Special case of surface integrals of scalar-valued functions Suppose that a surface S is defined as z = g ( x , y ) on a region R. Show that t x × t y = 〈– z x , – z y , 1〉 and that ∬ S f ( x , y , z ) d S = ∬ R f ( s , y , g ( x , y ) ) z x 2 + z y 2 + 1 d A .
Special case of surface integrals of scalar-valued functions Suppose that a surface S is defined as z = g ( x , y ) on a region R. Show that t x × t y = 〈– z x , – z y , 1〉 and that ∬ S f ( x , y , z ) d S = ∬ R f ( s , y , g ( x , y ) ) z x 2 + z y 2 + 1 d A .
Solution Summary: The author explains that the outward normal to a surface is z=g(x,y,hrangle) and the surface integral is
Special case of surface integrals of scalar-valued functions Suppose that a surface S is defined as z = g(x, y) on a region R. Show that tx × ty = 〈–zx, –zy, 1〉 and that
∬
S
f
(
x
,
y
,
z
)
d
S
=
∬
R
f
(
s
,
y
,
g
(
x
,
y
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)
z
x
2
+
z
y
2
+
1
d
A
.
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.
T
1
7. Fill in the blanks to write the calculus problem that would result in the following integral (do
not evaluate the interval). Draw a graph representing the problem.
So
π/2
2 2πxcosx dx
Find the volume of the solid obtained when the region under the curve
on the interval
is rotated about the
axis.
38,189
5. Draw a detailed graph to and set up, but do not evaluate, an integral for the volume of the
solid obtained by rotating the region bounded by the curve: y = cos²x_for_ |x|
≤
and the curve y
y =
about the line
x =
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2
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DG
Find f(x) and g(x) such that h(x) = (fog)(x) and g(x) = 3 - 5x.
h(x) = (3 –5x)3 – 7(3 −5x)2 + 3(3 −5x) – 1
-
-
-
f(x) = ☐
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