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
)
)
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.
Write the given third order linear equation as an equivalent system of first order equations with initial values.
Use
Y1 = Y, Y2 = y', and y3 = y".
-
-
√ (3t¹ + 3 − t³)y" — y" + (3t² + 3)y' + (3t — 3t¹) y = 1 − 3t²
\y(3) = 1, y′(3) = −2, y″(3) = −3
(8) - (888) -
with initial values
Y
=
If you don't get this in 3 tries, you can get a hint.
Question 2
1 pts
Let A be the value of the triple integral
SSS.
(x³ y² z) dV where D is the region
D
bounded by the planes 3z + 5y = 15, 4z — 5y = 20, x = 0, x = 1, and z = 0.
Then the value of sin(3A) is
-0.003
0.496
-0.408
-0.420
0.384
-0.162
0.367
0.364
Chapter 14 Solutions
Student Solutions Manual, Single Variable for Calculus: Early Transcendentals
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