Surface integral over graph of a function
$begingroup$
Q: Evaluate the surface integral $iint_S mathbf{F}cdot dmathbf{S}$, where $mathbf{F}(x,y,z)= 3x^2mathbf{i}-2xymathbf{j}+8mathbf{k}$, and $S$ is the graph of the function $z=f(x,y)=2x-y$ for $0leq xleq2$ and $0leq y leq2$.
Attempt: This question is from a previous years Vector Calculus exam. When attempting to solve, I had thought to parameterize the surface, and then use $int_a^b mathbf{F}(mathbf{c}(t))cdot mathbf{c}'(t),dt$, $a leq t leq b$. In the solutions, they have $$iint_S mathbf{F}cdot dmathbf{S} = int_0^2int_0^2 (-6x^2-2xy+8),dxdy= -8$$
I'm mostly looking for an explanation as to how they arrived at the expression inside the integral ($-6x^2-2xy+8$), any help would be greatly appreciated.
Thanks.
calculus multivariable-calculus
$endgroup$
add a comment |
$begingroup$
Q: Evaluate the surface integral $iint_S mathbf{F}cdot dmathbf{S}$, where $mathbf{F}(x,y,z)= 3x^2mathbf{i}-2xymathbf{j}+8mathbf{k}$, and $S$ is the graph of the function $z=f(x,y)=2x-y$ for $0leq xleq2$ and $0leq y leq2$.
Attempt: This question is from a previous years Vector Calculus exam. When attempting to solve, I had thought to parameterize the surface, and then use $int_a^b mathbf{F}(mathbf{c}(t))cdot mathbf{c}'(t),dt$, $a leq t leq b$. In the solutions, they have $$iint_S mathbf{F}cdot dmathbf{S} = int_0^2int_0^2 (-6x^2-2xy+8),dxdy= -8$$
I'm mostly looking for an explanation as to how they arrived at the expression inside the integral ($-6x^2-2xy+8$), any help would be greatly appreciated.
Thanks.
calculus multivariable-calculus
$endgroup$
add a comment |
$begingroup$
Q: Evaluate the surface integral $iint_S mathbf{F}cdot dmathbf{S}$, where $mathbf{F}(x,y,z)= 3x^2mathbf{i}-2xymathbf{j}+8mathbf{k}$, and $S$ is the graph of the function $z=f(x,y)=2x-y$ for $0leq xleq2$ and $0leq y leq2$.
Attempt: This question is from a previous years Vector Calculus exam. When attempting to solve, I had thought to parameterize the surface, and then use $int_a^b mathbf{F}(mathbf{c}(t))cdot mathbf{c}'(t),dt$, $a leq t leq b$. In the solutions, they have $$iint_S mathbf{F}cdot dmathbf{S} = int_0^2int_0^2 (-6x^2-2xy+8),dxdy= -8$$
I'm mostly looking for an explanation as to how they arrived at the expression inside the integral ($-6x^2-2xy+8$), any help would be greatly appreciated.
Thanks.
calculus multivariable-calculus
$endgroup$
Q: Evaluate the surface integral $iint_S mathbf{F}cdot dmathbf{S}$, where $mathbf{F}(x,y,z)= 3x^2mathbf{i}-2xymathbf{j}+8mathbf{k}$, and $S$ is the graph of the function $z=f(x,y)=2x-y$ for $0leq xleq2$ and $0leq y leq2$.
Attempt: This question is from a previous years Vector Calculus exam. When attempting to solve, I had thought to parameterize the surface, and then use $int_a^b mathbf{F}(mathbf{c}(t))cdot mathbf{c}'(t),dt$, $a leq t leq b$. In the solutions, they have $$iint_S mathbf{F}cdot dmathbf{S} = int_0^2int_0^2 (-6x^2-2xy+8),dxdy= -8$$
I'm mostly looking for an explanation as to how they arrived at the expression inside the integral ($-6x^2-2xy+8$), any help would be greatly appreciated.
Thanks.
calculus multivariable-calculus
calculus multivariable-calculus
edited Dec 26 '18 at 13:06
mechanodroid
27.8k62447
27.8k62447
asked Jun 13 '13 at 7:19
user68093user68093
603
603
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The surface integral is better written as
$$iint_S dS , hat{mathbf{n}} cdot mathbf{F}$$
where $hat{mathbf{n}}$ is the unit surface normal. For surfaces of the form $z=f(x,y)$, the unit normal is
$$hat{mathbf{n}} = frac{displaystyleleft(-frac{partial f}{partial x},-frac{partial f}{partial y},1right)}{displaystylesqrt{left(frac{partial f}{partial x}right)^2+left(frac{partial f}{partial y}right)^2+1}}$$
Note that the normalizing factor in the denominator is exactly the stretching factor relating the surface area element to the differential $dx,dy$:
$$dS = sqrt{left(frac{partial f}{partial x}right)^2+left(frac{partial f}{partial y}right)^2+1} , dx, dy$$
Now, $partial f/partial x = 2$ and $partial f/partial y = -1$, so that
$$iint_S dS , hat{mathbf{n}} cdot mathbf{F} = int_0^2 dx , int_0^2 dy , [-(2) cdot (3 x^2) - (-1) cdot (-2 x y) + (1)cdot (8)] $$
which is the result you wanted to see.
