limits of changing the order of a double integral
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I've solved this integral $int_{1}^3 int_{1-y}^{2-y}x^2y,dxdy$ which is 3 1/3. Now I'm trying it with different integral order but can't get the limits right. My try was, $int_{0}^1 int_{1-x}^{2-x}x^2y,dydx$.
integration multiple-integral
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add a comment |
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I've solved this integral $int_{1}^3 int_{1-y}^{2-y}x^2y,dxdy$ which is 3 1/3. Now I'm trying it with different integral order but can't get the limits right. My try was, $int_{0}^1 int_{1-x}^{2-x}x^2y,dydx$.
integration multiple-integral
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And the correct one is $int_{-2}^{1}int_{max{1-x,1}}^{min{2-x,3}}x^2y,dy,dx$ (not sure what you do it for).
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– metamorphy
Dec 22 '18 at 6:39
add a comment |
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I've solved this integral $int_{1}^3 int_{1-y}^{2-y}x^2y,dxdy$ which is 3 1/3. Now I'm trying it with different integral order but can't get the limits right. My try was, $int_{0}^1 int_{1-x}^{2-x}x^2y,dydx$.
integration multiple-integral
$endgroup$
I've solved this integral $int_{1}^3 int_{1-y}^{2-y}x^2y,dxdy$ which is 3 1/3. Now I'm trying it with different integral order but can't get the limits right. My try was, $int_{0}^1 int_{1-x}^{2-x}x^2y,dydx$.
integration multiple-integral
integration multiple-integral
asked Dec 22 '18 at 6:29
LLTLLT
32
32
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And the correct one is $int_{-2}^{1}int_{max{1-x,1}}^{min{2-x,3}}x^2y,dy,dx$ (not sure what you do it for).
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– metamorphy
Dec 22 '18 at 6:39
add a comment |
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And the correct one is $int_{-2}^{1}int_{max{1-x,1}}^{min{2-x,3}}x^2y,dy,dx$ (not sure what you do it for).
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– metamorphy
Dec 22 '18 at 6:39
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And the correct one is $int_{-2}^{1}int_{max{1-x,1}}^{min{2-x,3}}x^2y,dy,dx$ (not sure what you do it for).
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– metamorphy
Dec 22 '18 at 6:39
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And the correct one is $int_{-2}^{1}int_{max{1-x,1}}^{min{2-x,3}}x^2y,dy,dx$ (not sure what you do it for).
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– metamorphy
Dec 22 '18 at 6:39
add a comment |
3 Answers
3
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oldest
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A good drawing always helps for these sorts of questions.
As you can see, the region can be broken up into a triangle between $x=-2$ and $-1$, a pair of lines between $x=-1$ and $0$, and another triangle between $x=0$ and $1$.
So hopefully you can see that
$$int_1^3 int_{1-y}^{2-y} x^2y text{d}x text{d}y = int_{-2}^{-1}int_{1-x}^3 x^2y text{d}y text{d}x + int_{-1}^{0} int_{1-x}^{2-x}x^2y text{d}y text{d}x + int_{0}^{1} int_{1}^{2-x}x^2y text{d}y text{d}x$$
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Cool thx! I drew a graph but forgot the area in the negative area. Now everything’s clear. Cheers.
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– LLT
Dec 22 '18 at 6:57
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Sorry, should it be $$int_1^3 int_{1-y}^{2-y} x^2y text{d}x text{d}y = int_{-2}^{-1}int_{1-x}^3 x^2y text{d}y text{d}x + int_{-1}^{0} int_{1-x}^{2-x}x^2y text{d}y text{d}x + int_{0}^{1} int_{1}^{2-x}x^2y text{d}y text{d}x$$. i.e. the upper limit of dy of the first triangle should be 3?
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– LLT
Dec 23 '18 at 6:39
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Indeed it should, that was silly of me
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– Cade Reinberger
Dec 24 '18 at 8:53
add a comment |
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Hint: Try drawing the figure, bounded by
begin{align}
x=2-y\
x=1-y\
y=1\
y=3
end{align}
And then set the limits with y as a function of x first.
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add a comment |
$begingroup$
First of all you need to draw a graph of the region of integration which is a parallelogram.
You notice that you need three integrals to cover the region if you changed the order of integration.
The x runs from -2 to 1 and you upper and lower limits for y changes on each sub-intervals of length 1.
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add a comment |
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3 Answers
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active
oldest
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3 Answers
3
active
oldest
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$begingroup$
A good drawing always helps for these sorts of questions.
As you can see, the region can be broken up into a triangle between $x=-2$ and $-1$, a pair of lines between $x=-1$ and $0$, and another triangle between $x=0$ and $1$.
