Average Value of a Line Integral
I'm having quite a hard time calculating the average value of a line integral.
Given the surface $f(x,y) = sqrt{16 + 36y^{2/3}}$ and the curve $y = x^{3/2}$, I need to calculate the average value of the integral of the surface for $0 leq x leq 13$
I start by parameterizing the curve and the surface for $0 leq t leq 13:
$begin{align}
r(t) &= langle t, t^{3/2} rangle \
f(t) &= sqrt{16+36t} \
end{align}$
And calculate a few things I'll need later:
$begin{align}
r'(t) &= langle 1, frac{3}{2}t^{1/2} rangle \
left|r'(t)right| &= sqrt{ 1 + frac{9}{4}t } \
end{align}$
Next, I calculate the line integral:
$begin{align}
&int_0^{13} f(t) left|r'(t)right| dt \
&int_0^{13} sqrt{16+36t} sqrt{ 1 + frac{9}{4}t } dt \
2 &int_0^{13} sqrt{4+9t} sqrt{ 1 + frac{9}{4}t } dt \
2 &int_0^{13} sqrt{ frac{4}{4} (4+9t) } sqrt{ 1 + frac{9}{4}t } dt \
4 &int_0^{13} sqrt{ 1 + frac{9}{4}t } sqrt{ 1 + frac{9}{4}t } dt \
4 &int_0^{13} 1 + frac{9}{4}t dt \
4 &left( t + frac{9}{8}t^2right|_0^{13} \
4 &left( 13 + frac{9}{8}13^2right) \
&frac{1625}{2} \
end{align}$
Then, I calculate the length of the curve:
$begin{align}
L &= int_0^{13} sqrt{1 + [r'(t)]^2} dx \
L &= int_0^{13} sqrt{1 + 1 + frac{9}{4}t} dx \
L &= int_0^{13} sqrt{2 + frac{9}{4}t} dx \
end{align}$
Letting $u = 2 + 9/4 t$
$begin{align}
L &= frac{4}{9} int_2^{125 / 4} u^{1/2} du \
L &= frac{8}{27} left( u^{3/2} right|_{ 2}^{ 125 / 4} \
L &= frac{8}{27} left[ left(frac{125}{4}right)^{3/2} - 2^{3/2} right] \
L &= frac{625 sqrt{5} - 16 sqrt{2} }{27} \
end{align}$
Then, dividing the integral by the length of the curve gives a gnarly, incorrect mess.
What am I not understanding here? Are there algebra errors? Errors in the calculus?
integration multivariable-calculus average curves
asked Apr 15 '16 at 19:00
StudentsTea
7862722
add a comment |
I'm having quite a hard time calculating the average value of a line integral.
Given the surface $f(x,y) = sqrt{16 + 36y^{2/3}}$ and the curve $y = x^{3/2}$, I need to calculate the average value of the integral of the surface for $0 leq x leq 13$
I start by parameterizing the curve and the surface for $0 leq t leq 13:
$begin{align}
r(t) &= langle t, t^{3/2} rangle \
f(t) &= sqrt{16+36t} \
end{align}$
And calculate a few things I'll need later:
$begin{align}
r'(t) &= langle 1, frac{3}{2}t^{1/2} rangle \
left|r'(t)right| &= sqrt{ 1 + frac{9}{4}t } \
end{align}$
Next, I calculate the line integral:
$begin{align}
&int_0^{13} f(t) left|r'(t)right| dt \
&int_0^{13} sqrt{16+36t} sqrt{ 1 + frac{9}{4}t } dt \
2 &int_0^{13} sqrt{4+9t} sqrt{ 1 + frac{9}{4}t } dt \
2 &int_0^{13} sqrt{ frac{4}{4} (4+9t) } sqrt{ 1 + frac{9}{4}t } dt \
4 &int_0^{13} sqrt{ 1 + frac{9}{4}t } sqrt{ 1 + frac{9}{4}t } dt \
4 &int_0^{13} 1 + frac{9}{4}t dt \
4 &left( t + frac{9}{8}t^2right|_0^{13} \
4 &left( 13 + frac{9}{8}13^2right) \
&frac{1625}{2} \
end{align}$
Then, I calculate the length of the curve:
$begin{align}
L &= int_0^{13} sqrt{1 + [r'(t)]^2} dx \
L &= int_0^{13} sqrt{1 + 1 + frac{9}{4}t} dx \
L &= int_0^{13} sqrt{2 + frac{9}{4}t} dx \
end{align}$
Letting $u = 2 + 9/4 t$
$begin{align}
L &= frac{4}{9} int_2^{125 / 4} u^{1/2} du \
L &= frac{8}{27} left( u^{3/2} right|_{ 2}^{ 125 / 4} \
L &= frac{8}{27} left[ left(frac{125}{4}right)^{3/2} - 2^{3/2} right] \
L &= frac{625 sqrt{5} - 16 sqrt{2} }{27} \
end{align}$
Then, dividing the integral by the length of the curve gives a gnarly, incorrect mess.
What am I not understanding here? Are there algebra errors? Errors in the calculus?
integration multivariable-calculus average curves
asked Apr 15 '16 at 19:00
StudentsTea
7862722
add a comment |
I'm having quite a hard time calculating the average value of a line integral.
