Double integral circle limits

  • Thread starter LASmith
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  • #1
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Homework Statement


Evaluate
f(x,y)=y2[itex]\sqrt{1-x2}[/itex]
over the region
x2+y2< 1

Homework Equations





The Attempt at a Solution



using x limits between -1 & 1 followed by the y limits of 0 & [itex]\sqrt{1-x2}[/itex]

[itex]\int[/itex][itex]\int[/itex]y2[itex]\sqrt{1-x2}[/itex].dy.dx

Evaluating this and multiplying be 2 to get the whole circle I get 0 as the 1 and -1 limits both give zero as the answer. Therefore I used the x limits of 0 & 1 and multiplied by 4 instead, this gave an answer of 4/9, however I have been given the answer of 32/45, and I have no idea how this could possibly be wrong.
 

Answers and Replies

  • #2
tiny-tim
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Hi LASmith! :smile:
… Therefore I used the x limits of 0 & 1 and multiplied by 4 instead, this gave an answer of 4/9 …
Your intermediate integrand must be wrong. :redface:

Show us what you had inside the ∫ when it was just ∫ … dx​
 
  • #3
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Show us what you had inside the ∫ when it was just ∫ … dx​
[itex]\int[/itex] y3[itex]\sqrt{1-x2}[/itex]/3 .dx

Then substituting the limits [itex]\sqrt{1-x2}[/itex] and 0 for y I obtained

[itex]\int[/itex] (1-x2)2)/3 .dx

Limits between 0 and 1 for this final integral
 
  • #4
HallsofIvy
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The set [itex]x^2+ y^2< 1[/itex] is the interior of a circle with center at (0, 0) and radius 1. That means that a diameter, on the x-axis, goes from -1 to 1, and on the y-axis, from -1 to 1. By taking y from 0 (the x-axis) up to [itex]\sqrt{1- x^2}[/itex] (the circle) and x from -1 to 1, you are only integrating over 1/2 of the circle.

Your y needs to go from the point where the vertical line crosses the circle below the x-axis to the point where it crosses the circle above the x-axis.
 
  • #5
tiny-tim
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[itex]\int[/itex] (1-x2)2)/3 .dx
= ∫ (1 - 2x2 + x4)/3 dx …

how did you not get a factor 1/5 after integrating this? :confused:
 
  • #6
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= ∫ (1 - 2x2 + x4)/3 dx …

how did you not get a factor 1/5 after integrating this? :confused:
Expanding the brackets and then integrating I obtained the correct answer, so thanks for that.

However, looking back I had no idea how I did not get a factor of 1/5, I did not expand out the brackets, and just used the chain rule to obtain
(1-x2)3/-18x
Then putting in the limits 0 and 1 gets the answer 0-0, out of curiosity how does this method not work?
 
  • #7
tiny-tim
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Hi LASmith! :smile:
… looking back I had no idea how I did not get a factor of 1/5, I did not expand out the brackets, and just used the chain rule to obtain
(1-x2)3/-18x
Then putting in the limits 0 and 1 gets the answer 0-0, out of curiosity how does this method not work?
Just try differentiating (1-x2)3/-18x …

you'd have to use both the chain rule and the product (or quotient) rule, and they don't cancel (why should they?).
 

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