Mixing
Problem related to multiple integral
$begingroup$
$g : mathbb{R}^{3} to mathbb{R}^{3} $ be a function defined by $g(x,y,z) = (3y+4z, 2x-3z, x+3y).$
Also let $S = {(x,y,z)in mathbb{R}^{3}: 0leq x leq 1, 0leq y leq 1, 0leq zleq 1 }$
Now the question is
If $$intintint_{g(s)}(2x+y-2z)dxdydz = alphaintintint_{s}zdxdydz,$$ then $alpha = ?$
The way I was thinking was not taking me too far. I was thinking $(3y+4z,2x-3z,x+3y) =(u,v,w)$ and was considering the scenario to be a change of variables like scenario but it's fruitless.
Can anyone guide me ? Any suggestions will be highly appreciated. Thank you
integration
asked Jan 28 at 8:33
hiren_garaihiren_garai
515417
$endgroup$
add a comment |
$begingroup$
$g : mathbb{R}^{3} to mathbb{R}^{3} $ be a function defined by $g(x,y,z) = (3y+4z, 2x-3z, x+3y).$
Also let $S = {(x,y,z)in mathbb{R}^{3}: 0leq x leq 1, 0leq y leq 1, 0leq zleq 1 }$
Now the question is
If $$intintint_{g(s)}(2x+y-2z)dxdydz = alphaintintint_{s}zdxdydz,$$ then $alpha = ?$
The way I was thinking was not taking me too far. I was thinking $(3y+4z,2x-3z,x+3y) =(u,v,w)$ and was considering the scenario to be a change of variables like scenario but it's fruitless.
Can anyone guide me ? Any suggestions will be highly appreciated. Thank you
integration
asked Jan 28 at 8:33
hiren_garaihiren_garai
515417
$endgroup$
add a comment |
$begingroup$
$g : mathbb{R}^{3} to mathbb{R}^{3} $ be a function defined by $g(x,y,z) = (3y+4z, 2x-3z, x+3y).$
Also let $S = {(x,y,z)in mathbb{R}^{3}: 0leq x leq 1, 0leq y leq 1, 0leq zleq 1 }$
Now the question is
If $$intintint_{g(s)}(2x+y-2z)dxdydz = alphaintintint_{s}zdxdydz,$$ then $alpha = ?$
The way I was thinking was not taking me too far. I was thinking $(3y+4z,2x-3z,x+3y) =(u,v,w)$ and was considering the scenario to be a change of variables like scenario but it's fruitless.
Can anyone guide me ? Any suggestions will be highly appreciated. Thank you
integration
asked Jan 28 at 8:33
hiren_garaihiren_garai
515417
$endgroup$
$g : mathbb{R}^{3} to mathbb{R}^{3} $ be a function defined by $g(x,y,z) = (3y+4z, 2x-3z, x+3y).$
Also let $S = {(x,y,z)in mathbb{R}^{3}: 0leq x leq 1, 0leq y leq 1, 0leq zleq 1 }$
Now the question is
If $$intintint_{g(s)}(2x+y-2z)dxdydz = alphaintintint_{s}zdxdydz,$$ then $alpha = ?$
The way I was thinking was not taking me too far. I was thinking $(3y+4z,2x-3z,x+3y) =(u,v,w)$ and was considering the scenario to be a change of variables like scenario but it's fruitless.
Can anyone guide me ? Any suggestions will be highly appreciated. Thank you
integration
integration
asked Jan 28 at 8:33
hiren_garaihiren_garai
515417
asked Jan 28 at 8:33
hiren_garaihiren_garai
515417
asked Jan 28 at 8:33
hiren_garaihiren_garai
515417
asked Jan 28 at 8:33
hiren_garaihiren_garai
515417
asked Jan 28 at 8:33
hiren_garaihiren_garai
515417
515417
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Indeed, this is a nice change of variables problem. It's a bit confusing to use the same variables for both sides of the equation though.
After setting $(3y+4z,2x−3z,x+3y)=(u,v,w)$, the left hand side of the equation should be written using these coordinates. It is then very easy to see :
$ 2u + v -2w = 5z$. Do not forget to include the determinant of the transformation $g$ in your final answer.
answered Jan 28 at 9:00
ayberkayberk
33715
$endgroup$
$begingroup$
ah ! I see. O e thing, by equation you mean the integral one, right ?
$endgroup$
– hiren_garai
Jan 28 at 9:06
$begingroup$
Yes. The one that has triple integrals on both sides.
$endgroup$
– ayberk
Jan 28 at 9:25
$begingroup$
But then I would have to change the $dxdydz$ according to the settings.
$endgroup$
– hiren_garai
Jan 28 at 9:30
$begingroup$
Exactly, the left hand side should be: $int_{g(S)} (2u+v−2w) du dv dw $
$endgroup$
– ayberk
Jan 28 at 11:11
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Indeed, this is a nice change of variables problem. It's a bit confusing to use the same variables for both sides of the equation though.
After setting $(3y+4z,2x−3z,x+3y)=(u,v,w)$, the left hand side of the equation should be written using these coordinates. It is then very easy to see :
$ 2u + v -2w = 5z$. Do not forget to include the determinant of the transformation $g$ in your final answer.
answered Jan 28 at 9:00
ayberkayberk
33715
$endgroup$
$begingroup$
ah ! I see. O e thing, by equation you mean the integral one, right ?
