Mixing
Surface integral of a vector valued function
$begingroup$
The value of the surface integral $$ iint_S(xhat{i}+yhat{j})cdot hat{n}~dA $$ evaluated over the surface of a cube having sides of length $a$ is ($hat{n}$ is unit normal vector)
begin{align*}
iint_S(xhat{i}+yhat{j})cdot hat{n}~dA &= iiint_V 2 ~dV text{(divergence theorem)}\
&=2a^3
end{align*}
but the answer is $0$, I don't know where I am doing the mistake.
real-analysis surface-integrals
asked Jan 20 at 16:48
XYZABCXYZABC
344110
$endgroup$
add a comment |
$begingroup$
The value of the surface integral $$ iint_S(xhat{i}+yhat{j})cdot hat{n}~dA $$ evaluated over the surface of a cube having sides of length $a$ is ($hat{n}$ is unit normal vector)
begin{align*}
iint_S(xhat{i}+yhat{j})cdot hat{n}~dA &= iiint_V 2 ~dV text{(divergence theorem)}\
&=2a^3
end{align*}
but the answer is $0$, I don't know where I am doing the mistake.
real-analysis surface-integrals
asked Jan 20 at 16:48
XYZABCXYZABC
344110
$endgroup$
$begingroup$
Why do you think the answer should be $0$?
$endgroup$
– Andrei
Jan 20 at 16:57
$begingroup$
No I am not thinking, but the answer provided is zero and in the answer book it is written that $iint_S(xhat{i}+yhat{j})hat{n}~dA=iint_S(xhat{i}+yhat{j})(hat{i}~dx+hat{j}~dy)$ and proceeded.
$endgroup$
– XYZABC
Jan 20 at 16:59
add a comment |
$begingroup$
The value of the surface integral $$ iint_S(xhat{i}+yhat{j})cdot hat{n}~dA $$ evaluated over the surface of a cube having sides of length $a$ is ($hat{n}$ is unit normal vector)
begin{align*}
iint_S(xhat{i}+yhat{j})cdot hat{n}~dA &= iiint_V 2 ~dV text{(divergence theorem)}\
&=2a^3
end{align*}
but the answer is $0$, I don't know where I am doing the mistake.
real-analysis surface-integrals
asked Jan 20 at 16:48
XYZABCXYZABC
344110
$endgroup$
The value of the surface integral $$ iint_S(xhat{i}+yhat{j})cdot hat{n}~dA $$ evaluated over the surface of a cube having sides of length $a$ is ($hat{n}$ is unit normal vector)
begin{align*}
iint_S(xhat{i}+yhat{j})cdot hat{n}~dA &= iiint_V 2 ~dV text{(divergence theorem)}\
&=2a^3
end{align*}
but the answer is $0$, I don't know where I am doing the mistake.
real-analysis surface-integrals
real-analysis surface-integrals
asked Jan 20 at 16:48
XYZABCXYZABC
344110
asked Jan 20 at 16:48
XYZABCXYZABC
344110
asked Jan 20 at 16:48
XYZABCXYZABC
344110
asked Jan 20 at 16:48
XYZABCXYZABC
344110
asked Jan 20 at 16:48
XYZABCXYZABC
344110
344110
$begingroup$
Why do you think the answer should be $0$?
$endgroup$
– Andrei
Jan 20 at 16:57
$begingroup$
No I am not thinking, but the answer provided is zero and in the answer book it is written that $iint_S(xhat{i}+yhat{j})hat{n}~dA=iint_S(xhat{i}+yhat{j})(hat{i}~dx+hat{j}~dy)$ and proceeded.
$endgroup$
– XYZABC
Jan 20 at 16:59
add a comment |
$begingroup$
Why do you think the answer should be $0$?
$endgroup$
– Andrei
Jan 20 at 16:57
$begingroup$
No I am not thinking, but the answer provided is zero and in the answer book it is written that $iint_S(xhat{i}+yhat{j})hat{n}~dA=iint_S(xhat{i}+yhat{j})(hat{i}~dx+hat{j}~dy)$ and proceeded.
$endgroup$
– XYZABC
Jan 20 at 16:59
$begingroup$
Why do you think the answer should be $0$?
$endgroup$
– Andrei
Jan 20 at 16:57
$begingroup$
Why do you think the answer should be $0$?
$endgroup$
– Andrei
Jan 20 at 16:57
$begingroup$
No I am not thinking, but the answer provided is zero and in the answer book it is written that $iint_S(xhat{i}+yhat{j})hat{n}~dA=iint_S(xhat{i}+yhat{j})(hat{i}~dx+hat{j}~dy)$ and proceeded.
$endgroup$
– XYZABC
Jan 20 at 16:59
$begingroup$
No I am not thinking, but the answer provided is zero and in the answer book it is written that $iint_S(xhat{i}+yhat{j})hat{n}~dA=iint_S(xhat{i}+yhat{j})(hat{i}~dx+hat{j}~dy)$ and proceeded.
