Mixing
Solving an integral in polar coordinates
I am given the potential:
$$phi (x, y) = frac{k}{2}(x^2+y^2) + axy$$
Where a is a constant.
I have to compute:
$$int_{-infty}^infty int_{-infty}^infty mathrm e^{-betaphi}dxdy$$
Changing to polar coordinates:
$$int_{0}^{2pi}int_{0}^{b} e^{-betaphi}rdrdtheta$$
What I have done is:
$$iint e^{-betaphi}rdrdtheta = iint e^{-frac{beta Kr^2}{2}} e^{-beta a r^2 costheta sintheta}rdrdtheta = iint e^{-beta f(r, theta)}rdrdtheta$$
Where
$$f(r, theta) = frac{Kr^2}{2} [(1+frac{cos theta sin theta}{K})^2 -(frac{cos theta sin theta}{K})^2]$$
Now I guess I have to do a substitution, defining a new variable. But I do not get the right one, as I do not get a simpler form.
EDIT
Using polar coordinates here is not the best method.
Take squares and try:
$$x = u + v$$
$$y = u - v$$
calculus integration polar-coordinates
asked Nov 22 '18 at 11:36
JD_PMJD_PM
10410
add a comment |
I am given the potential:
$$phi (x, y) = frac{k}{2}(x^2+y^2) + axy$$
Where a is a constant.
I have to compute:
$$int_{-infty}^infty int_{-infty}^infty mathrm e^{-betaphi}dxdy$$
Changing to polar coordinates:
$$int_{0}^{2pi}int_{0}^{b} e^{-betaphi}rdrdtheta$$
What I have done is:
$$iint e^{-betaphi}rdrdtheta = iint e^{-frac{beta Kr^2}{2}} e^{-beta a r^2 costheta sintheta}rdrdtheta = iint e^{-beta f(r, theta)}rdrdtheta$$
Where
$$f(r, theta) = frac{Kr^2}{2} [(1+frac{cos theta sin theta}{K})^2 -(frac{cos theta sin theta}{K})^2]$$
Now I guess I have to do a substitution, defining a new variable. But I do not get the right one, as I do not get a simpler form.
EDIT
Using polar coordinates here is not the best method.
Take squares and try:
$$x = u + v$$
$$y = u - v$$
calculus integration polar-coordinates
asked Nov 22 '18 at 11:36
JD_PMJD_PM
10410
1
Is the integral meant to be indefinite?
– J.G.
Nov 22 '18 at 11:55
@J.G. I'm not provided with the extremes but I guessed them. Note that $b$ is the distance with respect to a reference point.
– JD_PM
Nov 22 '18 at 12:04
See math.stackexchange.com/questions/1096793/…
– JD_PM
Nov 23 '18 at 15:34
add a comment |
I am given the potential:
$$phi (x, y) = frac{k}{2}(x^2+y^2) + axy$$
Where a is a constant.
I have to compute:
$$int_{-infty}^infty int_{-infty}^infty mathrm e^{-betaphi}dxdy$$
Changing to polar coordinates:
$$int_{0}^{2pi}int_{0}^{b} e^{-betaphi}rdrdtheta$$
What I have done is:
$$iint e^{-betaphi}rdrdtheta = iint e^{-frac{beta Kr^2}{2}} e^{-beta a r^2 costheta sintheta}rdrdtheta = iint e^{-beta f(r, theta)}rdrdtheta$$
Where
$$f(r, theta) = frac{Kr^2}{2} [(1+frac{cos theta sin theta}{K})^2 -(frac{cos theta sin theta}{K})^2]$$
Now I guess I have to do a substitution, defining a new variable. But I do not get the right one, as I do not get a simpler form.
EDIT
Using polar coordinates here is not the best method.
Take squares and try:
$$x = u + v$$
$$y = u - v$$
calculus integration polar-coordinates
asked Nov 22 '18 at 11:36
JD_PMJD_PM
10410
I am given the potential:
$$phi (x, y) = frac{k}{2}(x^2+y^2) + axy$$
Where a is a constant.
I have to compute:
$$int_{-infty}^infty int_{-infty}^infty mathrm e^{-betaphi}dxdy$$
Changing to polar coordinates:
$$int_{0}^{2pi}int_{0}^{b} e^{-betaphi}rdrdtheta$$
What I have done is:
$$iint e^{-betaphi}rdrdtheta = iint e^{-frac{beta Kr^2}{2}} e^{-beta a r^2 costheta sintheta}rdrdtheta = iint e^{-beta f(r, theta)}rdrdtheta$$
Where
$$f(r, theta) = frac{Kr^2}{2} [(1+frac{cos theta sin theta}{K})^2 -(frac{cos theta sin theta}{K})^2]$$
Now I guess I have to do a substitution, defining a new variable. But I do not get the right one, as I do not get a simpler form.
EDIT
Using polar coordinates here is not the best method.
Take squares and try:
$$x = u + v$$
$$y = u - v$$
calculus integration polar-coordinates
calculus integration polar-coordinates
asked Nov 22 '18 at 11:36
JD_PMJD_PM
10410
asked Nov 22 '18 at 11:36
JD_PMJD_PM
10410
edited Nov 23 '18 at 15:33
JD_PM
asked Nov 22 '18 at 11:36
JD_PMJD_PM
10410
asked Nov 22 '18 at 11:36
JD_PMJD_PM
10410
asked Nov 22 '18 at 11:36
JD_PMJD_PM
10410
10410
1
Is the integral meant to be indefinite?
– J.G.
Nov 22 '18 at 11:55
@J.G. I'm not provided with the extremes but I guessed them. Note that $b$ is the distance with respect to a reference point.
– JD_PM
Nov 22 '18 at 12:04
See math.stackexchange.com/questions/1096793/…
– JD_PM
Nov 23 '18 at 15:34
add a comment |
1
Is the integral meant to be indefinite?
– J.G.
Nov 22 '18 at 11:55
@J.G. I'm not provided with the extremes but I guessed them. Note that $b$ is the distance with respect to a reference point.
– JD_PM
Nov 22 '18 at 12:04
See math.stackexchange.com/questions/1096793/…
– JD_PM
Nov 23 '18 at 15:34
1
1
Is the integral meant to be indefinite?
– J.G.
Nov 22 '18 at 11:55
Is the integral meant to be indefinite?
– J.G.
Nov 22 '18 at 11:55
@J.G. I'm not provided with the extremes but I guessed them. Note that $b$ is the distance with respect to a reference point.
– JD_PM
Nov 22 '18 at 12:04
@J.G. I'm not provided with the extremes but I guessed them. Note that $b$ is the distance with respect to a reference point.
– JD_PM
Nov 22 '18 at 12:04
See math.stackexchange.com/questions/1096793/…
– JD_PM
Nov 23 '18 at 15:34
See math.stackexchange.com/questions/1096793/…
– JD_PM
Nov 23 '18 at 15:34
add a comment |
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1
Is the integral meant to be indefinite?
– J.G.
Nov 22 '18 at 11:55
@J.G. I'm not provided with the extremes but I guessed them. Note that $b$ is the distance with respect to a reference point.
– JD_PM
Nov 22 '18 at 12:04
See math.stackexchange.com/questions/1096793/…
– JD_PM
Nov 23 '18 at 15:34