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How to Calculate $int_{S} xyz~d{sigma}$ where $S$ is the portion $x+y+z=1$ in the first octant $ 0 leq x , 0...
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Calculate $int_{S} xyz~d{sigma}$ where $S$ :
is the portion $x+y+z=1$ in the first octant $ 0 leq x , 0 leq y , 0 leq z$ .
should i calculate
$sqrt{3}int_0^1int_0^{1-x} (xy-x^2y-xy^2) dy~dx$ =
= $sqrt{3}~int_0^1int_0^1 (u-1)(u-v)(v)~du~dv$
where :$ ~~~~~u = x+y ~,~~~~ v=x ~~~, ~~~Jacobian = -1$
and so the full answer is $boxed{frac{sqrt{3}}{12}}$
multivariable-calculus surface-integrals
asked Jan 15 at 11:14
Mather Mather
3367
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add a comment |
$begingroup$
Calculate $int_{S} xyz~d{sigma}$ where $S$ :
is the portion $x+y+z=1$ in the first octant $ 0 leq x , 0 leq y , 0 leq z$ .
should i calculate
$sqrt{3}int_0^1int_0^{1-x} (xy-x^2y-xy^2) dy~dx$ =
= $sqrt{3}~int_0^1int_0^1 (u-1)(u-v)(v)~du~dv$
where :$ ~~~~~u = x+y ~,~~~~ v=x ~~~, ~~~Jacobian = -1$
and so the full answer is $boxed{frac{sqrt{3}}{12}}$
multivariable-calculus surface-integrals
asked Jan 15 at 11:14
Mather Mather
3367
$endgroup$
add a comment |
$begingroup$
Calculate $int_{S} xyz~d{sigma}$ where $S$ :
is the portion $x+y+z=1$ in the first octant $ 0 leq x , 0 leq y , 0 leq z$ .
should i calculate
$sqrt{3}int_0^1int_0^{1-x} (xy-x^2y-xy^2) dy~dx$ =
= $sqrt{3}~int_0^1int_0^1 (u-1)(u-v)(v)~du~dv$
where :$ ~~~~~u = x+y ~,~~~~ v=x ~~~, ~~~Jacobian = -1$
and so the full answer is $boxed{frac{sqrt{3}}{12}}$
multivariable-calculus surface-integrals
asked Jan 15 at 11:14
Mather Mather
3367
$endgroup$
Calculate $int_{S} xyz~d{sigma}$ where $S$ :
is the portion $x+y+z=1$ in the first octant $ 0 leq x , 0 leq y , 0 leq z$ .
should i calculate
$sqrt{3}int_0^1int_0^{1-x} (xy-x^2y-xy^2) dy~dx$ =
= $sqrt{3}~int_0^1int_0^1 (u-1)(u-v)(v)~du~dv$
where :$ ~~~~~u = x+y ~,~~~~ v=x ~~~, ~~~Jacobian = -1$
and so the full answer is $boxed{frac{sqrt{3}}{12}}$
multivariable-calculus surface-integrals
multivariable-calculus surface-integrals
asked Jan 15 at 11:14
Mather Mather
3367
asked Jan 15 at 11:14
Mather Mather
3367
edited Jan 15 at 11:24
Mather
asked Jan 15 at 11:14
Mather Mather
3367
asked Jan 15 at 11:14
Mather Mather
3367
asked Jan 15 at 11:14
Mather Mather
3367
3367
add a comment |
add a comment |
1 Answer
1
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The integrated region in the $xy$ plane is ${(x,y):x,yge0,x+yle1}$. $$0le xle1implies0le vle1\0le yle1implies0le u-vle1\0le x+yle1implies0le ule1$$
The integrated region in $uv$ plane is the intersection of the solution regions of the above three inequalities in $u,v$;$${(u,v):0le vle ule 1,vle1}$$which is not very different from the original integral. Also note that when applying change of coordinates, you are supposed to multiply by the absolute value of the jacobian. The integral should look like$$sqrt{3}int_0^1int_v^1v(1-u)(u-v)~du~dv$$which gives the answer as $dfrac{sqrt3}{120}$.