$endgroup$
– XYZABC
Jan 20 at 16:59
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The answer in the book is wrong. Without loss of generality, just to simplify calculations, let's choose one of the vertices at the origin, with faces perpendicular to the axes at $0$ and $a$. The faces perpendicular to $z$ axis have the normal either $hat j$ or $-hat j$, so they will not contribute to the surface integral. Let's choose the faces perpendicular to the $x$ axis. At $x=0$, the normal is $-hat i$, so the integrand is $-x=0$, therefore no contribution either. Now at $x=a$, the normal is $hat n=hat i$, and the integrand is $x=a$. Integral over the area of the side will yield $acdot a^2=a^3$. You can repeat the same for $y$ axis, with contributions $0$ and $a^3$, so the final integral will be the sum of contributions at $x=0$, $x=a$, $y=0$, $y=a$, $z=0$, $z=a$: $$ iint_S(xhat{i}+yhat{j})cdot hat{n}~dA =0+a^3+0+a^3+0+0=2a^3$$
answered Jan 20 at 17:13
AndreiAndrei
12.6k21128
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
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%2f3080809%2fsurface-integral-of-a-vector-valued-function%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 answer in the book is wrong. Without loss of generality, just to simplify calculations, let's choose one of the vertices at the origin, with faces perpendicular to the axes at $0$ and $a$. The faces perpendicular to $z$ axis have the normal either $hat j$ or $-hat j$, so they will not contribute to the surface integral. Let's choose the faces perpendicular to the $x$ axis. At $x=0$, the normal is $-hat i$, so the integrand is $-x=0$, therefore no contribution either. Now at $x=a$, the normal is $hat n=hat i$, and the integrand is $x=a$. Integral over the area of the side will yield $acdot a^2=a^3$. You can repeat the same for $y$ axis, with contributions $0$ and $a^3$, so the final integral will be the sum of contributions at $x=0$, $x=a$, $y=0$, $y=a$, $z=0$, $z=a$: $$ iint_S(xhat{i}+yhat{j})cdot hat{n}~dA =0+a^3+0+a^3+0+0=2a^3$$
answered Jan 20 at 17:13
AndreiAndrei
12.6k21128
$endgroup$
add a comment |
$begingroup$
The answer in the book is wrong. Without loss of generality, just to simplify calculations, let's choose one of the vertices at the origin, with faces perpendicular to the axes at $0$ and $a$. The faces perpendicular to $z$ axis have the normal either $hat j$ or $-hat j$, so they will not contribute to the surface integral. Let's choose the faces perpendicular to the $x$ axis. At $x=0$, the normal is $-hat i$, so the integrand is $-x=0$, therefore no contribution either. Now at $x=a$, the normal is $hat n=hat i$, and the integrand is $x=a$. Integral over the area of the side will yield $acdot a^2=a^3$. You can repeat the same for $y$ axis, with contributions $0$ and $a^3$, so the final integral will be the sum of contributions at $x=0$, $x=a$, $y=0$, $y=a$, $z=0$, $z=a$: $$ iint_S(xhat{i}+yhat{j})cdot hat{n}~dA =0+a^3+0+a^3+0+0=2a^3$$
answered Jan 20 at 17:13
AndreiAndrei
12.6k21128
$endgroup$
add a comment |
$begingroup$
The answer in the book is wrong. Without loss of generality, just to simplify calculations, let's choose one of the vertices at the origin, with faces perpendicular to the axes at $0$ and $a$. The faces perpendicular to $z$ axis have the normal either $hat j$ or $-hat j$, so they will not contribute to the surface integral. Let's choose the faces perpendicular to the $x$ axis. At $x=0$, the normal is $-hat i$, so the integrand is $-x=0$, therefore no contribution either. Now at $x=a$, the normal is $hat n=hat i$, and the integrand is $x=a$. Integral over the area of the side will yield $acdot a^2=a^3$. You can repeat the same for $y$ axis, with contributions $0$ and $a^3$, so the final integral will be the sum of contributions at $x=0$, $x=a$, $y=0$, $y=a$, $z=0$, $z=a$: $$ iint_S(xhat{i}+yhat{j})cdot hat{n}~dA =0+a^3+0+a^3+0+0=2a^3$$
answered Jan 20 at 17:13
AndreiAndrei
12.6k21128
$endgroup$
The answer in the book is wrong. Without loss of generality, just to simplify calculations, let's choose one of the vertices at the origin, with faces perpendicular to the axes at $0$ and $a$. The faces perpendicular to $z$ axis have the normal either $hat j$ or $-hat j$, so they will not contribute to the surface integral. Let's choose the faces perpendicular to the $x$ axis. At $x=0$, the normal is $-hat i$, so the integrand is $-x=0$, therefore no contribution either. Now at $x=a$, the normal is $hat n=hat i$, and the integrand is $x=a$. Integral over the area of the side will yield $acdot a^2=a^3$. You can repeat the same for $y$ axis, with contributions $0$ and $a^3$, so the final integral will be the sum of contributions at $x=0$, $x=a$, $y=0$, $y=a$, $z=0$, $z=a$: $$ iint_S(xhat{i}+yhat{j})cdot hat{n}~dA =0+a^3+0+a^3+0+0=2a^3$$
answered Jan 20 at 17:13
AndreiAndrei
12.6k21128
answered Jan 20 at 17:13
AndreiAndrei
12.6k21128
answered Jan 20 at 17:13
AndreiAndrei
12.6k21128
answered Jan 20 at 17:13
AndreiAndrei
12.6k21128
12.6k21128
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%2f3080809%2fsurface-integral-of-a-vector-valued-function%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
$begingroup$
Why do you think the answer should be $0$?
$endgroup$
– Andrei
Jan 20 at 16:57
$begingroup$
No I am not thinking, but the answer provided is zero and in the answer book it is written that $iint_S(xhat{i}+yhat{j})hat{n}~dA=iint_S(xhat{i}+yhat{j})(hat{i}~dx+hat{j}~dy)$ and proceeded.
$endgroup$
– XYZABC
Jan 20 at 16:59