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determine the limits and evaluate.. ∫∫x2y2dA where f is a figure bounded by F lines y = 1, y = 2 and x = 0, x = y

Mathematics
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|dw:1364052128248:dw| that gives you the limits
it is easier to travel on y-axis because there is no sudden breaks
so \[0\le x\le y \qquad\text{&}\qquad1\le y\le2\]

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Other answers:

if you are on the x-axis, you have to consider two pieces A) \(1\le y\le2 \qquad\text{&}\qquad0\le x\le1\) B) \(x\le y\le2 \qquad\text{&}\qquad1\le x\le2\) and sum them
follow?
in x=y, the value of y will become a and 2?
"a"?
all points on the "x=y" line will have the same abscissa and ordinate
ah ok..
yup, in the problem, all you need are the limits
will i add these A) 1≤y≤2&0≤x≤1 B) x≤y≤2&1≤x≤2 ??
use the first method its easier.
yes those are your two regions for the second method
x=0 to 1 and y= 1 to 2 ?
no x=0 to y and y=1 to 2
|dw:1364053324530:dw|
i will integrate x^2y^2 using the limits x=0 to y and y=1 to 2 ?
methid "1" is the horizontal element method "2" is the vertical
yep. you got it
ah ok.. in \[\int\limits_{1}^{2}\int\limits_{0}^{y} x^2y^2 dxdy\], got 127/36.. is it right?
\[ \int_{y=1}^2y^2\left[x^3\over3\right]_0^ydy={1\over3\times6}\left[y^6\right]_1^2 \]
7/2?
yep
.. wait..
ok..
false alarm
\[dA=dxdy\]
kapeesh?
127/36 = 3.53 7/2 = 3.5 my answer is 127/36 :)) its almost the same :)
lol
|dw:1364054668113:dw|
@electrokid ? look at my answer?
great job :D I'd personally not like to have many terms while integrating ... it gives migraine without it!
thank you!so is my answer also correct?
yep...
thank you! tnx for ur time :)
yer welcome :)

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