Answer:
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By Stokes' theorem,

where

is the circular boundary of the hemisphere

in the

-

plane. We can parameterize the boundary via the "standard" choice of polar coordinates, setting

where

. Then the line integral is


We can check this result by evaluating the equivalent surface integral. We have

and we can parameterize

by

so that

where

and

. Then,

as expected.
Answer:
Expression
Step-by-step explanation:
Remember, an expression is a mathematical phrase that contains numbers, variables, or both. Expressions never have an equal sign. An equation is a mathematical sentence that says two expressions are equal. My work don't copy®️
A point that bisects a segment would be its midpoint. This is a case where the vocabulary (bisect as opposed to midpoint) makes it harder.
To find the midpoint, we use the midpoint formula. The midpoint formula is:
midpoint = (x₁ + x₂/2, y₁ + y/2). To find it, you add the x coordinates and then divide them by 2. Repeat for the y coordinates.
x: (-2 + 6)/2 = 4/2 = 2
y: (5 +1)/2 = 6/2 = 3
Thus the point B bisecting AC is at (2, 3).