About surface of integral of type one
$begingroup$
Compute F(x,y,z,t)=$iint _Sf(xi ,eta ,zeta) dS$,S is surface
$$(xi -x)^2+(eta-y)^2+(zeta-z)^2=t^2$$,suppose $ r^2={x^2+y^2+z^2}>a^2>0$
$f(xi ,eta ,zeta)=1$
,when $xi^ 2+eta ^2+zeta ^2<a^2$,
$f=0$ , when $\ xi^ 2+eta ^2+zeta ^2ge a^2$
My attempt.
If r$ge t+a$,then the integral is zero,but when r$<t+a$ ,the integral is difficult .i think maybe I should rotate the Rectangular Coordinates.
real-analysis calculus integration limits analysis
asked Jan 8 at 10:14
jacksonjackson
1499
$endgroup$
add a comment |
$begingroup$
Compute F(x,y,z,t)=$iint _Sf(xi ,eta ,zeta) dS$,S is surface
$$(xi -x)^2+(eta-y)^2+(zeta-z)^2=t^2$$,suppose $ r^2={x^2+y^2+z^2}>a^2>0$
$f(xi ,eta ,zeta)=1$
,when $xi^ 2+eta ^2+zeta ^2<a^2$,
$f=0$ , when $\ xi^ 2+eta ^2+zeta ^2ge a^2$
My attempt.
If r$ge t+a$,then the integral is zero,but when r$<t+a$ ,the integral is difficult .i think maybe I should rotate the Rectangular Coordinates.
real-analysis calculus integration limits analysis
asked Jan 8 at 10:14
jacksonjackson
1499
$endgroup$
$begingroup$
ooyes thanks a lot
$endgroup$
– jackson
Jan 8 at 16:34
add a comment |
$begingroup$
Compute F(x,y,z,t)=$iint _Sf(xi ,eta ,zeta) dS$,S is surface
$$(xi -x)^2+(eta-y)^2+(zeta-z)^2=t^2$$,suppose $ r^2={x^2+y^2+z^2}>a^2>0$
$f(xi ,eta ,zeta)=1$
,when $xi^ 2+eta ^2+zeta ^2<a^2$,
$f=0$ , when $\ xi^ 2+eta ^2+zeta ^2ge a^2$
My attempt.
If r$ge t+a$,then the integral is zero,but when r$<t+a$ ,the integral is difficult .i think maybe I should rotate the Rectangular Coordinates.
real-analysis calculus integration limits analysis
asked Jan 8 at 10:14
jacksonjackson
1499
$endgroup$
Compute F(x,y,z,t)=$iint _Sf(xi ,eta ,zeta) dS$,S is surface
$$(xi -x)^2+(eta-y)^2+(zeta-z)^2=t^2$$,suppose $ r^2={x^2+y^2+z^2}>a^2>0$
$f(xi ,eta ,zeta)=1$
,when $xi^ 2+eta ^2+zeta ^2<a^2$,
$f=0$ , when $\ xi^ 2+eta ^2+zeta ^2ge a^2$
My attempt.
If r$ge t+a$,then the integral is zero,but when r$<t+a$ ,the integral is difficult .i think maybe I should rotate the Rectangular Coordinates.
real-analysis calculus integration limits analysis
real-analysis calculus integration limits analysis
asked Jan 8 at 10:14
jacksonjackson
1499
asked Jan 8 at 10:14
jacksonjackson
1499
edited Jan 8 at 16:34
jackson
asked Jan 8 at 10:14
jacksonjackson
1499
asked Jan 8 at 10:14
jacksonjackson
1499
asked Jan 8 at 10:14
jacksonjackson
1499
1499
$begingroup$
ooyes thanks a lot
$endgroup$
– jackson
Jan 8 at 16:34
add a comment |
$begingroup$
ooyes thanks a lot
$endgroup$
– jackson
Jan 8 at 16:34
$begingroup$
ooyes thanks a lot
$endgroup$
– jackson
Jan 8 at 16:34
$begingroup$
ooyes thanks a lot
$endgroup$
– jackson
Jan 8 at 16:34
add a comment |
1 Answer
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The situation is rotation symmetric with respect to the axis through $O=(0,0,0)$ and $P=(x,y,z)$. It follows that the value $F(x,y,z,t)$ depends only on $a$, $r>a$, and $t$. You may assume $P=(0,0,r)$, and then have to compute the surface area of the part of the large sphere $S_t: xi^2+eta^2+(zeta-r)^2=t^2$ lying within the small sphere $xi^2+eta^2+zeta^2=a^2$. To this end it is sufficient to draw two circles in the $(rho,zeta)$-half plane with $rho:=sqrt{xi^2+eta^2}geq0$, and to do some elementary geometry. Note that the surface area of a sphere cap is equal to the height of the cap times the length of the equator.
