x² + 8.5x+1
Find an equation of the line tangent to the graph of f(x) = (5x-9)(x+4) at (2,6).
Find the point on the graph of the given function at which the slope of the tangent line is the given slope.
2
f(x)=8x²+4x-7; slope of the tangent line = -3
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