So hopefully you can see that
$$int_1^3 int_{1-y}^{2-y} x^2y text{d}x text{d}y = int_{-2}^{-1}int_{1-x}^3 x^2y text{d}y text{d}x + int_{-1}^{0} int_{1-x}^{2-x}x^2y text{d}y text{d}x + int_{0}^{1} int_{1}^{2-x}x^2y text{d}y text{d}x$$
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Cool thx! I drew a graph but forgot the area in the negative area. Now everything’s clear. Cheers.
$endgroup$
– LLT
Dec 22 '18 at 6:57
$begingroup$
Sorry, should it be $$int_1^3 int_{1-y}^{2-y} x^2y text{d}x text{d}y = int_{-2}^{-1}int_{1-x}^3 x^2y text{d}y text{d}x + int_{-1}^{0} int_{1-x}^{2-x}x^2y text{d}y text{d}x + int_{0}^{1} int_{1}^{2-x}x^2y text{d}y text{d}x$$. i.e. the upper limit of dy of the first triangle should be 3?
$endgroup$
– LLT
Dec 23 '18 at 6:39
$begingroup$
Indeed it should, that was silly of me
$endgroup$
– Cade Reinberger
Dec 24 '18 at 8:53
add a comment |
$begingroup$
A good drawing always helps for these sorts of questions.
As you can see, the region can be broken up into a triangle between $x=-2$ and $-1$, a pair of lines between $x=-1$ and $0$, and another triangle between $x=0$ and $1$.
So hopefully you can see that
$$int_1^3 int_{1-y}^{2-y} x^2y text{d}x text{d}y = int_{-2}^{-1}int_{1-x}^3 x^2y text{d}y text{d}x + int_{-1}^{0} int_{1-x}^{2-x}x^2y text{d}y text{d}x + int_{0}^{1} int_{1}^{2-x}x^2y text{d}y text{d}x$$
$endgroup$
$begingroup$
Cool thx! I drew a graph but forgot the area in the negative area. Now everything’s clear. Cheers.
$endgroup$
– LLT
Dec 22 '18 at 6:57
$begingroup$
Sorry, should it be $$int_1^3 int_{1-y}^{2-y} x^2y text{d}x text{d}y = int_{-2}^{-1}int_{1-x}^3 x^2y text{d}y text{d}x + int_{-1}^{0} int_{1-x}^{2-x}x^2y text{d}y text{d}x + int_{0}^{1} int_{1}^{2-x}x^2y text{d}y text{d}x$$. i.e. the upper limit of dy of the first triangle should be 3?
$endgroup$
– LLT
Dec 23 '18 at 6:39
$begingroup$
Indeed it should, that was silly of me
$endgroup$
– Cade Reinberger
Dec 24 '18 at 8:53
add a comment |
$begingroup$
A good drawing always helps for these sorts of questions.
As you can see, the region can be broken up into a triangle between $x=-2$ and $-1$, a pair of lines between $x=-1$ and $0$, and another triangle between $x=0$ and $1$.
So hopefully you can see that
$$int_1^3 int_{1-y}^{2-y} x^2y text{d}x text{d}y = int_{-2}^{-1}int_{1-x}^3 x^2y text{d}y text{d}x + int_{-1}^{0} int_{1-x}^{2-x}x^2y text{d}y text{d}x + int_{0}^{1} int_{1}^{2-x}x^2y text{d}y text{d}x$$
$endgroup$
A good drawing always helps for these sorts of questions.
As you can see, the region can be broken up into a triangle between $x=-2$ and $-1$, a pair of lines between $x=-1$ and $0$, and another triangle between $x=0$ and $1$.
So hopefully you can see that
$$int_1^3 int_{1-y}^{2-y} x^2y text{d}x text{d}y = int_{-2}^{-1}int_{1-x}^3 x^2y text{d}y text{d}x + int_{-1}^{0} int_{1-x}^{2-x}x^2y text{d}y text{d}x + int_{0}^{1} int_{1}^{2-x}x^2y text{d}y text{d}x$$
edited Dec 24 '18 at 8:53
answered Dec 22 '18 at 6:41
Cade ReinbergerCade Reinberger
38317
38317
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Cool thx! I drew a graph but forgot the area in the negative area. Now everything’s clear. Cheers.
$endgroup$
– LLT
Dec 22 '18 at 6:57
$begingroup$
Sorry, should it be $$int_1^3 int_{1-y}^{2-y} x^2y text{d}x text{d}y = int_{-2}^{-1}int_{1-x}^3 x^2y text{d}y text{d}x + int_{-1}^{0} int_{1-x}^{2-x}x^2y text{d}y text{d}x + int_{0}^{1} int_{1}^{2-x}x^2y text{d}y text{d}x$$. i.e. the upper limit of dy of the first triangle should be 3?
$endgroup$
– LLT
Dec 23 '18 at 6:39
$begingroup$
Indeed it should, that was silly of me
$endgroup$
– Cade Reinberger
Dec 24 '18 at 8:53
add a comment |
$begingroup$
Cool thx! I drew a graph but forgot the area in the negative area. Now everything’s clear. Cheers.
