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Derivative of a double integral over a circular region

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8

2

$begingroup$

Calculate the following derivative

$$fracddtiint_D_tF(x,y,t) , mathrm d x mathrm d y$$

where

$$D_t = lbrace (x,y) mid (x-t)^2+(y-t)^2leq r^2 rbrace$$

I've read here that the answer to the above question is in the form of:
beginalignfracddtiint_D_tF(x,y,t)dxdy &=int_partial D_t F(udy-vdx) + iint_D_tfracpartial Fpartial tdx dy \
&=iint_D_tleft[textdiv(Fmathbfv)+fracpartial Fpartial tright]dx dyendalign
which is a generalization of Leibniz integral rule.

I don't know how I can calculate $mathbfv$ and $mathbfn$ or $u$ and $v$ in my problem. The paper states that $mathbfv$ with components $u$ and $v$ is the velocity and $mathbfn$ is the normal vector.

Also, I would be grateful if someone could introduce some other references on the derivative of such double integrals to me.

calculus integration multivariable-calculus derivatives reference-request

share|cite|improve this question

edited 8 hours ago

Rodrigo de Azevedo

13.4k42165

asked 8 hours ago

SMA.DSMA.D

506421

$endgroup$

add a comment | 

8

2

$begingroup$

Calculate the following derivative

$$fracddtiint_D_tF(x,y,t) , mathrm d x mathrm d y$$

where

$$D_t = lbrace (x,y) mid (x-t)^2+(y-t)^2leq r^2 rbrace$$

I've read here that the answer to the above question is in the form of:
beginalignfracddtiint_D_tF(x,y,t)dxdy &=int_partial D_t F(udy-vdx) + iint_D_tfracpartial Fpartial tdx dy \
&=iint_D_tleft[textdiv(Fmathbfv)+fracpartial Fpartial tright]dx dyendalign
which is a generalization of Leibniz integral rule.

I don't know how I can calculate $mathbfv$ and $mathbfn$ or $u$ and $v$ in my problem. The paper states that $mathbfv$ with components $u$ and $v$ is the velocity and $mathbfn$ is the normal vector.

Also, I would be grateful if someone could introduce some other references on the derivative of such double integrals to me.

calculus integration multivariable-calculus derivatives reference-request

share|cite|improve this question

edited 8 hours ago

Rodrigo de Azevedo

13.4k42165

asked 8 hours ago

SMA.DSMA.D

506421

$endgroup$

add a comment | 

$begingroup$

Calculate the following derivative

$$fracddtiint_D_tF(x,y,t) , mathrm d x mathrm d y$$

where

$$D_t = lbrace (x,y) mid (x-t)^2+(y-t)^2leq r^2 rbrace$$

I've read here that the answer to the above question is in the form of:
beginalignfracddtiint_D_tF(x,y,t)dxdy &=int_partial D_t F(udy-vdx) + iint_D_tfracpartial Fpartial tdx dy \
&=iint_D_tleft[textdiv(Fmathbfv)+fracpartial Fpartial tright]dx dyendalign
which is a generalization of Leibniz integral rule.

I don't know how I can calculate $mathbfv$ and $mathbfn$ or $u$ and $v$ in my problem. The paper states that $mathbfv$ with components $u$ and $v$ is the velocity and $mathbfn$ is the normal vector.

Also, I would be grateful if someone could introduce some other references on the derivative of such double integrals to me.

calculus integration multivariable-calculus derivatives reference-request

share|cite|improve this question

edited 8 hours ago

Rodrigo de Azevedo

13.4k42165

asked 8 hours ago

SMA.DSMA.D

506421

$endgroup$

share|cite|improve this question

edited 8 hours ago

Rodrigo de Azevedo

13.4k42165

asked 8 hours ago

SMA.DSMA.D

506421

share|cite|improve this question

edited 8 hours ago

Rodrigo de Azevedo

13.4k42165

asked 8 hours ago

SMA.DSMA.D

506421

edited 8 hours ago

Rodrigo de Azevedo

13.4k42165

edited 8 hours ago

Rodrigo de Azevedo

13.4k42165

asked 8 hours ago

SMA.DSMA.D

506421

asked 8 hours ago

SMA.DSMA.D

506421

1 Answer
1

active

oldest

votes

4

$begingroup$

Your domain $D_t$ is quite simple, you should be able to compute whatever you need. But also because of how simple it is you can just use a simple change of variables
$$ fracdd t iint_D_t F(x, y, t) d x d y= fracdd tiint_D_0 F(x+t,y+t,t)dxdy = iint_D_0 fracdd t(F(x+t,y+t,t))dxdy= iint_D_0 (partial_xF+partial_yF+partial_t F)(x+t,y+t,t)dxdy = iint_D_tpartial_xF+partial_yF+partial_t Fdxdy $$

share|cite|improve this answer

answered 8 hours ago

Calvin KhorCalvin Khor

13.4k21540

$endgroup$

add a comment | 

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4

$begingroup$

Your domain $D_t$ is quite simple, you should be able to compute whatever you need. But also because of how simple it is you can just use a simple change of variables
$$ fracdd t iint_D_t F(x, y, t) d x d y= fracdd tiint_D_0 F(x+t,y+t,t)dxdy = iint_D_0 fracdd t(F(x+t,y+t,t))dxdy= iint_D_0 (partial_xF+partial_yF+partial_t F)(x+t,y+t,t)dxdy = iint_D_tpartial_xF+partial_yF+partial_t Fdxdy $$

share|cite|improve this answer

answered 8 hours ago

Calvin KhorCalvin Khor

13.4k21540

$endgroup$

add a comment | 

4

$begingroup$

Your domain $D_t$ is quite simple, you should be able to compute whatever you need. But also because of how simple it is you can just use a simple change of variables
$$ fracdd t iint_D_t F(x, y, t) d x d y= fracdd tiint_D_0 F(x+t,y+t,t)dxdy = iint_D_0 fracdd t(F(x+t,y+t,t))dxdy= iint_D_0 (partial_xF+partial_yF+partial_t F)(x+t,y+t,t)dxdy = iint_D_tpartial_xF+partial_yF+partial_t Fdxdy $$

share|cite|improve this answer

answered 8 hours ago

Calvin KhorCalvin Khor

13.4k21540

$endgroup$

add a comment | 

$begingroup$

Your domain $D_t$ is quite simple, you should be able to compute whatever you need. But also because of how simple it is you can just use a simple change of variables
$$ fracdd t iint_D_t F(x, y, t) d x d y= fracdd tiint_D_0 F(x+t,y+t,t)dxdy = iint_D_0 fracdd t(F(x+t,y+t,t))dxdy= iint_D_0 (partial_xF+partial_yF+partial_t F)(x+t,y+t,t)dxdy = iint_D_tpartial_xF+partial_yF+partial_t Fdxdy $$

share|cite|improve this answer

answered 8 hours ago

Calvin KhorCalvin Khor

13.4k21540

$endgroup$

share|cite|improve this answer

answered 8 hours ago

Calvin KhorCalvin Khor

13.4k21540

answered 8 hours ago

Calvin KhorCalvin Khor

13.4k21540

answered 8 hours ago

Calvin KhorCalvin Khor

13.4k21540