Lateral Surface Area In Exercises 65-72, find the area of the lateral surface (see figure) over the curve C in the xy-plane and under the surface z = f ( x , y ) where Lateral surface area = ∫ C f ( x , y ) d s . f ( x , y ) = x y , C: x 2 + y 2 = 1 line from (1, 0) to (0, 1)
Lateral Surface Area In Exercises 65-72, find the area of the lateral surface (see figure) over the curve C in the xy-plane and under the surface z = f ( x , y ) where Lateral surface area = ∫ C f ( x , y ) d s . f ( x , y ) = x y , C: x 2 + y 2 = 1 line from (1, 0) to (0, 1)
Solution Summary: The author calculates the lateral surface area over the curve C and under the given surface. The parametrization form is r(t)=mathrm
Lateral Surface Area In Exercises 65-72, find the area of the lateral surface (see figure) over the curve C in the xy-plane and under the surface
z
=
f
(
x
,
y
)
where Lateral surface
area
=
∫
C
f
(
x
,
y
)
d
s
.
f
(
x
,
y
)
=
x
y
,
C:
x
2
+
y
2
=
1
line from (1, 0) to (0, 1)
Check that the point (−1,−1,1) lies on the given surface. Then, viewing the surface as a level surface for a function f(x,y,z) find a vector normal to the surface and an equation for the tangent plane to the surface at (−1,−1,1)
x^2−3y^2+z^2=−1
Sketch the surface of f(x,y)
R(a, b) = (2b + cosa, 2a + sin b, ab)
Determine the equations for the (a) tangent plane and (b) the normal line to the surface S at the point
(-1,2TT, 0)
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