Question

[CALCULUS III—DOUBLE INTEGRALS] Set up an integral for both orders of integration, and use the more convenient order to evaluate the double integral. (figure inside)

Answer 1

\[\int \int_R f(x,y)\ dA\]\[f(x,y)=\frac{y}{x^2+y^2}\]R: triangle bounded by y = x, y = 2x, x = 2.

Answer 2

@calmat01

Answer 3

I get how the boundaries work, but is f(x,y) arbitrary or does it relate to the bounds?

Answer 4

Or is this R2 image a countour of f(x,y) = y/(x^2+y^2)

Answer 5

Like a cross section of f(x,y) and we want the area

Answer 6

You will need to calculate the slope of the two lines, and then find their equations.

Answer 7

lines are given in bounds

you would want to integrate between the 2 lines first

simply because y is given in terms of x

once you integrate with respect to y, then you want to integrate with respect to x

the bounds of x would be from the intersection of hte 2 lines (0) to x=2 which is given

Answer 8

OOPs, didn't see the equations up there till just now.

Answer 9

Thanks @completeidiot

Answer 10

i would not suggest trying to integrate with respect to x first
you would have to convert the lines to x in terms of y but even after doing so, you might have to break up the entire integral into parts

Answer 11

on second thought, if you wanted to integrate in terms of x first, you would definitely need to break the area into 2 parts

Answer 12

yeah,use the fact that your boundaries are in terms of x so integrate in terms of y first with your limits of integration being y=x and y=2x. After you get your result, the resulting integral will be in terms of x and you can integrate that from a lower limit of x=0 to an upper limit of x=2.

Answer 13

\[\int\int_R\frac{y}{x^2+y^2}dA\]

Answer 14

I am still seing only commands, but it looks like y over (x^2+y^2)

Answer 15

seeing*

Answer 16

yes thats right

Answer 17

@calmat01

Answer 18

ok, I see it now.

Answer 19

So completeidiot, are you still maintaining that we try splitting the graph into two parts?

Answer 20

sorry did we ever address if this R2 graph is a contour of f(x, y)

Answer 21

I am good at evaluating integrals, single and double, but conceptually I am not where I want to be in terms of visualizing and setting up these equations.

Answer 22

notes — http://tutorial.math.lamar.edu/Classes/CalcIII/DoubleIntegrals.aspx

Answer 23

I also have a textbook I have yet to refer to, but usually lamar.edu gets me through these topics.

Answer 24

can you help expound this concept?

Answer 25

Yes, you are just looking at the graph from a topographical view. It's a map created to represent the changes in elevation, or how it dips and rises.

Answer 26

So f(x, y) is the 3 dimensional graph, and R is the triangle bounded by the three equations?

Answer 27

I should say f(x, y) = S and triangle = R

Answer 28

Exactly.

Answer 29

naturally my next question is why do we need a double integral? I am going to keep reading and see if i find an explanation for this

Answer 30

alright i read more and I am less skeptical of the double integral, there is an x and a y dimension so we need to integrate in both respects.

That idea is sitting better with me, even though I do not think I would have independently derived all of these thoughts

Answer 31

I am trying to envision our f(x, y) in 3d and the triangular projection, but it's eluding me a bit. Regardless, now that I am more comfortable with the idea of double integrals dA, how do we set one up for this problem

Answer 32

Could we split it up like completeidiot said, and envision rectangles? and take half the area?

Answer 33

that way sounds like it would work but it is a bit lengthy

Answer 34

here is an example that matches our specific problem — http://tutorial.math.lamar.edu/Classes/CalcIII/DIGeneralRegion.aspx#MultInt_GenReg_Ex1c

Answer 35

I would suggest using the the fact that the boundaries of this region are your y-equations. y=x and y=2x My concern is that you would have to integrate y over x^2+ y^2. Let me take a look. Hang on.

Answer 36

\[\int\int_D \frac{y}{x^2+y^2}dA\]
D is the triangle with vertices (0,0), (2, 2), and (2,4).

Answer 37

yes, that problem you just posted looks identical to the one we are working on. Before I throw a monkey wrench into things, let me go reference a textbook I use all the time. brb

Answer 38

ok, I am at work so I will be in and out. Thank you for your time so far and time in the future

Answer 39

Would it be the integral from 0 to x dx integral from 0 2x dy

Answer 40

\[\int_0^{2x}\int_0^x\frac{y}{x^2+y^2}\ dxdy\]

Answer 41

That will not give you a constant area though.

x = 2 comes into play perhaps?

Answer 42

that looks closer to a proper integral but still not one that will provide an exact area :(

Answer 43

not quite. First, if we integrate WRT y, your lower limit is the function y=x and your upper limit is y=2x. After obtaining a function in terms of x, we will integrate WRT x with our lower limit being x=0 and our upper limit being x=2.

Answer 44

\[\int_0^2 \int_x^{2x}\frac{y}{x^2+y^2}dxdy\]

Answer 45

That's it.

Answer 46

Why is it 0 to 2? The range of y is from 0 to 4

Answer 47

Yes, but your limits in terms of x are the vertical lines x=0 and x=2.

Answer 48

Perhaps one of my buddies can help shed some more light. @SithsAndGiggles @jim_thompson5910 @zepdrix , @phi @experimentX

Answer 49

I like how you phrased "Yes, but your limits in terms of x are the vertical lines x=0 and x=2."

Limits was the key word missing for me.

How can we describe the limits in terms of y?