The weight of an object on the moon varies directly as its weight on Earth. With all of his gear on, Neil Armstrong weighed 360
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As you can see in this diagram, the answer is A, the ten-thousands place.
I'll only look at (37) here, since
• (38) was addressed in 24438105
• (39) was addressed in 24434477
• (40) and (41) were both addressed in 24434541
In both parts, we're considering the line integral
and I assume <em>C</em> has a positive orientation in both cases
(a) It looks like the region has the curves <em>y</em> = <em>x</em> and <em>y</em> = <em>x</em> ² as its boundary***, so that the interior of <em>C</em> is the set <em>D</em> given by
• Compute the line integral directly by splitting up <em>C</em> into two component curves,
<em>C₁ </em>: <em>x</em> = <em>t</em> and <em>y</em> = <em>t</em> ² with 0 ≤ <em>t</em> ≤ 1
<em>C₂</em> : <em>x</em> = 1 - <em>t</em> and <em>y</em> = 1 - <em>t</em> with 0 ≤ <em>t</em> ≤ 1
Then
*** Obviously this interpretation is incorrect if the solution is supposed to be 3/2, so make the appropriate adjustment when you work this out for yourself.
• Compute the same integral using Green's theorem:
(b) <em>C</em> is the boundary of the region
• Compute the line integral directly, splitting up <em>C</em> into 3 components,
<em>C₁</em> : <em>x</em> = <em>t</em> and <em>y</em> = 0 with 0 ≤ <em>t</em> ≤ 1
<em>C₂</em> : <em>x</em> = 1 - <em>t</em> and <em>y</em> = <em>t</em> with 0 ≤ <em>t</em> ≤ 1
<em>C₃</em> : <em>x</em> = 0 and <em>y</em> = 1 - <em>t</em> with 0 ≤ <em>t</em> ≤ 1
Then
• Using Green's theorem: