Mixing
Find $lambda^{2}(A)$ while $A:={(x,y)in mathbb R^{2}: x<y,yfrac{1}{x}}$
$begingroup$
Find $lambda^{2}(A)$ whereby $A:={(x,y)in mathbb R^{2}: x<y,y<2,y>frac{1}{x}}$
I am struggling to find suitable restrictions.
Possible ideas:
Looking at restrictions:
$frac{1}{x} < y < 2$ and therfore $1 < x < 4$
$int_{1}^{4}int_{frac{1}{x}}^{2}1dydx=(2-frac{1}{x})(4-1)$
Is this correct?
real-analysis measure-theory multivariable-calculus lebesgue-measure
asked Jan 14 at 10:46
MinaThumaMinaThuma
1558
$endgroup$
add a comment |
$begingroup$
Find $lambda^{2}(A)$ whereby $A:={(x,y)in mathbb R^{2}: x<y,y<2,y>frac{1}{x}}$
I am struggling to find suitable restrictions.
Possible ideas:
Looking at restrictions:
$frac{1}{x} < y < 2$ and therfore $1 < x < 4$
$int_{1}^{4}int_{frac{1}{x}}^{2}1dydx=(2-frac{1}{x})(4-1)$
Is this correct?
real-analysis measure-theory multivariable-calculus lebesgue-measure
asked Jan 14 at 10:46
MinaThumaMinaThuma
1558
$endgroup$
$begingroup$
What is $lambda$?
$endgroup$
– Ingix
Jan 14 at 10:47
$begingroup$
Lebesgue Borel Measure
$endgroup$
– MinaThuma
Jan 14 at 10:49
$begingroup$
The answer shouldn't be $x$-dependent since it's basically an area of a shape in $Bbb R^2$.
$endgroup$
– BigbearZzz
Jan 14 at 10:53
$begingroup$
Try to make a sketch of $A$, using the given delimiting conditions. That should clear up where you made a mistake in the integral. Also, it seems something like $x,y>0$ is missing, as otherwise the whole area $(-infty,0) times (0,2)$ (with obvious inifinite measure) is a subset of $A$.
$endgroup$
– Ingix
Jan 14 at 11:05
add a comment |
$begingroup$
Find $lambda^{2}(A)$ whereby $A:={(x,y)in mathbb R^{2}: x<y,y<2,y>frac{1}{x}}$
I am struggling to find suitable restrictions.
Possible ideas:
Looking at restrictions:
$frac{1}{x} < y < 2$ and therfore $1 < x < 4$
$int_{1}^{4}int_{frac{1}{x}}^{2}1dydx=(2-frac{1}{x})(4-1)$
Is this correct?
real-analysis measure-theory multivariable-calculus lebesgue-measure
asked Jan 14 at 10:46
MinaThumaMinaThuma
1558
$endgroup$
Find $lambda^{2}(A)$ whereby $A:={(x,y)in mathbb R^{2}: x<y,y<2,y>frac{1}{x}}$
I am struggling to find suitable restrictions.
Possible ideas:
Looking at restrictions:
$frac{1}{x} < y < 2$ and therfore $1 < x < 4$
$int_{1}^{4}int_{frac{1}{x}}^{2}1dydx=(2-frac{1}{x})(4-1)$
Is this correct?
real-analysis measure-theory multivariable-calculus lebesgue-measure
real-analysis measure-theory multivariable-calculus lebesgue-measure
asked Jan 14 at 10:46
MinaThumaMinaThuma
1558
asked Jan 14 at 10:46
MinaThumaMinaThuma
1558
asked Jan 14 at 10:46
MinaThumaMinaThuma
1558
asked Jan 14 at 10:46
MinaThumaMinaThuma
1558
asked Jan 14 at 10:46
MinaThumaMinaThuma
1558
1558
$begingroup$
What is $lambda$?
$endgroup$
– Ingix
Jan 14 at 10:47
$begingroup$
Lebesgue Borel Measure
$endgroup$
– MinaThuma
Jan 14 at 10:49
$begingroup$
The answer shouldn't be $x$-dependent since it's basically an area of a shape in $Bbb R^2$.
