How to change from a double to a single integral changing variables?
$begingroup$
So I have the following integral:
$$I_1 = iint u(x,y)dR$$
where $u(x,y)= e^{-(x^2 + y^2)}$
and the region $R$ is the rectangle $[-M,M]times[-M,M]$.
I need to prove that $I_1$ equals:
$$I_2 = left(∫_{-M}^Muleft(frac l{sqrt2},frac l{sqrt2}right)dlright)^2.$$
I have been thinking for a long time and surfing the internet for integral properties that might help without any luck. I am really confused because I don't see how an area times an area will give a volume.
The only thing I'm positive about using is the fact that the integration variable is mute.
Where do I start? What can I try? or What hints can you give me?
calculus integration multivariable-calculus proof-writing
edited Dec 6 '18 at 6:28
Tianlalu
3,08621038
asked Dec 6 '18 at 6:10
ReginaRegina
55
$endgroup$
add a comment |
$begingroup$
So I have the following integral:
$$I_1 = iint u(x,y)dR$$
where $u(x,y)= e^{-(x^2 + y^2)}$
and the region $R$ is the rectangle $[-M,M]times[-M,M]$.
I need to prove that $I_1$ equals:
$$I_2 = left(∫_{-M}^Muleft(frac l{sqrt2},frac l{sqrt2}right)dlright)^2.$$
I have been thinking for a long time and surfing the internet for integral properties that might help without any luck. I am really confused because I don't see how an area times an area will give a volume.
The only thing I'm positive about using is the fact that the integration variable is mute.
Where do I start? What can I try? or What hints can you give me?
calculus integration multivariable-calculus proof-writing
edited Dec 6 '18 at 6:28
Tianlalu
3,08621038
asked Dec 6 '18 at 6:10
ReginaRegina
55
$endgroup$
add a comment |
$begingroup$
So I have the following integral:
$$I_1 = iint u(x,y)dR$$
where $u(x,y)= e^{-(x^2 + y^2)}$
and the region $R$ is the rectangle $[-M,M]times[-M,M]$.
I need to prove that $I_1$ equals:
$$I_2 = left(∫_{-M}^Muleft(frac l{sqrt2},frac l{sqrt2}right)dlright)^2.$$
I have been thinking for a long time and surfing the internet for integral properties that might help without any luck. I am really confused because I don't see how an area times an area will give a volume.
The only thing I'm positive about using is the fact that the integration variable is mute.
Where do I start? What can I try? or What hints can you give me?
calculus integration multivariable-calculus proof-writing
edited Dec 6 '18 at 6:28
Tianlalu
3,08621038
asked Dec 6 '18 at 6:10
ReginaRegina
55
$endgroup$
So I have the following integral:
$$I_1 = iint u(x,y)dR$$
where $u(x,y)= e^{-(x^2 + y^2)}$
and the region $R$ is the rectangle $[-M,M]times[-M,M]$.
I need to prove that $I_1$ equals:
$$I_2 = left(∫_{-M}^Muleft(frac l{sqrt2},frac l{sqrt2}right)dlright)^2.$$
I have been thinking for a long time and surfing the internet for integral properties that might help without any luck. I am really confused because I don't see how an area times an area will give a volume.
The only thing I'm positive about using is the fact that the integration variable is mute.
Where do I start? What can I try? or What hints can you give me?
calculus integration multivariable-calculus proof-writing
calculus integration multivariable-calculus proof-writing
edited Dec 6 '18 at 6:28
Tianlalu
3,08621038
asked Dec 6 '18 at 6:10
ReginaRegina
55
edited Dec 6 '18 at 6:28
Tianlalu
3,08621038
asked Dec 6 '18 at 6:10
ReginaRegina
55
edited Dec 6 '18 at 6:28
Tianlalu
3,08621038
edited Dec 6 '18 at 6:28
Tianlalu
3,08621038
edited Dec 6 '18 at 6:28
Tianlalu
3,08621038
3,08621038
asked Dec 6 '18 at 6:10
ReginaRegina
55
asked Dec 6 '18 at 6:10
ReginaRegina
55
asked Dec 6 '18 at 6:10
ReginaRegina
55
55
add a comment |
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
$$int_{-M}^{M} int_{-M}^{M} e^{{-(x^{2}+y^{2}})} dxdy\
=(int_{-M}^{M} e^{-x^{2}}dx )(int_{-M}^{M} e^{-y^{2}}dy)\
=((int_{-M}^{M} e^{-x^{2}}dx )^{2}$$ and
$$int_{-M}^{M} u(frac l {sqrt 2} ,frac l {sqrt 2})dl=int_{-M}^{M} e^{-l^{2}}dl$$
edited Dec 6 '18 at 6:20
Arthur
113k7110193
answered Dec 6 '18 at 6:19
Kavi Rama MurthyKavi Rama Murthy
55.9k42158
$endgroup$
$begingroup$
Thank you so much for your help. I got caught up trying to solve the integrals and looking at the theory that I didn't realice how easy it was to separate the integral.