Remark. You could just as well avoid the change of coordinates and solve it directly.$$sqrt3int_0^1int_0^{1-x} xy(1-x-y)dy~dx=sqrt3int_0^1int_0^{1-x}x(1-x)y-xy^2dy~dx\=sqrt3int_0^1frac{x(1-x)^3}6dx=sqrt3int_0^1frac{x^3(1-x)}6dx=frac{sqrt3}{120}$$
answered Jan 15 at 12:01
Shubham JohriShubham Johri
5,177717
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$begingroup$
i didn't understand why $0le yle1implies0le u-vle1$
$endgroup$
– Mather
Jan 15 at 12:14
$begingroup$
It is because $y$ is $(x+y)-x=u-v$
$endgroup$
– Shubham Johri
Jan 15 at 12:15
$begingroup$
but $ 0 leq y leq (1-x) $ not $ 0 leq y leq 1$
$endgroup$
– Mather
Jan 15 at 12:18
$begingroup$
Yes, that is covered in the third statement. $x+yle1implies ule1$
$endgroup$
– Shubham Johri
Jan 15 at 12:19
$begingroup$
if $y=1$ and $x=1$ then $z < 0$ and they want $0 leq z $
$endgroup$
– Mather
Jan 15 at 12:20
|
show 8 more comments
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The integrated region in the $xy$ plane is ${(x,y):x,yge0,x+yle1}$. $$0le xle1implies0le vle1\0le yle1implies0le u-vle1\0le x+yle1implies0le ule1$$The integrated region in $uv$ plane is the intersection of the solution regions of the above three inequalities in $u,v$;$${(u,v):0le vle ule 1,vle1}$$which is not very different from the original integral. Also note that when applying change of coordinates, you are supposed to multiply by the absolute value of the jacobian. The integral should look like$$sqrt{3}int_0^1int_v^1v(1-u)(u-v)~du~dv$$which gives the answer as $dfrac{sqrt3}{120}$.
Remark. You could just as well avoid the change of coordinates and solve it directly.$$sqrt3int_0^1int_0^{1-x} xy(1-x-y)dy~dx=sqrt3int_0^1int_0^{1-x}x(1-x)y-xy^2dy~dx\=sqrt3int_0^1frac{x(1-x)^3}6dx=sqrt3int_0^1frac{x^3(1-x)}6dx=frac{sqrt3}{120}$$
answered Jan 15 at 12:01
Shubham JohriShubham Johri
5,177717
$endgroup$
$begingroup$
i didn't understand why $0le yle1implies0le u-vle1$
$endgroup$
– Mather
Jan 15 at 12:14
$begingroup$
It is because $y$ is $(x+y)-x=u-v$
$endgroup$
– Shubham Johri
Jan 15 at 12:15
$begingroup$
but $ 0 leq y leq (1-x) $ not $ 0 leq y leq 1$
$endgroup$
– Mather
Jan 15 at 12:18
$begingroup$
Yes, that is covered in the third statement. $x+yle1implies ule1$
$endgroup$
– Shubham Johri
Jan 15 at 12:19
$begingroup$
if $y=1$ and $x=1$ then $z < 0$ and they want $0 leq z $
$endgroup$
– Mather
Jan 15 at 12:20
|
show 8 more comments
$begingroup$
The integrated region in the $xy$ plane is ${(x,y):x,yge0,x+yle1}$. $$0le xle1implies0le vle1\0le yle1implies0le u-vle1\0le x+yle1implies0le ule1$$The integrated region in $uv$ plane is the intersection of the solution regions of the above three inequalities in $u,v$;$${(u,v):0le vle ule 1,vle1}$$which is not very different from the original integral. Also note that when applying change of coordinates, you are supposed to multiply by the absolute value of the jacobian. The integral should look like$$sqrt{3}int_0^1int_v^1v(1-u)(u-v)~du~dv$$which gives the answer as $dfrac{sqrt3}{120}$.
Remark. You could just as well avoid the change of coordinates and solve it directly.$$sqrt3int_0^1int_0^{1-x} xy(1-x-y)dy~dx=sqrt3int_0^1int_0^{1-x}x(1-x)y-xy^2dy~dx\=sqrt3int_0^1frac{x(1-x)^3}6dx=sqrt3int_0^1frac{x^3(1-x)}6dx=frac{sqrt3}{120}$$
answered Jan 15 at 12:01
Shubham JohriShubham Johri
5,177717
$endgroup$
$begingroup$
i didn't understand why $0le yle1implies0le u-vle1$
$endgroup$
– Mather
Jan 15 at 12:14
$begingroup$
It is because $y$ is $(x+y)-x=u-v$
$endgroup$
– Shubham Johri
Jan 15 at 12:15
$begingroup$
but $ 0 leq y leq (1-x) $ not $ 0 leq y leq 1$
$endgroup$
– Mather
Jan 15 at 12:18
$begingroup$
Yes, that is covered in the third statement. $x+yle1implies ule1$
$endgroup$
– Shubham Johri
Jan 15 at 12:19
$begingroup$
if $y=1$ and $x=1$ then $z < 0$ and they want $0 leq z $
$endgroup$
– Mather
Jan 15 at 12:20
|
show 8 more comments
$begingroup$
The integrated region in the $xy$ plane is ${(x,y):x,yge0,x+yle1}$. $$0le xle1implies0le vle1\0le yle1implies0le u-vle1\0le x+yle1implies0le ule1$$The integrated region in $uv$ plane is the intersection of the solution regions of the above three inequalities in $u,v$;$${(u,v):0le vle ule 1,vle1}$$which is not very different from the original integral. Also note that when applying change of coordinates, you are supposed to multiply by the absolute value of the jacobian. The integral should look like$$sqrt{3}int_0^1int_v^1v(1-u)(u-v)~du~dv$$which gives the answer as $dfrac{sqrt3}{120}$.