$endgroup$
– LLT
Dec 22 '18 at 6:57
$begingroup$
Sorry, should it be $$int_1^3 int_{1-y}^{2-y} x^2y text{d}x text{d}y = int_{-2}^{-1}int_{1-x}^3 x^2y text{d}y text{d}x + int_{-1}^{0} int_{1-x}^{2-x}x^2y text{d}y text{d}x + int_{0}^{1} int_{1}^{2-x}x^2y text{d}y text{d}x$$. i.e. the upper limit of dy of the first triangle should be 3?
$endgroup$
– LLT
Dec 23 '18 at 6:39
$begingroup$
Indeed it should, that was silly of me
$endgroup$
– Cade Reinberger
Dec 24 '18 at 8:53
$begingroup$
Cool thx! I drew a graph but forgot the area in the negative area. Now everything’s clear. Cheers.
$endgroup$
– LLT
Dec 22 '18 at 6:57
$begingroup$
Cool thx! I drew a graph but forgot the area in the negative area. Now everything’s clear. Cheers.
$endgroup$
– LLT
Dec 22 '18 at 6:57
$begingroup$
Sorry, should it be $$int_1^3 int_{1-y}^{2-y} x^2y text{d}x text{d}y = int_{-2}^{-1}int_{1-x}^3 x^2y text{d}y text{d}x + int_{-1}^{0} int_{1-x}^{2-x}x^2y text{d}y text{d}x + int_{0}^{1} int_{1}^{2-x}x^2y text{d}y text{d}x$$. i.e. the upper limit of dy of the first triangle should be 3?
$endgroup$
– LLT
Dec 23 '18 at 6:39
$begingroup$
Sorry, should it be $$int_1^3 int_{1-y}^{2-y} x^2y text{d}x text{d}y = int_{-2}^{-1}int_{1-x}^3 x^2y text{d}y text{d}x + int_{-1}^{0} int_{1-x}^{2-x}x^2y text{d}y text{d}x + int_{0}^{1} int_{1}^{2-x}x^2y text{d}y text{d}x$$. i.e. the upper limit of dy of the first triangle should be 3?
$endgroup$
– LLT
Dec 23 '18 at 6:39
$begingroup$
Indeed it should, that was silly of me
$endgroup$
– Cade Reinberger
Dec 24 '18 at 8:53
$begingroup$
Indeed it should, that was silly of me
$endgroup$
– Cade Reinberger
Dec 24 '18 at 8:53
add a comment |
$begingroup$
Hint: Try drawing the figure, bounded by
begin{align}
x=2-y\
x=1-y\
y=1\
y=3
end{align}
And then set the limits with y as a function of x first.
$endgroup$
add a comment |
$begingroup$
Hint: Try drawing the figure, bounded by
begin{align}
x=2-y\
x=1-y\
y=1\
y=3
end{align}
And then set the limits with y as a function of x first.
$endgroup$
add a comment |
$begingroup$
Hint: Try drawing the figure, bounded by
begin{align}
x=2-y\
x=1-y\
y=1\
y=3
end{align}
And then set the limits with y as a function of x first.
$endgroup$
Hint: Try drawing the figure, bounded by
begin{align}
x=2-y\
x=1-y\
y=1\
y=3
end{align}
And then set the limits with y as a function of x first.
answered Dec 22 '18 at 6:39
mm-crjmm-crj
425213
425213
add a comment |
add a comment |
$begingroup$
First of all you need to draw a graph of the region of integration which is a parallelogram.
You notice that you need three integrals to cover the region if you changed the order of integration.
The x runs from -2 to 1 and you upper and lower limits for y changes on each sub-intervals of length 1.
$endgroup$
add a comment |
$begingroup$
First of all you need to draw a graph of the region of integration which is a parallelogram.
You notice that you need three integrals to cover the region if you changed the order of integration.
The x runs from -2 to 1 and you upper and lower limits for y changes on each sub-intervals of length 1.
$endgroup$
add a comment |
$begingroup$
First of all you need to draw a graph of the region of integration which is a parallelogram.
You notice that you need three integrals to cover the region if you changed the order of integration.
The x runs from -2 to 1 and you upper and lower limits for y changes on each sub-intervals of length 1.
$endgroup$
First of all you need to draw a graph of the region of integration which is a parallelogram.
You notice that you need three integrals to cover the region if you changed the order of integration.
The x runs from -2 to 1 and you upper and lower limits for y changes on each sub-intervals of length 1.
answered Dec 22 '18 at 6:49
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
41.6k42061
add a comment |
add a comment |
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$begingroup$
And the correct one is $int_{-2}^{1}int_{max{1-x,1}}^{min{2-x,3}}x^2y,dy,dx$ (not sure what you do it for).
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– metamorphy
Dec 22 '18 at 6:39