Area of region between circle and curve

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Homework Help Overview

The discussion revolves around finding the area between a circle and a curve, specifically a parabola. Participants are analyzing the setup of the problem and the integration methods to compute the area of the shaded region depicted in the provided graphs.

Discussion Character

  • Exploratory, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss using double integrals to find the area, with one participant suggesting a single integral might be simpler. There is mention of exploiting symmetry to simplify calculations. Others raise questions about the limits of integration and the points of intersection between the curves.

Discussion Status

The discussion is active, with participants providing insights and suggestions for correcting limits and approaches. One participant acknowledges a misunderstanding of the limits and indicates that they found the correct answer after receiving guidance.

Contextual Notes

There are references to the intersection points of the curves and the necessity of accurate graphical representation to determine the correct limits for integration. Some participants express uncertainty about their initial setups and calculations.

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.
 
Last edited:
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sharks said:
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.
 
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
 
Notice where the circle & parabola intersect.

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

Attachments

  • wolframalpha-circ-parab.gif
    wolframalpha-circ-parab.gif
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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.
 
Sometimes a picture is worth a thousand words! :smile:
 

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