$endgroup$
– BigbearZzz
Jan 14 at 10:53
$begingroup$
Try to make a sketch of $A$, using the given delimiting conditions. That should clear up where you made a mistake in the integral. Also, it seems something like $x,y>0$ is missing, as otherwise the whole area $(-infty,0) times (0,2)$ (with obvious inifinite measure) is a subset of $A$.
$endgroup$
– Ingix
Jan 14 at 11:05
add a comment |
$begingroup$
What is $lambda$?
$endgroup$
– Ingix
Jan 14 at 10:47
$begingroup$
Lebesgue Borel Measure
$endgroup$
– MinaThuma
Jan 14 at 10:49
$begingroup$
The answer shouldn't be $x$-dependent since it's basically an area of a shape in $Bbb R^2$.
$endgroup$
– BigbearZzz
Jan 14 at 10:53
$begingroup$
Try to make a sketch of $A$, using the given delimiting conditions. That should clear up where you made a mistake in the integral. Also, it seems something like $x,y>0$ is missing, as otherwise the whole area $(-infty,0) times (0,2)$ (with obvious inifinite measure) is a subset of $A$.
$endgroup$
– Ingix
Jan 14 at 11:05
$begingroup$
What is $lambda$?
$endgroup$
– Ingix
Jan 14 at 10:47
$begingroup$
What is $lambda$?
$endgroup$
– Ingix
Jan 14 at 10:47
$begingroup$
Lebesgue Borel Measure
$endgroup$
– MinaThuma
Jan 14 at 10:49
$begingroup$
Lebesgue Borel Measure
$endgroup$
– MinaThuma
Jan 14 at 10:49
$begingroup$
The answer shouldn't be $x$-dependent since it's basically an area of a shape in $Bbb R^2$.
$endgroup$
– BigbearZzz
Jan 14 at 10:53
$begingroup$
The answer shouldn't be $x$-dependent since it's basically an area of a shape in $Bbb R^2$.
$endgroup$
– BigbearZzz
Jan 14 at 10:53
$begingroup$
Try to make a sketch of $A$, using the given delimiting conditions. That should clear up where you made a mistake in the integral. Also, it seems something like $x,y>0$ is missing, as otherwise the whole area $(-infty,0) times (0,2)$ (with obvious inifinite measure) is a subset of $A$.
$endgroup$
– Ingix
Jan 14 at 11:05
$begingroup$
Try to make a sketch of $A$, using the given delimiting conditions. That should clear up where you made a mistake in the integral. Also, it seems something like $x,y>0$ is missing, as otherwise the whole area $(-infty,0) times (0,2)$ (with obvious inifinite measure) is a subset of $A$.
$endgroup$
– Ingix
Jan 14 at 11:05
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Your answer is wrong because it should be a constant which does not depend on $x$ as @BigbearZzz pointed out. Note that we can write $$A={(x,y)inmathbb{R}^2;|;max{x,frac{1}{x}}<y<2}.$$ As @Ingix pointed out, if we allow $x$ to be negative, then $(-infty,0)times [0,2)$ is contained in $A$ and thus we have $lambda_2(A)=infty$. So I'll presume that $x$ is always positive.
To evaluate the volume of $A$, we can write
$$begin{eqnarray}
lambda_2(A)&=&int_A1; dxdy\
&=&int_{x>0, max{x,frac{1}{x}}<y<2}1; dxdy.
end{eqnarray}$$ Note that the range of $x$ for which $max{x,frac{1}{x}}<2$ is $frac{1}{2}<x<2$. We can see this by solving $0<x<2$ and $0<frac{1}{x}<2$. By Fubini-Tonelli's theorem the given integral is
$$begin{eqnarray}
int_{x>0, max{x,frac{1}{x}}<y<2}1; dxdy&=&int_{frac{1}{2}<x<2}left[int_{max{x,frac{1}{x}}}^2 1; dyright]dx\
&=&int_{frac{1}{2}<x<2}left(2-max{x,frac{1}{x}}right)dx\
&=&int_frac{1}{2}^1left(2-max{x,frac{1}{x}}right)dx+int_1^2left(2-max{x,frac{1}{x}}right)dx\
&=&int_frac{1}{2}^1left(2-frac{1}{x}right)dx+int_1^2left(2-xright)dx\
&=&1-left[log xright]Big|^1_{frac{1}{2}}+2-left[frac{1}{2}x^2right]Big|^2_1\
&=&frac{3}{2}-log 2.
