# Double IntegrationWatch

#1
Using the method in the attachment, the area of the region is but if you tackle the problem by thinking using the area of a circle...

The two answers are different but both appear correct?
0
4 years ago
#2
(Original post by Serendreamers)
Using the method in the attachment, the area of the region is
That's not the area.

If you wanted the area, you'd do

but if you tackle the problem by thinking using the area of a circle...

The two answers are different but both appear correct?
They're different, as they're different things.
4 years ago
#3
(Original post by Serendreamers)
Using the method in the attachment, the area of the region is but if you tackle the problem by thinking using the area of a circle...

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
0
#4
(Original post by ghostwalker)
That's not the area.

If you wanted the area, you'd do

They're different, as they're different things.
What am I working out in my attached picture?
0
4 years ago
#5
(Original post by Serendreamers)
What am I working out in my attached picture?
You're integrating the function over the region D.
#6
(Original post by ghostwalker)
You're integrating the function over the region D.
I see, that's cleared things up. Thanks as always.
0
4 years ago
#7
(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"
0
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