When s is the open hemisphere x 2 + y 2 + z 2 = 1, z ≤ 0 , oriented by the inward normal pointing to the origin, then the bounda
ry orientation on ∂s is clockwise. true or false?
1 answer:
The figure below shows a diagram of this problem. First of all we graph the hemisphere. This one has a radius equal to 1. Given that z ≤ 0 a sphere will be valid only in the negative z-axis, that is, we will get a half of a sphere that is the hemisphere shown in the figure. We know that this hemisphere is oriented by the inward normal pointing to the origin, then we have a Differential Surface Vector called N, using the Right-hand rule <span>the boundary orientation is </span>counterclockwise.
Therefore, the answer above False
Therefore, the answer above False
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Answer:
see explanation
Step-by-step explanation:
Using the trigonometric identity
tan²x + 1 = sec²x ⇒ tan²x = sec²x - 1
Consider the left side
sec²x × tan²x + sec²x
= sec²x(sec²x - 1) + sec²x ← distribute and simplify
= x - sec²x + sec²x
= x = right side ⇒ verified
Answer:
Step-by-step explanation:
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