end{eqnarray}$$ Thus $lambda_2(A)=frac{3}{2}-log 2$.
answered Jan 14 at 11:26
SongSong
13.6k632
$endgroup$
add a comment |
$begingroup$
The left picture shows the integral computed first dy and then dx (this is represented by the direction of the segments coloring the domain). It is necessary to split the integral in two parts $intlimits_{1/2}^1;intlimits_{1/x}^2 dy;dx + intlimits_{1}^2intlimits_{x}^2 dy;dx.$
The right one presents the same area but the integral is easier to manage, $$int_{1}^2intlimits_{1/y}^y dx;dy=int_{1}^2 left(y-frac1yright) dy =left[frac{y^2}{2}-log yright]_{1}^{2}=frac32 - log 2. $$
answered Jan 14 at 11:37
user376343user376343
3,7883828
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Your answer is wrong because it should be a constant which does not depend on $x$ as @BigbearZzz pointed out. Note that we can write $$A={(x,y)inmathbb{R}^2;|;max{x,frac{1}{x}}<y<2}.$$ As @Ingix pointed out, if we allow $x$ to be negative, then $(-infty,0)times [0,2)$ is contained in $A$ and thus we have $lambda_2(A)=infty$. So I'll presume that $x$ is always positive.
To evaluate the volume of $A$, we can write
$$begin{eqnarray}
lambda_2(A)&=&int_A1; dxdy\
&=&int_{x>0, max{x,frac{1}{x}}<y<2}1; dxdy.
end{eqnarray}$$ Note that the range of $x$ for which $max{x,frac{1}{x}}<2$ is $frac{1}{2}<x<2$. We can see this by solving $0<x<2$ and $0<frac{1}{x}<2$. By Fubini-Tonelli's theorem the given integral is
$$begin{eqnarray}
int_{x>0, max{x,frac{1}{x}}<y<2}1; dxdy&=&int_{frac{1}{2}<x<2}left[int_{max{x,frac{1}{x}}}^2 1; dyright]dx\
&=&int_{frac{1}{2}<x<2}left(2-max{x,frac{1}{x}}right)dx\
&=&int_frac{1}{2}^1left(2-max{x,frac{1}{x}}right)dx+int_1^2left(2-max{x,frac{1}{x}}right)dx\
&=&int_frac{1}{2}^1left(2-frac{1}{x}right)dx+int_1^2left(2-xright)dx\
&=&1-left[log xright]Big|^1_{frac{1}{2}}+2-left[frac{1}{2}x^2right]Big|^2_1\
&=&frac{3}{2}-log 2.
end{eqnarray}$$ Thus $lambda_2(A)=frac{3}{2}-log 2$.
answered Jan 14 at 11:26
SongSong
13.6k632
$endgroup$
add a comment |
$begingroup$
Your answer is wrong because it should be a constant which does not depend on $x$ as @BigbearZzz pointed out. Note that we can write $$A={(x,y)inmathbb{R}^2;|;max{x,frac{1}{x}}<y<2}.$$ As @Ingix pointed out, if we allow $x$ to be negative, then $(-infty,0)times [0,2)$ is contained in $A$ and thus we have $lambda_2(A)=infty$. So I'll presume that $x$ is always positive.
To evaluate the volume of $A$, we can write
$$begin{eqnarray}
lambda_2(A)&=&int_A1; dxdy\
&=&int_{x>0, max{x,frac{1}{x}}<y<2}1; dxdy.
end{eqnarray}$$ Note that the range of $x$ for which $max{x,frac{1}{x}}<2$ is $frac{1}{2}<x<2$. We can see this by solving $0<x<2$ and $0<frac{1}{x}<2$. By Fubini-Tonelli's theorem the given integral is
$$begin{eqnarray}
int_{x>0, max{x,frac{1}{x}}<y<2}1; dxdy&=&int_{frac{1}{2}<x<2}left[int_{max{x,frac{1}{x}}}^2 1; dyright]dx\
&=&int_{frac{1}{2}<x<2}left(2-max{x,frac{1}{x}}right)dx\
&=&int_frac{1}{2}^1left(2-max{x,frac{1}{x}}right)dx+int_1^2left(2-max{x,frac{1}{x}}right)dx\
&=&int_frac{1}{2}^1left(2-frac{1}{x}right)dx+int_1^2left(2-xright)dx\
&=&1-left[log xright]Big|^1_{frac{1}{2}}+2-left[frac{1}{2}x^2right]Big|^2_1\
&=&frac{3}{2}-log 2.
