Compute the integral >$int (x-y) dx +xy dy$ over the circle of radius 2 and centre at origin.
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Compute the integral
$int (x-y) dx +xy dy$ over the circle of radius 2 and centre at origin.
My try:
Put $x=2cos theta,y=2sin theta $
$int (x-y) dx +xy dy$ becomes
$-4int (cos theta -sin theta) sin theta dtheta+int 8sin theta cos^2 theta dtheta$
$=2 int 1-cos 2theta dtheta$
$=4pi$
Is my answer correct?
Please help.
integration multivariable-calculus
edited Nov 21 at 15:32
Nosrati
26.3k62353
asked Nov 21 at 15:30
Join_PhD
947
add a comment |
up vote
0
down vote
favorite
Compute the integral
$int (x-y) dx +xy dy$ over the circle of radius 2 and centre at origin.
My try:
Put $x=2cos theta,y=2sin theta $
$int (x-y) dx +xy dy$ becomes
$-4int (cos theta -sin theta) sin theta dtheta+int 8sin theta cos^2 theta dtheta$
$=2 int 1-cos 2theta dtheta$
$=4pi$
Is my answer correct?
Please help.
integration multivariable-calculus
edited Nov 21 at 15:32
Nosrati
26.3k62353
asked Nov 21 at 15:30
Join_PhD
947
1
Use Green's theorem.
– Nosrati
Nov 21 at 15:33
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Compute the integral
$int (x-y) dx +xy dy$ over the circle of radius 2 and centre at origin.
My try:
Put $x=2cos theta,y=2sin theta $
$int (x-y) dx +xy dy$ becomes
$-4int (cos theta -sin theta) sin theta dtheta+int 8sin theta cos^2 theta dtheta$
$=2 int 1-cos 2theta dtheta$
$=4pi$
Is my answer correct?
Please help.
integration multivariable-calculus
edited Nov 21 at 15:32
Nosrati
26.3k62353
asked Nov 21 at 15:30
Join_PhD
947
Compute the integral
$int (x-y) dx +xy dy$ over the circle of radius 2 and centre at origin.
My try:
Put $x=2cos theta,y=2sin theta $
$int (x-y) dx +xy dy$ becomes
$-4int (cos theta -sin theta) sin theta dtheta+int 8sin theta cos^2 theta dtheta$
$=2 int 1-cos 2theta dtheta$
$=4pi$
Is my answer correct?
Please help.
integration multivariable-calculus
integration multivariable-calculus
edited Nov 21 at 15:32
Nosrati
26.3k62353
asked Nov 21 at 15:30
Join_PhD
947
edited Nov 21 at 15:32
Nosrati
26.3k62353
asked Nov 21 at 15:30
Join_PhD
947
edited Nov 21 at 15:32
Nosrati
26.3k62353
edited Nov 21 at 15:32
Nosrati
26.3k62353
edited Nov 21 at 15:32
Nosrati
26.3k62353
26.3k62353
asked Nov 21 at 15:30
Join_PhD
947
asked Nov 21 at 15:30
Join_PhD
947
asked Nov 21 at 15:30
Join_PhD
947
947
1
Use Green's theorem.
– Nosrati
Nov 21 at 15:33
add a comment |
1
Use Green's theorem.
– Nosrati
Nov 21 at 15:33
1
1
Use Green's theorem.
– Nosrati
Nov 21 at 15:33
Use Green's theorem.
– Nosrati
Nov 21 at 15:33
add a comment |
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
Seems good. You could have also applied Green's theorem:
begin{align}
oint_{x^2+y^2 = 4} (x-y), mathrm dx + xy , mathrm dy &= iint_{x^2+y^2 leq 4} frac{partial}{partial x} (xy) - frac{partial}{partial y} (x-y) , mathrm dx ,mathrm dy\
&= iint_{x^2+y^2 leq 4} y+1 , mathrm dx ,mathrm dy\
&= iint_{x^2+y^2 leq 4} 1 , mathrm dx ,mathrm dy = 4pi,
end{align}
since
$$ iint_{x^2+y^2 leq 4} y , mathrm dx ,mathrm dy = 0 $$
by symmetry.
answered Nov 21 at 15:36
MisterRiemann
5,7041624
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Seems good. You could have also applied Green's theorem:
begin{align}
oint_{x^2+y^2 = 4} (x-y), mathrm dx + xy , mathrm dy &= iint_{x^2+y^2 leq 4} frac{partial}{partial x} (xy) - frac{partial}{partial y} (x-y) , mathrm dx ,mathrm dy\
&= iint_{x^2+y^2 leq 4} y+1 , mathrm dx ,mathrm dy\
&= iint_{x^2+y^2 leq 4} 1 , mathrm dx ,mathrm dy = 4pi,
end{align}
since
$$ iint_{x^2+y^2 leq 4} y , mathrm dx ,mathrm dy = 0 $$
by symmetry.
answered Nov 21 at 15:36
MisterRiemann
5,7041624
add a comment |
up vote
2
down vote
accepted
Seems good. You could have also applied Green's theorem:
begin{align}
oint_{x^2+y^2 = 4} (x-y), mathrm dx + xy , mathrm dy &= iint_{x^2+y^2 leq 4} frac{partial}{partial x} (xy) - frac{partial}{partial y} (x-y) , mathrm dx ,mathrm dy\
&= iint_{x^2+y^2 leq 4} y+1 , mathrm dx ,mathrm dy\
&= iint_{x^2+y^2 leq 4} 1 , mathrm dx ,mathrm dy = 4pi,
end{align}
since
$$ iint_{x^2+y^2 leq 4} y , mathrm dx ,mathrm dy = 0 $$
by symmetry.
answered Nov 21 at 15:36
MisterRiemann
5,7041624
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Seems good. You could have also applied Green's theorem:
begin{align}
oint_{x^2+y^2 = 4} (x-y), mathrm dx + xy , mathrm dy &= iint_{x^2+y^2 leq 4} frac{partial}{partial x} (xy) - frac{partial}{partial y} (x-y) , mathrm dx ,mathrm dy\
&= iint_{x^2+y^2 leq 4} y+1 , mathrm dx ,mathrm dy\
&= iint_{x^2+y^2 leq 4} 1 , mathrm dx ,mathrm dy = 4pi,
end{align}
since
$$ iint_{x^2+y^2 leq 4} y , mathrm dx ,mathrm dy = 0 $$
by symmetry.
answered Nov 21 at 15:36
MisterRiemann
5,7041624
Seems good. You could have also applied Green's theorem:
begin{align}
oint_{x^2+y^2 = 4} (x-y), mathrm dx + xy , mathrm dy &= iint_{x^2+y^2 leq 4} frac{partial}{partial x} (xy) - frac{partial}{partial y} (x-y) , mathrm dx ,mathrm dy\
&= iint_{x^2+y^2 leq 4} y+1 , mathrm dx ,mathrm dy\
&= iint_{x^2+y^2 leq 4} 1 , mathrm dx ,mathrm dy = 4pi,
end{align}
since
$$ iint_{x^2+y^2 leq 4} y , mathrm dx ,mathrm dy = 0 $$
by symmetry.
answered Nov 21 at 15:36
MisterRiemann
5,7041624
answered Nov 21 at 15:36
MisterRiemann
5,7041624
answered Nov 21 at 15:36
MisterRiemann
5,7041624
answered Nov 21 at 15:36
MisterRiemann
5,7041624
5,7041624
add a comment |
add a comment |
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Use Green's theorem.
– Nosrati
Nov 21 at 15:33