Given the surface $f(x,y) = sqrt{16 + 36y^{2/3}}$ and the curve $y = x^{3/2}$, I need to calculate the average value of the integral of the surface for $0 leq x leq 13$
I start by parameterizing the curve and the surface for $0 leq t leq 13:
$begin{align}
r(t) &= langle t, t^{3/2} rangle \
f(t) &= sqrt{16+36t} \
end{align}$
And calculate a few things I'll need later:
$begin{align}
r'(t) &= langle 1, frac{3}{2}t^{1/2} rangle \
left|r'(t)right| &= sqrt{ 1 + frac{9}{4}t } \
end{align}$
Next, I calculate the line integral:
$begin{align}
&int_0^{13} f(t) left|r'(t)right| dt \
&int_0^{13} sqrt{16+36t} sqrt{ 1 + frac{9}{4}t } dt \
2 &int_0^{13} sqrt{4+9t} sqrt{ 1 + frac{9}{4}t } dt \
2 &int_0^{13} sqrt{ frac{4}{4} (4+9t) } sqrt{ 1 + frac{9}{4}t } dt \
4 &int_0^{13} sqrt{ 1 + frac{9}{4}t } sqrt{ 1 + frac{9}{4}t } dt \
4 &int_0^{13} 1 + frac{9}{4}t dt \
4 &left( t + frac{9}{8}t^2right|_0^{13} \
4 &left( 13 + frac{9}{8}13^2right) \
&frac{1625}{2} \
end{align}$
Then, I calculate the length of the curve:
$begin{align}
L &= int_0^{13} sqrt{1 + [r'(t)]^2} dx \
L &= int_0^{13} sqrt{1 + 1 + frac{9}{4}t} dx \
L &= int_0^{13} sqrt{2 + frac{9}{4}t} dx \
end{align}$
Letting $u = 2 + 9/4 t$
$begin{align}
L &= frac{4}{9} int_2^{125 / 4} u^{1/2} du \
L &= frac{8}{27} left( u^{3/2} right|_{ 2}^{ 125 / 4} \
L &= frac{8}{27} left[ left(frac{125}{4}right)^{3/2} - 2^{3/2} right] \
L &= frac{625 sqrt{5} - 16 sqrt{2} }{27} \
end{align}$
Then, dividing the integral by the length of the curve gives a gnarly, incorrect mess.
What am I not understanding here? Are there algebra errors? Errors in the calculus?
integration multivariable-calculus average curves
asked Apr 15 '16 at 19:00
StudentsTea
7862722
I'm having quite a hard time calculating the average value of a line integral.
Given the surface $f(x,y) = sqrt{16 + 36y^{2/3}}$ and the curve $y = x^{3/2}$, I need to calculate the average value of the integral of the surface for $0 leq x leq 13$
I start by parameterizing the curve and the surface for $0 leq t leq 13:
$begin{align}
r(t) &= langle t, t^{3/2} rangle \
f(t) &= sqrt{16+36t} \
end{align}$
And calculate a few things I'll need later:
$begin{align}
r'(t) &= langle 1, frac{3}{2}t^{1/2} rangle \
left|r'(t)right| &= sqrt{ 1 + frac{9}{4}t } \
end{align}$
Next, I calculate the line integral:
$begin{align}
&int_0^{13} f(t) left|r'(t)right| dt \
&int_0^{13} sqrt{16+36t} sqrt{ 1 + frac{9}{4}t } dt \
2 &int_0^{13} sqrt{4+9t} sqrt{ 1 + frac{9}{4}t } dt \
2 &int_0^{13} sqrt{ frac{4}{4} (4+9t) } sqrt{ 1 + frac{9}{4}t } dt \
4 &int_0^{13} sqrt{ 1 + frac{9}{4}t } sqrt{ 1 + frac{9}{4}t } dt \
4 &int_0^{13} 1 + frac{9}{4}t dt \
4 &left( t + frac{9}{8}t^2right|_0^{13} \
4 &left( 13 + frac{9}{8}13^2right) \
&frac{1625}{2} \
end{align}$
Then, I calculate the length of the curve:
$begin{align}
L &= int_0^{13} sqrt{1 + [r'(t)]^2} dx \
L &= int_0^{13} sqrt{1 + 1 + frac{9}{4}t} dx \
L &= int_0^{13} sqrt{2 + frac{9}{4}t} dx \
end{align}$
Letting $u = 2 + 9/4 t$
$begin{align}
L &= frac{4}{9} int_2^{125 / 4} u^{1/2} du \
L &= frac{8}{27} left( u^{3/2} right|_{ 2}^{ 125 / 4} \
L &= frac{8}{27} left[ left(frac{125}{4}right)^{3/2} - 2^{3/2} right] \
L &= frac{625 sqrt{5} - 16 sqrt{2} }{27} \
end{align}$
Then, dividing the integral by the length of the curve gives a gnarly, incorrect mess.
What am I not understanding here? Are there algebra errors? Errors in the calculus?
integration multivariable-calculus average curves
integration multivariable-calculus average curves
asked Apr 15 '16 at 19:00
StudentsTea
7862722
asked Apr 15 '16 at 19:00
StudentsTea
7862722
asked Apr 15 '16 at 19:00
StudentsTea
7862722
asked Apr 15 '16 at 19:00
StudentsTea
7862722
asked Apr 15 '16 at 19:00
StudentsTea
7862722
7862722
add a comment |
add a comment |
0
active
oldest
votes
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%2f1744155%2faverage-value-of-a-line-integral%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2f1744155%2faverage-value-of-a-line-integral%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
