Double Integrals: Solving Homework in 1st Quadrant

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The discussion focuses on solving a double integral over a region in the first quadrant bounded by the curves x = y^2 and x = 8 - y^2. A user initially considers evaluating two separate integrals and subtracting them but later decides to add the areas of the two integrals instead. They calculate the area for each integral, finding both to be 4, resulting in a total area of 8. Another participant suggests that the problem's symmetry could have simplified the process by allowing the evaluation of just one integral and doubling the result. The conversation highlights the importance of thoroughness in solving integrals while also recognizing opportunities for simplification.
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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
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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