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
1
active
oldest
votes
$begingroup$
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
176k8115328
$endgroup$
add a comment |
Your Answer
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3066003%2fabout-surface-of-integral-of-type-one%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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
176k8115328
$endgroup$
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
176k8115328
$endgroup$
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
176k8115328
$endgroup$
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
176k8115328
edited Jan 9 at 19:35
answered Jan 8 at 16:53
Christian BlatterChristian Blatter
176k8115328
answered Jan 8 at 16:53
Christian BlatterChristian Blatter
176k8115328
answered Jan 8 at 16:53
Christian BlatterChristian Blatter
176k8115328
176k8115328
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3066003%2fabout-surface-of-integral-of-type-one%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
ooyes thanks a lot
$endgroup$
– jackson
Jan 8 at 16:34