- #1

Tony11235

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D={(x,y)|0<=x<=3...is this even half way right?

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- Thread starter Tony11235
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- #1

Tony11235

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D={(x,y)|0<=x<=3...is this even half way right?

- #2

Maxos

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e. g. {0<x<1, 2x<y<-x+3}

- #3

cronxeh

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Just add [tex]y^3[/tex] into the double integral and evaluate it. Keep in mind for simple structures like a triangle you can verify that your answer is correct simply but using trigonometric formulas for area.

In this case area of a triangle is= [tex]\frac{(b*h)}{2}=\frac{3*1}{2}[/tex]

In this case area of a triangle is= [tex]\frac{(b*h)}{2}=\frac{3*1}{2}[/tex]

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- #4

HallsofIvy

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Tony11235 said:

D={(x,y)|0<=x<=3...is this even half way right?

No, because the "3" in (0, 3) is a y-value not x. The first thing you should do is

For the line from (0,3) to (1,2), you can argue that the equation of a line is always of the form y- y

y- 3= 1(x-0) or y= x+ 3. If you take (x

Now, since x can range from 0 to 2, the "outer integral" will be [tex]\int_0^3 dx[/tex].

Since the integrand is y

[tex]\int_{x=0}^2\int_{y= 2x}^{3-x} y^3 dy dx[/tex].

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- #5

HallsofIvy

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You could, by the way, do this "the other way around" (I wouldn't recommend it here- it's much more complicated. My point is to show how you can get the limits of integration.)

Overall, y ranges from 0 to 3. If we choose to use the y integral as the "outer integral", the limits would be from 0 to 3.**For each y**, the left sides is always x= 0 and so the lower limit for the "inner integral" is x= 0. However, there are two different lines that make up the right side: y= 2x and y= 3-x or, since we are using x as the variable for the inner integral and the limits of the inner integral cannot depend on x, x= (1/2)y and x= 3- y. Which of those we use as the upper depends on what y is. If y< 2, the right side is x= (1/2)y and if y> 2, the right side is x= 3- y. The simplest thing to do is to do that as two separate integrals and add them:

[tex]\int_{y=0}^2\int_{x=0}^{\frac{y}{2}}y^3 dxdy+ \int_{y=2}^3\int_{x=0}^{3-y}y^3dxdy[/tex].

Overall, y ranges from 0 to 3. If we choose to use the y integral as the "outer integral", the limits would be from 0 to 3.

[tex]\int_{y=0}^2\int_{x=0}^{\frac{y}{2}}y^3 dxdy+ \int_{y=2}^3\int_{x=0}^{3-y}y^3dxdy[/tex].

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- #6

Tony11235

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Thanks. I really didn't think about it enough. :grumpy:

- #7

okigetitnow

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is there a #1 answer rating?

- #8

Rubik

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Why does x range from 0 to 2 and why does this give the outer integral from 0 to 3?

- #9

SammyS

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No doubt, that was a typo (In post #4).Why does x range from 0 to 2 and why does this give the outer integral from 0 to 3?

The limits of integration for x should be from 0 to 1.

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