Integrating with respect to area? Past paper question

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The discussion centers on evaluating the double integral ∫A x³ dA over the rectangle defined by -2 ≤ x ≤ 2 and -1 ≤ y ≤ 1. The original poster attempted to solve the integral by parametrizing x and y with a new variable t, resulting in an answer of 64/5, which was incorrect. After further consideration and discussion, it was concluded that using a double integral is appropriate, and the correct evaluation of the integral yields a result of zero.

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TheGreatCabbage
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New member has been reminded to fill out the Homework Help Template in future schoolwork threads
This isn't exactly homework or coursework, it is a past paper question that I cannot find a solution to (my university doesn't like releasing answers for some reason unknown to me).
The question is attached as an image (edit: the image displays while editing but not in the post, so I'll try to type it out):

uc?export=download&id=0B2XR9EE7g17JSFA5R3B6eTQ0LXc.png

Question: Let A be the rectangle -2 ≤ x ≤ 2, -1 ≤ y ≤ 1. Evaluate ∫A x3 dA.

I attempted to solve this by parametising x and y in terms of a new variable t, such that x = 2t and y = t for -1 ≤ t ≤ 1. Then I used the fact that A = xy to obtain A = xy = 2t2 and dA = 4t dt. Substituting into the integral with limits -1 and 1 gave my a final answer of 64/5, but I lack confidence in my method and answer.
Should I use a double integral? Any help would be much appreciated.
 
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TheGreatCabbage said:
This isn't exactly homework or coursework, it is a past paper question that I cannot find a solution to (my university doesn't like releasing answers for some reason unknown to me).
The question is attached as an image (edit: the image displays while editing but not in the post, so I'll try to type it out):

uc?export=download&id=0B2XR9EE7g17JSFA5R3B6eTQ0LXc.png

Question: Let A be the rectangle -2 ≤ x ≤ 2, -1 ≤ y ≤ 1. Evaluate ∫A x3 dA.

I attempted to solve this by parametising x and y in terms of a new variable t, such that x = 2t and y = t for -1 ≤ t ≤ 1. Then I used the fact that A = xy to obtain A = xy = 2t2 and dA = 4t dt. Substituting into the integral with limits -1 and 1 gave my a final answer of 64/5, but I lack confidence in my method and answer.
Should I use a double integral? Any help would be much appreciated.
I don't see any reason NOT to use a double integral.

BTW, what would you get from this integral? ##\int_{-1}^1 x^3 dx##
 
Mark44 said:
I don't see any reason NOT to use a double integral.

BTW, what would you get from this integral? ##\int_{-1}^1 x^3 dx##

Zero :)
I think I understand it now, I tried a double integral method earlier and it gave me zero and I thought that it was wrong, but after considering it properly I think that it's right.
 

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