end{eqnarray}$$ Thus $lambda_2(A)=frac{3}{2}-log 2$.
answered Jan 14 at 11:26
SongSong
13.6k632
$endgroup$
add a comment |
$begingroup$
Your answer is wrong because it should be a constant which does not depend on $x$ as @BigbearZzz pointed out. Note that we can write $$A={(x,y)inmathbb{R}^2;|;max{x,frac{1}{x}}<y<2}.$$ As @Ingix pointed out, if we allow $x$ to be negative, then $(-infty,0)times [0,2)$ is contained in $A$ and thus we have $lambda_2(A)=infty$. So I'll presume that $x$ is always positive.
To evaluate the volume of $A$, we can write
$$begin{eqnarray}
lambda_2(A)&=&int_A1; dxdy\
&=&int_{x>0, max{x,frac{1}{x}}<y<2}1; dxdy.
end{eqnarray}$$ Note that the range of $x$ for which $max{x,frac{1}{x}}<2$ is $frac{1}{2}<x<2$. We can see this by solving $0<x<2$ and $0<frac{1}{x}<2$. By Fubini-Tonelli's theorem the given integral is
$$begin{eqnarray}
int_{x>0, max{x,frac{1}{x}}<y<2}1; dxdy&=&int_{frac{1}{2}<x<2}left[int_{max{x,frac{1}{x}}}^2 1; dyright]dx\
&=&int_{frac{1}{2}<x<2}left(2-max{x,frac{1}{x}}right)dx\
&=&int_frac{1}{2}^1left(2-max{x,frac{1}{x}}right)dx+int_1^2left(2-max{x,frac{1}{x}}right)dx\
&=&int_frac{1}{2}^1left(2-frac{1}{x}right)dx+int_1^2left(2-xright)dx\
&=&1-left[log xright]Big|^1_{frac{1}{2}}+2-left[frac{1}{2}x^2right]Big|^2_1\
&=&frac{3}{2}-log 2.
end{eqnarray}$$ Thus $lambda_2(A)=frac{3}{2}-log 2$.
answered Jan 14 at 11:26
SongSong
13.6k632
$endgroup$
Your answer is wrong because it should be a constant which does not depend on $x$ as @BigbearZzz pointed out. Note that we can write $$A={(x,y)inmathbb{R}^2;|;max{x,frac{1}{x}}<y<2}.$$ As @Ingix pointed out, if we allow $x$ to be negative, then $(-infty,0)times [0,2)$ is contained in $A$ and thus we have $lambda_2(A)=infty$. So I'll presume that $x$ is always positive.
To evaluate the volume of $A$, we can write
$$begin{eqnarray}
lambda_2(A)&=&int_A1; dxdy\
&=&int_{x>0, max{x,frac{1}{x}}<y<2}1; dxdy.
end{eqnarray}$$ Note that the range of $x$ for which $max{x,frac{1}{x}}<2$ is $frac{1}{2}<x<2$. We can see this by solving $0<x<2$ and $0<frac{1}{x}<2$. By Fubini-Tonelli's theorem the given integral is
$$begin{eqnarray}
int_{x>0, max{x,frac{1}{x}}<y<2}1; dxdy&=&int_{frac{1}{2}<x<2}left[int_{max{x,frac{1}{x}}}^2 1; dyright]dx\
&=&int_{frac{1}{2}<x<2}left(2-max{x,frac{1}{x}}right)dx\
&=&int_frac{1}{2}^1left(2-max{x,frac{1}{x}}right)dx+int_1^2left(2-max{x,frac{1}{x}}right)dx\
&=&int_frac{1}{2}^1left(2-frac{1}{x}right)dx+int_1^2left(2-xright)dx\
&=&1-left[log xright]Big|^1_{frac{1}{2}}+2-left[frac{1}{2}x^2right]Big|^2_1\
&=&frac{3}{2}-log 2.
