Area of region between circle and curve

  • Thread starter DryRun
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  • #1
DryRun
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Homework Statement
http://s2.ipicture.ru/uploads/20120107/67Ag24Qb.jpg

The attempt at a solution
So, i plotted the graphs of the circle and the curve:
http://s2.ipicture.ru/uploads/20120107/x32KTV6y.jpg

The shaded area is what i need to find. My plan to solve this problem is to find the area with respect to the y-axis of the curve minus the right-half of the circle.

Area of left-side of curve:
For y fixed, x varies from x=0 to x=4-y^2
y varies from y=-2 to y=2

I did double integral with the limits above, w.r.t. dxdy and got 32/3

Area of right-half of circle:
For y fixed, x varies from x=0 to x=√(4-y^2)
y varies from y=-2 to y=2

Again, i've done double integral and ended up with 2pi

I don't know which of these 2 integrals are wrong, as the difference doesn't give me the correct answer, although it's close.
 
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Answers and Replies

  • #2
33,722
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Homework Statement
http://s2.ipicture.ru/uploads/20120107/67Ag24Qb.jpg

The attempt at a solution
So, i plotted the graphs of the circle and the curve:
http://s2.ipicture.ru/uploads/20120107/x32KTV6y.jpg

The shaded area is what i need to find. My plan to solve this problem is to find the area with respect to the y-axis of the curve minus the right-half of the circle.

Area of left-side of curve:
For y fixed, x varies from x=0 to x=4-y^2
y varies from y=-2 to y=2

I did double integral with the limits above, w.r.t. dxdy and got 32/3

Area of right-half of circle:
For y fixed, x varies from x=0 to x=√(4-y^2)
y varies from y=-2 to y=2

Again, i've done double integral and ended up with 2pi

I don't know which of these 2 integrals are wrong, as the difference doesn't give me the correct answer, although it's close.
Why don't you use just a single integral? Determining an iterated integral seems more complicated than is necessary. Also, I would exploit the symmetry in your curves by calculating the area of the upper half and then doubling it.

I would use horizontal slices (of thickness Δy) and of length xparabola - xcircle, and integrate from y = 0 to y = 2.
 
  • #3
ehild
Homework Helper
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The parabola and the circle intersect at x=0 and at x=1. I think the shaded area should extend from x=1 to x=4, while y varies between-√3 and √3.

ehild
 
  • #4
SammyS
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Notice where the circle & parabola intersect.

attachment.php?attachmentid=42520&stc=1&d=1325900049.gif
(from WolframAlpha)​
 

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  • #5
DryRun
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838
4
I obviously had the wrong limits. I had drawn the graph roughly in my copybook so it wasn't clear at first. I used the correct limits as advised and got the answer. Thanks for the help.
 
  • #6
SammyS
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Sometimes a picture is worth a thousand words! :smile:
 

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