## anonymous 3 years ago determine the limits and evaluate.. ∫∫x2y2dA where f is a figure bounded by F lines y = 1, y = 2 and x = 0, x = y

1. anonymous

|dw:1364052128248:dw| that gives you the limits

2. anonymous

it is easier to travel on y-axis because there is no sudden breaks

3. anonymous

so $0\le x\le y \qquad\text{&}\qquad1\le y\le2$

4. 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

5. anonymous

follow?

6. anonymous

in x=y, the value of y will become a and 2?

7. anonymous

"a"?

8. anonymous

all points on the "x=y" line will have the same abscissa and ordinate

9. anonymous

ah ok..

10. anonymous

yup, in the problem, all you need are the limits

11. anonymous

will i add these A) 1≤y≤2&0≤x≤1 B) x≤y≤2&1≤x≤2 ??

12. anonymous

use the first method its easier.

13. anonymous

yes those are your two regions for the second method

14. anonymous

x=0 to 1 and y= 1 to 2 ?

15. anonymous

no x=0 to y and y=1 to 2

16. anonymous

|dw:1364053324530:dw|

17. anonymous

i will integrate x^2y^2 using the limits x=0 to y and y=1 to 2 ?

18. anonymous

methid "1" is the horizontal element method "2" is the vertical

19. anonymous

yep. you got it

20. anonymous

ah ok.. in $\int\limits_{1}^{2}\int\limits_{0}^{y} x^2y^2 dxdy$, got 127/36.. is it right?

21. anonymous

$\int_{y=1}^2y^2\left[x^3\over3\right]_0^ydy={1\over3\times6}\left[y^6\right]_1^2$

22. anonymous

7/2?

23. anonymous

yep

24. anonymous

.. wait..

25. anonymous

ok..

26. anonymous

false alarm

27. anonymous

$dA=dxdy$

28. anonymous

kapeesh?

29. anonymous

127/36 = 3.53 7/2 = 3.5 my answer is 127/36 :)) its almost the same :)

30. anonymous

lol

31. anonymous

|dw:1364054668113:dw|

32. anonymous

@electrokid ? look at my answer?

33. anonymous

great job :D I'd personally not like to have many terms while integrating ... it gives migraine without it!

34. anonymous

thank you!so is my answer also correct?

35. anonymous

yep...

36. anonymous

thank you! tnx for ur time :)

37. anonymous

yer welcome :)