anonymous
  • anonymous
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
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
|dw:1364052128248:dw| that gives you the limits
anonymous
  • anonymous
it is easier to travel on y-axis because there is no sudden breaks
anonymous
  • anonymous
so \[0\le x\le y \qquad\text{&}\qquad1\le y\le2\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.