Double Integration Watch

Serendreamers
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#1
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#1
Using the method in the attachment, the area of the region is \frac{\pi}{8}(b^4-a^4) but if you tackle the problem by thinking using the area of a circle...A = \frac{\pi b^2}{4} - \frac{\pi a^2}{4}

The two answers are different but both appear correct?
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ghostwalker
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#2
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(Original post by Serendreamers)
Using the method in the attachment, the area of the region is \frac{\pi}{8}(b^4-a^4)
That's not the area.

If you wanted the area, you'd do \displaystyle\int_D 1\;dD

but if you tackle the problem by thinking using the area of a circle...A = \frac{\pi b^2}{4} - \frac{\pi a^2}{4}

The two answers are different but both appear correct?
They're different, as they're different things.
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Doctor_Einstein
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#3
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(Original post by Serendreamers)
Using the method in the attachment, the area of the region is \frac{\pi}{8}(b^4-a^4) but if you tackle the problem by thinking using the area of a circle...A = \frac{\pi b^2}{4} - \frac{\pi a^2}{4}

The two answers are different but both appear correct?
The working out in your attached picture is wrong.

Area = integral(from 0 to 2pi) integral(from r=a to b) of r d(theta)dr

= integral(from 0 to 2pi) (b^2 - a^2)/2 d(theta)
= (b^2 - a^2)pi/4
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Serendreamers
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#4
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(Original post by ghostwalker)
That's not the area.

If you wanted the area, you'd do \displaystyle\int_D 1\;dD



They're different, as they're different things.
What am I working out in my attached picture?
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ghostwalker
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(Original post by Serendreamers)
What am I working out in my attached picture?
You're integrating the function x^2+y^2 over the region D.
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Serendreamers
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(Original post by ghostwalker)
You're integrating the function x^2+y^2 over the region D.
I see, that's cleared things up. Thanks as always.
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TeeEm
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#7
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(Original post by Serendreamers)
I see, that's cleared things up. Thanks as always.
to add to ghostwalker's comment (all he said was correct)

your integral can be though as

the mass of a plate whose shape is that of the "quarter - annulus" and variable density pho(x,y) = x2+y2


OR

the volume of a right prism which is located under the surface z = x2+y2 and the xy plane on the "quarter - annulus"
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