$endgroup$
– hiren_garai
Jan 28 at 9:06
$begingroup$
Yes. The one that has triple integrals on both sides.
$endgroup$
– ayberk
Jan 28 at 9:25
$begingroup$
But then I would have to change the $dxdydz$ according to the settings.
$endgroup$
– hiren_garai
Jan 28 at 9:30
$begingroup$
Exactly, the left hand side should be: $int_{g(S)} (2u+v−2w) du dv dw $
$endgroup$
– ayberk
Jan 28 at 11:11
add a comment |
$begingroup$
Indeed, this is a nice change of variables problem. It's a bit confusing to use the same variables for both sides of the equation though.
After setting $(3y+4z,2x−3z,x+3y)=(u,v,w)$, the left hand side of the equation should be written using these coordinates. It is then very easy to see :
$ 2u + v -2w = 5z$. Do not forget to include the determinant of the transformation $g$ in your final answer.
answered Jan 28 at 9:00
ayberkayberk
33715
$endgroup$
$begingroup$
ah ! I see. O e thing, by equation you mean the integral one, right ?
$endgroup$
– hiren_garai
Jan 28 at 9:06
$begingroup$
Yes. The one that has triple integrals on both sides.
$endgroup$
– ayberk
Jan 28 at 9:25
$begingroup$
But then I would have to change the $dxdydz$ according to the settings.
$endgroup$
– hiren_garai
Jan 28 at 9:30
$begingroup$
Exactly, the left hand side should be: $int_{g(S)} (2u+v−2w) du dv dw $
$endgroup$
– ayberk
Jan 28 at 11:11
add a comment |
$begingroup$
Indeed, this is a nice change of variables problem. It's a bit confusing to use the same variables for both sides of the equation though.
After setting $(3y+4z,2x−3z,x+3y)=(u,v,w)$, the left hand side of the equation should be written using these coordinates. It is then very easy to see :
$ 2u + v -2w = 5z$. Do not forget to include the determinant of the transformation $g$ in your final answer.
answered Jan 28 at 9:00
ayberkayberk
33715
$endgroup$
Indeed, this is a nice change of variables problem. It's a bit confusing to use the same variables for both sides of the equation though.
After setting $(3y+4z,2x−3z,x+3y)=(u,v,w)$, the left hand side of the equation should be written using these coordinates. It is then very easy to see :
$ 2u + v -2w = 5z$. Do not forget to include the determinant of the transformation $g$ in your final answer.
answered Jan 28 at 9:00
ayberkayberk
33715
answered Jan 28 at 9:00
ayberkayberk
33715
answered Jan 28 at 9:00
ayberkayberk
33715
answered Jan 28 at 9:00
ayberkayberk
33715
33715
$begingroup$
ah ! I see. O e thing, by equation you mean the integral one, right ?
$endgroup$
– hiren_garai
Jan 28 at 9:06
$begingroup$
Yes. The one that has triple integrals on both sides.
$endgroup$
– ayberk
Jan 28 at 9:25
$begingroup$
But then I would have to change the $dxdydz$ according to the settings.
$endgroup$
– hiren_garai
Jan 28 at 9:30
$begingroup$
Exactly, the left hand side should be: $int_{g(S)} (2u+v−2w) du dv dw $
$endgroup$
– ayberk
Jan 28 at 11:11
add a comment |
$begingroup$
ah ! I see. O e thing, by equation you mean the integral one, right ?
$endgroup$
– hiren_garai
Jan 28 at 9:06
$begingroup$
Yes. The one that has triple integrals on both sides.
$endgroup$
– ayberk
Jan 28 at 9:25
$begingroup$
But then I would have to change the $dxdydz$ according to the settings.
$endgroup$
– hiren_garai
Jan 28 at 9:30
$begingroup$
Exactly, the left hand side should be: $int_{g(S)} (2u+v−2w) du dv dw $
$endgroup$
– ayberk
Jan 28 at 11:11
$begingroup$
ah ! I see. O e thing, by equation you mean the integral one, right ?
$endgroup$
– hiren_garai
Jan 28 at 9:06
$begingroup$
ah ! I see. O e thing, by equation you mean the integral one, right ?
$endgroup$
– hiren_garai
Jan 28 at 9:06
$begingroup$
Yes. The one that has triple integrals on both sides.
$endgroup$
– ayberk
Jan 28 at 9:25
$begingroup$
Yes. The one that has triple integrals on both sides.
$endgroup$
– ayberk
Jan 28 at 9:25
$begingroup$
But then I would have to change the $dxdydz$ according to the settings.
$endgroup$
– hiren_garai
Jan 28 at 9:30
$begingroup$
But then I would have to change the $dxdydz$ according to the settings.
$endgroup$
– hiren_garai
Jan 28 at 9:30
$begingroup$
Exactly, the left hand side should be: $int_{g(S)} (2u+v−2w) du dv dw $
$endgroup$
– ayberk
Jan 28 at 11:11
$begingroup$
Exactly, the left hand side should be: $int_{g(S)} (2u+v−2w) du dv dw $
$endgroup$
– ayberk
Jan 28 at 11:11
add a comment |
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