Integrating with respect to area? Past paper question

In summary, the conversation discusses a past paper question about evaluating an integral over a given rectangle. The question is attached as an image and the conversation includes attempts at solving it by parametrizing and using a double integral. The final answer is 64/5, and it is determined that using a double integral is appropriate. The integral of x^3 over the given limits is also discussed.
  • #1
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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  • #2
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##
 
  • #3
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.
 

1. What is the purpose of integrating with respect to area?

Integrating with respect to area is used to calculate the total area under a curve or between two curves. It is an important tool in mathematics and physics for finding the total value of a function over a given interval.

2. How is integration with respect to area different from integration with respect to x or y?

When integrating with respect to area, the integration variable is the area instead of x or y. This allows for a more general and versatile approach to calculating the area under a curve, as the boundaries of integration can be any shape or size instead of just a straight line.

3. What is the relationship between integration with respect to area and the Fundamental Theorem of Calculus?

The Fundamental Theorem of Calculus states that integration and differentiation are inverse operations. When integrating with respect to area, the result is the antiderivative of the function being integrated. This antiderivative can then be used to calculate the total area under the curve.

4. Can integration with respect to area be used to find the area of irregular shapes?

Yes, integration with respect to area can be used to find the area of irregular shapes. This is because the boundaries of integration can be any shape, allowing for the calculation of the area under a non-linear curve or between two non-linear curves.

5. What are some real-world applications of integrating with respect to area?

Integrating with respect to area has many real-world applications, such as calculating the volume of a 3D object, finding the work done by a variable force, and determining the total amount of water in a given area. It is also commonly used in physics, economics, and engineering to analyze and solve problems involving changing quantities over time.

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