We have to compute $$h:=zeta_*-(r-t)={1over2r}bigl(a^2-(r-t)^2bigr) ,$$
and then
$$F(r,t)=int_{S_t}f(xi,eta,zeta)>{rm d}S=2pi t> h .$$
answered Jan 8 at 16:53
Christian BlatterChristian Blatter
176k9115328
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
The situation is rotation symmetric with respect to the axis through $O=(0,0,0)$ and $P=(x,y,z)$. It follows that the value $F(x,y,z,t)$ depends only on $a$, $r>a$, and $t$. You may assume $P=(0,0,r)$, and then have to compute the surface area of the part of the large sphere $S_t: xi^2+eta^2+(zeta-r)^2=t^2$ lying within the small sphere $xi^2+eta^2+zeta^2=a^2$. To this end it is sufficient to draw two circles in the $(rho,zeta)$-half plane with $rho:=sqrt{xi^2+eta^2}geq0$, and to do some elementary geometry. Note that the surface area of a sphere cap is equal to the height of the cap times the length of the equator.
We have to compute $$h:=zeta_*-(r-t)={1over2r}bigl(a^2-(r-t)^2bigr) ,$$
and then
$$F(r,t)=int_{S_t}f(xi,eta,zeta)>{rm d}S=2pi t> h .$$
answered Jan 8 at 16:53
Christian BlatterChristian Blatter
176k9115328
$endgroup$
add a comment |
$begingroup$
The situation is rotation symmetric with respect to the axis through $O=(0,0,0)$ and $P=(x,y,z)$. It follows that the value $F(x,y,z,t)$ depends only on $a$, $r>a$, and $t$. You may assume $P=(0,0,r)$, and then have to compute the surface area of the part of the large sphere $S_t: xi^2+eta^2+(zeta-r)^2=t^2$ lying within the small sphere $xi^2+eta^2+zeta^2=a^2$. To this end it is sufficient to draw two circles in the $(rho,zeta)$-half plane with $rho:=sqrt{xi^2+eta^2}geq0$, and to do some elementary geometry. Note that the surface area of a sphere cap is equal to the height of the cap times the length of the equator.
We have to compute $$h:=zeta_*-(r-t)={1over2r}bigl(a^2-(r-t)^2bigr) ,$$
and then
$$F(r,t)=int_{S_t}f(xi,eta,zeta)>{rm d}S=2pi t> h .$$
answered Jan 8 at 16:53
Christian BlatterChristian Blatter
176k9115328
$endgroup$
add a comment |
$begingroup$
The situation is rotation symmetric with respect to the axis through $O=(0,0,0)$ and $P=(x,y,z)$. It follows that the value $F(x,y,z,t)$ depends only on $a$, $r>a$, and $t$. You may assume $P=(0,0,r)$, and then have to compute the surface area of the part of the large sphere $S_t: xi^2+eta^2+(zeta-r)^2=t^2$ lying within the small sphere $xi^2+eta^2+zeta^2=a^2$. To this end it is sufficient to draw two circles in the $(rho,zeta)$-half plane with $rho:=sqrt{xi^2+eta^2}geq0$, and to do some elementary geometry. Note that the surface area of a sphere cap is equal to the height of the cap times the length of the equator.
We have to compute $$h:=zeta_*-(r-t)={1over2r}bigl(a^2-(r-t)^2bigr) ,$$
and then
$$F(r,t)=int_{S_t}f(xi,eta,zeta)>{rm d}S=2pi t> h .$$
answered Jan 8 at 16:53
Christian BlatterChristian Blatter
176k9115328
$endgroup$
The situation is rotation symmetric with respect to the axis through $O=(0,0,0)$ and $P=(x,y,z)$. It follows that the value $F(x,y,z,t)$ depends only on $a$, $r>a$, and $t$. You may assume $P=(0,0,r)$, and then have to compute the surface area of the part of the large sphere $S_t: xi^2+eta^2+(zeta-r)^2=t^2$ lying within the small sphere $xi^2+eta^2+zeta^2=a^2$. To this end it is sufficient to draw two circles in the $(rho,zeta)$-half plane with $rho:=sqrt{xi^2+eta^2}geq0$, and to do some elementary geometry. Note that the surface area of a sphere cap is equal to the height of the cap times the length of the equator.
We have to compute $$h:=zeta_*-(r-t)={1over2r}bigl(a^2-(r-t)^2bigr) ,$$
and then
$$F(r,t)=int_{S_t}f(xi,eta,zeta)>{rm d}S=2pi t> h .$$
answered Jan 8 at 16:53
Christian BlatterChristian Blatter
176k9115328
edited Jan 9 at 19:35
answered Jan 8 at 16:53
Christian BlatterChristian Blatter
176k9115328
answered Jan 8 at 16:53
Christian BlatterChristian Blatter
176k9115328
answered Jan 8 at 16:53
Christian BlatterChristian Blatter
176k9115328
176k9115328
add a comment |
add a comment |
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$begingroup$
ooyes thanks a lot
$endgroup$
– jackson
Jan 8 at 16:34