Remark. You could just as well avoid the change of coordinates and solve it directly.$$sqrt3int_0^1int_0^{1-x} xy(1-x-y)dy~dx=sqrt3int_0^1int_0^{1-x}x(1-x)y-xy^2dy~dx\=sqrt3int_0^1frac{x(1-x)^3}6dx=sqrt3int_0^1frac{x^3(1-x)}6dx=frac{sqrt3}{120}$$
answered Jan 15 at 12:01
Shubham JohriShubham Johri
5,177717
$endgroup$
The integrated region in the $xy$ plane is ${(x,y):x,yge0,x+yle1}$. $$0le xle1implies0le vle1\0le yle1implies0le u-vle1\0le x+yle1implies0le ule1$$The integrated region in $uv$ plane is the intersection of the solution regions of the above three inequalities in $u,v$;$${(u,v):0le vle ule 1,vle1}$$which is not very different from the original integral. Also note that when applying change of coordinates, you are supposed to multiply by the absolute value of the jacobian. The integral should look like$$sqrt{3}int_0^1int_v^1v(1-u)(u-v)~du~dv$$which gives the answer as $dfrac{sqrt3}{120}$.
Remark. You could just as well avoid the change of coordinates and solve it directly.$$sqrt3int_0^1int_0^{1-x} xy(1-x-y)dy~dx=sqrt3int_0^1int_0^{1-x}x(1-x)y-xy^2dy~dx\=sqrt3int_0^1frac{x(1-x)^3}6dx=sqrt3int_0^1frac{x^3(1-x)}6dx=frac{sqrt3}{120}$$
answered Jan 15 at 12:01
Shubham JohriShubham Johri
5,177717
edited Jan 15 at 12:35
answered Jan 15 at 12:01
Shubham JohriShubham Johri
5,177717
answered Jan 15 at 12:01
Shubham JohriShubham Johri
5,177717
answered Jan 15 at 12:01
Shubham JohriShubham Johri
5,177717
5,177717
$begingroup$
i didn't understand why $0le yle1implies0le u-vle1$
$endgroup$
– Mather
Jan 15 at 12:14
$begingroup$
It is because $y$ is $(x+y)-x=u-v$
$endgroup$
– Shubham Johri
Jan 15 at 12:15
$begingroup$
but $ 0 leq y leq (1-x) $ not $ 0 leq y leq 1$
$endgroup$
– Mather
Jan 15 at 12:18
$begingroup$
Yes, that is covered in the third statement. $x+yle1implies ule1$
$endgroup$
– Shubham Johri
Jan 15 at 12:19
$begingroup$
if $y=1$ and $x=1$ then $z < 0$ and they want $0 leq z $
$endgroup$
– Mather
Jan 15 at 12:20
|
show 8 more comments
$begingroup$
i didn't understand why $0le yle1implies0le u-vle1$
$endgroup$
– Mather
Jan 15 at 12:14
$begingroup$
It is because $y$ is $(x+y)-x=u-v$
$endgroup$
– Shubham Johri
Jan 15 at 12:15
$begingroup$
but $ 0 leq y leq (1-x) $ not $ 0 leq y leq 1$
$endgroup$
– Mather
Jan 15 at 12:18
$begingroup$
Yes, that is covered in the third statement. $x+yle1implies ule1$
$endgroup$
– Shubham Johri
Jan 15 at 12:19
$begingroup$
if $y=1$ and $x=1$ then $z < 0$ and they want $0 leq z $
$endgroup$
– Mather
Jan 15 at 12:20
$begingroup$
i didn't understand why $0le yle1implies0le u-vle1$
$endgroup$
– Mather
Jan 15 at 12:14
$begingroup$
i didn't understand why $0le yle1implies0le u-vle1$
$endgroup$
– Mather
Jan 15 at 12:14
$begingroup$
It is because $y$ is $(x+y)-x=u-v$
$endgroup$
– Shubham Johri
Jan 15 at 12:15
$begingroup$
It is because $y$ is $(x+y)-x=u-v$
$endgroup$
– Shubham Johri
Jan 15 at 12:15
$begingroup$
but $ 0 leq y leq (1-x) $ not $ 0 leq y leq 1$
$endgroup$
– Mather
Jan 15 at 12:18
$begingroup$
but $ 0 leq y leq (1-x) $ not $ 0 leq y leq 1$
$endgroup$
– Mather
Jan 15 at 12:18
$begingroup$
Yes, that is covered in the third statement. $x+yle1implies ule1$
$endgroup$
– Shubham Johri
Jan 15 at 12:19
$begingroup$
Yes, that is covered in the third statement. $x+yle1implies ule1$
$endgroup$
– Shubham Johri
Jan 15 at 12:19
$begingroup$
if $y=1$ and $x=1$ then $z < 0$ and they want $0 leq z $
$endgroup$
– Mather
Jan 15 at 12:20
$begingroup$
if $y=1$ and $x=1$ then $z < 0$ and they want $0 leq z $
$endgroup$
– Mather
Jan 15 at 12:20
|
show 8 more comments
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