$endgroup$
– Regina
Dec 6 '18 at 6:42
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$$int_{-M}^{M} int_{-M}^{M} e^{{-(x^{2}+y^{2}})} dxdy\
=(int_{-M}^{M} e^{-x^{2}}dx )(int_{-M}^{M} e^{-y^{2}}dy)\
=((int_{-M}^{M} e^{-x^{2}}dx )^{2}$$ and
$$int_{-M}^{M} u(frac l {sqrt 2} ,frac l {sqrt 2})dl=int_{-M}^{M} e^{-l^{2}}dl$$
edited Dec 6 '18 at 6:20
Arthur
113k7110193
answered Dec 6 '18 at 6:19
Kavi Rama MurthyKavi Rama Murthy
55.9k42158
$endgroup$
$begingroup$
Thank you so much for your help. I got caught up trying to solve the integrals and looking at the theory that I didn't realice how easy it was to separate the integral.
$endgroup$
– Regina
Dec 6 '18 at 6:42
add a comment |
$begingroup$
$$int_{-M}^{M} int_{-M}^{M} e^{{-(x^{2}+y^{2}})} dxdy\
=(int_{-M}^{M} e^{-x^{2}}dx )(int_{-M}^{M} e^{-y^{2}}dy)\
=((int_{-M}^{M} e^{-x^{2}}dx )^{2}$$ and
$$int_{-M}^{M} u(frac l {sqrt 2} ,frac l {sqrt 2})dl=int_{-M}^{M} e^{-l^{2}}dl$$
edited Dec 6 '18 at 6:20
Arthur
113k7110193
answered Dec 6 '18 at 6:19
Kavi Rama MurthyKavi Rama Murthy
55.9k42158
$endgroup$
$begingroup$
Thank you so much for your help. I got caught up trying to solve the integrals and looking at the theory that I didn't realice how easy it was to separate the integral.
$endgroup$
– Regina
Dec 6 '18 at 6:42
add a comment |
$begingroup$
$$int_{-M}^{M} int_{-M}^{M} e^{{-(x^{2}+y^{2}})} dxdy\
=(int_{-M}^{M} e^{-x^{2}}dx )(int_{-M}^{M} e^{-y^{2}}dy)\
=((int_{-M}^{M} e^{-x^{2}}dx )^{2}$$ and
$$int_{-M}^{M} u(frac l {sqrt 2} ,frac l {sqrt 2})dl=int_{-M}^{M} e^{-l^{2}}dl$$
edited Dec 6 '18 at 6:20
Arthur
113k7110193
answered Dec 6 '18 at 6:19
Kavi Rama MurthyKavi Rama Murthy
55.9k42158
$endgroup$
$$int_{-M}^{M} int_{-M}^{M} e^{{-(x^{2}+y^{2}})} dxdy\
=(int_{-M}^{M} e^{-x^{2}}dx )(int_{-M}^{M} e^{-y^{2}}dy)\
=((int_{-M}^{M} e^{-x^{2}}dx )^{2}$$ and
$$int_{-M}^{M} u(frac l {sqrt 2} ,frac l {sqrt 2})dl=int_{-M}^{M} e^{-l^{2}}dl$$
edited Dec 6 '18 at 6:20
Arthur
113k7110193
answered Dec 6 '18 at 6:19
Kavi Rama MurthyKavi Rama Murthy
55.9k42158
edited Dec 6 '18 at 6:20
Arthur
113k7110193
edited Dec 6 '18 at 6:20
Arthur
113k7110193
edited Dec 6 '18 at 6:20
Arthur
113k7110193
113k7110193
answered Dec 6 '18 at 6:19
Kavi Rama MurthyKavi Rama Murthy
55.9k42158
answered Dec 6 '18 at 6:19
Kavi Rama MurthyKavi Rama Murthy
55.9k42158
answered Dec 6 '18 at 6:19
Kavi Rama MurthyKavi Rama Murthy
55.9k42158
55.9k42158
$begingroup$
Thank you so much for your help. I got caught up trying to solve the integrals and looking at the theory that I didn't realice how easy it was to separate the integral.
$endgroup$
– Regina
Dec 6 '18 at 6:42
add a comment |
$begingroup$
Thank you so much for your help. I got caught up trying to solve the integrals and looking at the theory that I didn't realice how easy it was to separate the integral.
$endgroup$
– Regina
Dec 6 '18 at 6:42
$begingroup$
Thank you so much for your help. I got caught up trying to solve the integrals and looking at the theory that I didn't realice how easy it was to separate the integral.
$endgroup$
– Regina
Dec 6 '18 at 6:42
$begingroup$
Thank you so much for your help. I got caught up trying to solve the integrals and looking at the theory that I didn't realice how easy it was to separate the integral.
$endgroup$
– Regina
Dec 6 '18 at 6:42
add a comment |
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