end{eqnarray}$$ Thus $lambda_2(A)=frac{3}{2}-log 2$.
answered Jan 14 at 11:26
SongSong
13.6k632
answered Jan 14 at 11:26
SongSong
13.6k632
answered Jan 14 at 11:26
SongSong
13.6k632
answered Jan 14 at 11:26
SongSong
13.6k632
13.6k632
add a comment |
add a comment |
$begingroup$
The left picture shows the integral computed first dy and then dx (this is represented by the direction of the segments coloring the domain). It is necessary to split the integral in two parts $intlimits_{1/2}^1;intlimits_{1/x}^2 dy;dx + intlimits_{1}^2intlimits_{x}^2 dy;dx.$
The right one presents the same area but the integral is easier to manage, $$int_{1}^2intlimits_{1/y}^y dx;dy=int_{1}^2 left(y-frac1yright) dy =left[frac{y^2}{2}-log yright]_{1}^{2}=frac32 - log 2. $$
answered Jan 14 at 11:37
user376343user376343
3,7883828
$endgroup$
add a comment |
$begingroup$
The left picture shows the integral computed first dy and then dx (this is represented by the direction of the segments coloring the domain). It is necessary to split the integral in two parts $intlimits_{1/2}^1;intlimits_{1/x}^2 dy;dx + intlimits_{1}^2intlimits_{x}^2 dy;dx.$
The right one presents the same area but the integral is easier to manage, $$int_{1}^2intlimits_{1/y}^y dx;dy=int_{1}^2 left(y-frac1yright) dy =left[frac{y^2}{2}-log yright]_{1}^{2}=frac32 - log 2. $$
answered Jan 14 at 11:37
user376343user376343
3,7883828
$endgroup$
add a comment |
$begingroup$
The left picture shows the integral computed first dy and then dx (this is represented by the direction of the segments coloring the domain). It is necessary to split the integral in two parts $intlimits_{1/2}^1;intlimits_{1/x}^2 dy;dx + intlimits_{1}^2intlimits_{x}^2 dy;dx.$
The right one presents the same area but the integral is easier to manage, $$int_{1}^2intlimits_{1/y}^y dx;dy=int_{1}^2 left(y-frac1yright) dy =left[frac{y^2}{2}-log yright]_{1}^{2}=frac32 - log 2. $$
answered Jan 14 at 11:37
user376343user376343
3,7883828
$endgroup$
The left picture shows the integral computed first dy and then dx (this is represented by the direction of the segments coloring the domain). It is necessary to split the integral in two parts $intlimits_{1/2}^1;intlimits_{1/x}^2 dy;dx + intlimits_{1}^2intlimits_{x}^2 dy;dx.$
The right one presents the same area but the integral is easier to manage, $$int_{1}^2intlimits_{1/y}^y dx;dy=int_{1}^2 left(y-frac1yright) dy =left[frac{y^2}{2}-log yright]_{1}^{2}=frac32 - log 2. $$
answered Jan 14 at 11:37
user376343user376343
3,7883828
answered Jan 14 at 11:37
user376343user376343
3,7883828
answered Jan 14 at 11:37
user376343user376343
3,7883828
answered Jan 14 at 11:37
user376343user376343
3,7883828
3,7883828
add a comment |
add a comment |
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$begingroup$
What is $lambda$?
$endgroup$
– Ingix
Jan 14 at 10:47
$begingroup$
Lebesgue Borel Measure
$endgroup$
– MinaThuma
Jan 14 at 10:49
$begingroup$
The answer shouldn't be $x$-dependent since it's basically an area of a shape in $Bbb R^2$.
$endgroup$
– BigbearZzz
Jan 14 at 10:53
$begingroup$
Try to make a sketch of $A$, using the given delimiting conditions. That should clear up where you made a mistake in the integral. Also, it seems something like $x,y>0$ is missing, as otherwise the whole area $(-infty,0) times (0,2)$ (with obvious inifinite measure) is a subset of $A$.
$endgroup$
– Ingix
Jan 14 at 11:05