$endgroup$
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%2f419119%2fsurface-integral-over-graph-of-a-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 surface integral is better written as
$$iint_S dS , hat{mathbf{n}} cdot mathbf{F}$$
where $hat{mathbf{n}}$ is the unit surface normal. For surfaces of the form $z=f(x,y)$, the unit normal is
$$hat{mathbf{n}} = frac{displaystyleleft(-frac{partial f}{partial x},-frac{partial f}{partial y},1right)}{displaystylesqrt{left(frac{partial f}{partial x}right)^2+left(frac{partial f}{partial y}right)^2+1}}$$
Note that the normalizing factor in the denominator is exactly the stretching factor relating the surface area element to the differential $dx,dy$:
$$dS = sqrt{left(frac{partial f}{partial x}right)^2+left(frac{partial f}{partial y}right)^2+1} , dx, dy$$
Now, $partial f/partial x = 2$ and $partial f/partial y = -1$, so that
$$iint_S dS , hat{mathbf{n}} cdot mathbf{F} = int_0^2 dx , int_0^2 dy , [-(2) cdot (3 x^2) - (-1) cdot (-2 x y) + (1)cdot (8)] $$
which is the result you wanted to see.
$endgroup$
add a comment |
$begingroup$
The surface integral is better written as
$$iint_S dS , hat{mathbf{n}} cdot mathbf{F}$$
where $hat{mathbf{n}}$ is the unit surface normal. For surfaces of the form $z=f(x,y)$, the unit normal is
$$hat{mathbf{n}} = frac{displaystyleleft(-frac{partial f}{partial x},-frac{partial f}{partial y},1right)}{displaystylesqrt{left(frac{partial f}{partial x}right)^2+left(frac{partial f}{partial y}right)^2+1}}$$
Note that the normalizing factor in the denominator is exactly the stretching factor relating the surface area element to the differential $dx,dy$:
$$dS = sqrt{left(frac{partial f}{partial x}right)^2+left(frac{partial f}{partial y}right)^2+1} , dx, dy$$
Now, $partial f/partial x = 2$ and $partial f/partial y = -1$, so that
$$iint_S dS , hat{mathbf{n}} cdot mathbf{F} = int_0^2 dx , int_0^2 dy , [-(2) cdot (3 x^2) - (-1) cdot (-2 x y) + (1)cdot (8)] $$
which is the result you wanted to see.
$endgroup$
add a comment |
$begingroup$
The surface integral is better written as
$$iint_S dS , hat{mathbf{n}} cdot mathbf{F}$$
where $hat{mathbf{n}}$ is the unit surface normal. For surfaces of the form $z=f(x,y)$, the unit normal is
$$hat{mathbf{n}} = frac{displaystyleleft(-frac{partial f}{partial x},-frac{partial f}{partial y},1right)}{displaystylesqrt{left(frac{partial f}{partial x}right)^2+left(frac{partial f}{partial y}right)^2+1}}$$
Note that the normalizing factor in the denominator is exactly the stretching factor relating the surface area element to the differential $dx,dy$:
$$dS = sqrt{left(frac{partial f}{partial x}right)^2+left(frac{partial f}{partial y}right)^2+1} , dx, dy$$
Now, $partial f/partial x = 2$ and $partial f/partial y = -1$, so that
$$iint_S dS , hat{mathbf{n}} cdot mathbf{F} = int_0^2 dx , int_0^2 dy , [-(2) cdot (3 x^2) - (-1) cdot (-2 x y) + (1)cdot (8)] $$
which is the result you wanted to see.
$endgroup$
The surface integral is better written as
$$iint_S dS , hat{mathbf{n}} cdot mathbf{F}$$
where $hat{mathbf{n}}$ is the unit surface normal. For surfaces of the form $z=f(x,y)$, the unit normal is
$$hat{mathbf{n}} = frac{displaystyleleft(-frac{partial f}{partial x},-frac{partial f}{partial y},1right)}{displaystylesqrt{left(frac{partial f}{partial x}right)^2+left(frac{partial f}{partial y}right)^2+1}}$$
Note that the normalizing factor in the denominator is exactly the stretching factor relating the surface area element to the differential $dx,dy$:
$$dS = sqrt{left(frac{partial f}{partial x}right)^2+left(frac{partial f}{partial y}right)^2+1} , dx, dy$$
Now, $partial f/partial x = 2$ and $partial f/partial y = -1$, so that
$$iint_S dS , hat{mathbf{n}} cdot mathbf{F} = int_0^2 dx , int_0^2 dy , [-(2) cdot (3 x^2) - (-1) cdot (-2 x y) + (1)cdot (8)] $$
which is the result you wanted to see.
answered Jun 13 '13 at 8:16
Ron GordonRon Gordon
122k14155265
122k14155265
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%2f419119%2fsurface-integral-over-graph-of-a-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