Question

Evaluate the integral ∫R2xydA, where R is the triangle 2x+y≤2, x≥0, y≥0

Answer 1

Did you draw the region as recommended to you 12 hours ago?

Answer 2

What then are the limits of integration?

Answer 3

lol..i posed that last night..then had to leave

Answer 4

so yea....i get the limits from 0 to 2, and 0 to 2x if that's correct?

Answer 5

@james: lol, you remember the problems you solved last night?!

Answer 6

No. Look carefully at the equation of that line. It's not y = 2x

Answer 7

@M: it would seem so.

Answer 8

oh it's 2x + y....so then x = 2/y?

Answer 9

;-)

Answer 10

Nooo. Draw the region. What are the three straight lines that form the boundary of the region R.

Answer 11

okay hang on lemme look

Answer 12

R triangle where 2x+y≤2, x≥0, y≥0

Answer 13

okay..so if i'm reading this correctly...just elaborating...x is greater than 0...y is greater than zero putting this in quadrant 1....2x+y has to be less than 2?

Answer 14

Yes, hence ....

Answer 15

the three boundary lines of the triangle are ... what? And what are the coordinates of the vertices?

Answer 16

how would i find that?

Answer 17

i mean...it's probably simple and i'm overthinking it...but i don't see it

Answer 18

Try to draw the line 2y+x=2.

Answer 19

What does the straight line 2x + y = 2 look like?

Answer 20

just a straight diagonal line

Answer 21

Crossing the axes where?
R = { (x,y) | 2x+y≤2, x≥0, y≥0 }

Answer 22

Then focus on the region "below" this line, to the top of the x-axis, and to the right of the y-axis. That's what the given boundaries mean.

Answer 23

at 2 and 1

Answer 24

At (1,0) and (0,2)

Answer 25

you've got to be more precise. Now, where is the third vertex/corner of the triangle?

Answer 26

origin?

Answer 27

Yes. That's where the lines x = 0 and y = 0 intersect.

Answer 28

Given all that then, what are the limits of integration. You can do either y first then x, or vice versa. They give different answers. Be clear when you give your answer which order you are using.

Answer 29

Actually, before you do that, draw yourself a diagram so it is clear to you what R looks like.

Answer 30

i graphed it on my calculator

Answer 31

and have it in front of me

Answer 32

Ok. I recommend integrating y first; it's more intuitive in way. So y goes from what to what?

Answer 33

0 to 2

Answer 34

No. It does when x = 0 but not anywhere else.

Answer 35

y = 0 to (some function of x)

Answer 36

2x-y?

Answer 37

y can't appear as a limit of a y integration. It's just a function of x.

What is the equation of the straight line, y = f(x) = .... what?

Answer 38

mx+b

Answer 39

and what are the values of m and b?
R = { (x,y) | 2x+y≤2, x≥0, y≥0 }

Answer 40

2x-2

Answer 41

can't be right. The line has negative slope.

Answer 42

The equation of the line is
2x + y = 2.

hence y = ...

Answer 43

-2x+2

Answer 44

Yes. Hence the limit of integration in the y-variable is
y = 0 to (-2x + 2)

Now, what's the limits of integration in the x-variable?

Answer 45

This should have no variables; just numbers.

Answer 46

x = .... to ....

Answer 47

look at your diagram

Answer 48

-1

Answer 49

What? x = -1 is completely outside the region R.

Answer 50

x is from 0 to 2

Answer 51

sorry this server is getting laggy

Answer 52

No. x only goes from 0 to 1.

Answer 53

oh right...and y goes from 0 to 2

Answer 54

Yes.

Notice that if you had integrated over y = 0 to 2 and x = 0 to 1, that would be a rectangle, not a triangular region.

Answer 55

This is why the first variable of integration in this case has to pick up on that triangular boundary.

Answer 56

Now, given all that, your integral is
\[ \int_0^1 \int_0^{-2x +2} 2xy \ dy \ dx \]

Answer 57

Now integrate carefully and you'll have your answer.

Answer 58

As an exercise, after you've found the answer, change the order of integration and make sure you get the same answer. The limits of integration will be
x = 0 to (some function of y)
y = 0 to 2

Answer 59

i'm getting errors on my bounds

Answer 60

i have to plug a b c and d for limits of integration

Answer 61

What do you mean? In your calculator?

Answer 62

no..i have to submit my answers online

Answer 63

Well, I don't know what's up with your system, but the limits we have are definitely correct.

Answer 64

...or I should say definitely correct given the order in which we're integrating. What order to they ask for in the question?

Answer 65

sonofa posted the question again in a new post and I helped her out, I didn't know where exactly you've stopped.

Answer 66

let me find it