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bizoid
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Homework Statement
The question verbatim from the text is as follows:For every two-dimensional set C contained in R2 for which the integral exists, let Q(C) = ∫∫c x2 + y2 dxdy.
If C1 = {(x,y): -1 ≤ x ≤ 1, -1 ≤ y ≤ 1},
C2 = {(x,y): -1 ≤ x=y ≤ 1},
C3 = {(x,y): x2 + y2 ≤ 1},
Find Q(C1), Q(C2), Q(C3).
My post describes what I did for C2.
I used itterated integrals for C1 with an answer of 4/3, and
I switch to polar coordinates for C1 with an answer of 2pi
Homework Equations
None
The Attempt at a Solution
OK. So the domain appears to be the line x=y. Thus I will try a line integral.
Parametrize the curve x = t, thus y= t. So f(x,y) = f(t) = 2t2.
The Line integral states that I take the function * position of the curve
and we end up with
[-1,1][tex]\int[/tex] 2t2 * Sqrt( (t dx/dt)^2 + (t dy/dt)^2)
= [-1,1][tex]\int[/tex] 2t2 * Sqrt(2)
= (2Sqrt(2)*t3)/3 from [-1,1]
= 2Sqrt(2) / 3 + 2Sqrt(2) / 3
= 4Sqrt(2) / 3
My questions:
(1) I assume this is incorrect, because if C = {(x,y): -1 <= x <= 1, -1<= y <= 1} we get 4/3. And it seems unintuitive in this case that the entire area in a square region is less than the area along one path. Since we have a parabolic bowl shape.
Is this intuition wrong. Why?
(2) I understand the line integral to be the sum of the areas along each point in the curve's path. And I intuitively see how this works, if I have a non-constant curve. But in this case I have a scalar increasing the area. I see that the length of the curve is clearly 2Sqrt(2), from basic geometry, but it doesn't "feel" right. However, it is synonymous to the case of a circle cos^2 + sin^2 which will give an arc length of 1, noting that the radius is 1.
What am I missing, is it because I am parametrizing the function that changes the perspective?
(3) What is the correct answer? What other ways can this be solved.
Thank you in advance for reviewing my over analysis of a basic calc II question and any input that you provide.
Also, are there any good links to the PHP codes for all math symbols? This way my next question can be better formed.
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