Double Integrals: Solving Homework in 1st Quadrant

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Homework Help Overview

The problem involves evaluating a double integral of the function y over a region in the first quadrant, specifically bounded by the curves x = y² and x = 8 - y². The discussion centers around the setup and evaluation of this integral.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss breaking the integral into two parts and whether it is appropriate to evaluate them separately with the variable y present. There are questions about the method of subtraction and the interpretation of the area under the curves.

Discussion Status

The discussion includes attempts to evaluate the integral by considering symmetry and the possibility of simplifying the process. Some participants express uncertainty about their methods, while others provide affirmations of the approaches taken. There is no explicit consensus on the best method, but guidance is offered regarding the evaluation process.

Contextual Notes

Participants mention concerns about thoroughness in their calculations and the expectations of their teacher, indicating a focus on understanding the problem setup rather than rushing to a solution.

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Homework Statement



integrate y DA over the regions s, where s is in the first quadrant bounded by x = y^2 and x = 8 - y^2

Homework Equations





The Attempt at a Solution



If the y wasnt there i would evaluate two integrals and subtract them to get the area.
But since the y is there, can i still evaluate them seperatly with the y in place and subtract.
 
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Hi joemama69! :smile:
joemama69 said:
integrate y DA over the regions s, where s is in the first quadrant bounded by x = y^2 and x = 8 - y^2

and which axis??

If the y wasnt there i would evaluate two integrals and subtract them to get the area.
But since the y is there, can i still evaluate them seperatly with the y in place and subtract.

(not sure what you're subtracting, but:) yes … what's worrying you about that? :smile:
 
ya i was thinkin to hard

heres what i did, i broke it into two intigrals and added them

first intigral

i differentiated y dy from 0 to x^(1/2)
then i did in trems of x from 0 to 4 and got an area for 4

2nd integral

i differentiated y dy from 0 to (8-x)^(1/2)
then i did in trems of x from 4 to 8 and got an area for 4

4 + 4 = 8 is that right
 
joemama69 said:
ya i was thinkin to hard

heres what i did, i broke it into two intigrals and added them

4 + 4 = 8 is that right

Yup, that's fine! :biggrin:

(though you could have saved a little work by pointing out that the region was obviously symmetric about x = 4, so all you had to do was evaluate one integral, and then double it. :wink:)
 
ya i noticed that, but i figured for my teachers sake i should do it thoroughly
 

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