Simplest Way to Evaluate Lengthy Integration Equation with Succinct Answer
$begingroup$
Recently, I determined the volume of the solid of revolution created by rotating the parabola, $y=ax^2$, about the axis of revolution, $y=bx$, as a function of $a$ and $b$, where both variables are positive, real numbers to be
$$V=frac{pi b^5}{30a^3sqrt{1+b^2}}tag{1}$$
I proved this several different ways, but I got stuck at the conclusion of my application of the disk method. The final, integrable equation that I derived was
$$V=int_0^{frac{b}{a}sqrt{1+b^2}}pileft(-frac{1}{b}x-frac{sqrt{1+b^2}}{2ab^2}+frac{sqrt{1+b^2}}{2ab^2}sqrt{4abxsqrt{1+b^2}+1}right)^2dxtag{2}$$
I happen to know that equation 2 does simplify to equation 1 (thank you, Desmos). However, I could not think of a way of performing this simplification without a full-blown expansion of the integrand, separation at the plus/minus signs, integration, returning of limits, and simplification, which (I began but did not finish) took a head-exploding amount of care.
In effect, my question is: What is the most conservative, simplest, best way of evaluating equation 2? All solutions are welcome, but one that uses relatively elementary calculus concepts would be best (I'm a junior in high school taking AP BC Calculus).
Thanks!
calculus algebra-precalculus volume
asked Jan 1 at 20:00
Shady PuckShady Puck
1285
$endgroup$
add a comment |
$begingroup$
Recently, I determined the volume of the solid of revolution created by rotating the parabola, $y=ax^2$, about the axis of revolution, $y=bx$, as a function of $a$ and $b$, where both variables are positive, real numbers to be
$$V=frac{pi b^5}{30a^3sqrt{1+b^2}}tag{1}$$
I proved this several different ways, but I got stuck at the conclusion of my application of the disk method. The final, integrable equation that I derived was
$$V=int_0^{frac{b}{a}sqrt{1+b^2}}pileft(-frac{1}{b}x-frac{sqrt{1+b^2}}{2ab^2}+frac{sqrt{1+b^2}}{2ab^2}sqrt{4abxsqrt{1+b^2}+1}right)^2dxtag{2}$$
I happen to know that equation 2 does simplify to equation 1 (thank you, Desmos). However, I could not think of a way of performing this simplification without a full-blown expansion of the integrand, separation at the plus/minus signs, integration, returning of limits, and simplification, which (I began but did not finish) took a head-exploding amount of care.
In effect, my question is: What is the most conservative, simplest, best way of evaluating equation 2? All solutions are welcome, but one that uses relatively elementary calculus concepts would be best (I'm a junior in high school taking AP BC Calculus).
Thanks!
calculus algebra-precalculus volume
asked Jan 1 at 20:00
Shady PuckShady Puck
1285
$endgroup$
$begingroup$
In your link to "proved this several ways", $b$ was $1.$
$endgroup$
– coffeemath
Jan 1 at 20:22
1
$begingroup$
@coffeemath Yes — I just meant that I applied the cone, shell, and disk methods described in that question and its answers. I apologize for the confusion.
$endgroup$
– Shady Puck
Jan 1 at 20:46
$begingroup$
Shady Puck-- Just finished getting the rotation volume using Pappus' Theorem [find centroid, get distance $d$ from there to rotation axis, then it's $2pi d A$ with $A$ the area of rotated region]. I got exactly your formula (1). I'm curious how you set up your disc method-- looks like an involved expression to integrate; I assume discs are perpendicular to rotation axis?
$endgroup$
– coffeemath
Jan 2 at 19:45
add a comment |
$begingroup$
Recently, I determined the volume of the solid of revolution created by rotating the parabola, $y=ax^2$, about the axis of revolution, $y=bx$, as a function of $a$ and $b$, where both variables are positive, real numbers to be
$$V=frac{pi b^5}{30a^3sqrt{1+b^2}}tag{1}$$
I proved this several different ways, but I got stuck at the conclusion of my application of the disk method. The final, integrable equation that I derived was
$$V=int_0^{frac{b}{a}sqrt{1+b^2}}pileft(-frac{1}{b}x-frac{sqrt{1+b^2}}{2ab^2}+frac{sqrt{1+b^2}}{2ab^2}sqrt{4abxsqrt{1+b^2}+1}right)^2dxtag{2}$$
I happen to know that equation 2 does simplify to equation 1 (thank you, Desmos). However, I could not think of a way of performing this simplification without a full-blown expansion of the integrand, separation at the plus/minus signs, integration, returning of limits, and simplification, which (I began but did not finish) took a head-exploding amount of care.
In effect, my question is: What is the most conservative, simplest, best way of evaluating equation 2? All solutions are welcome, but one that uses relatively elementary calculus concepts would be best (I'm a junior in high school taking AP BC Calculus).
Thanks!
calculus algebra-precalculus volume
asked Jan 1 at 20:00
Shady PuckShady Puck
1285
$endgroup$
Recently, I determined the volume of the solid of revolution created by rotating the parabola, $y=ax^2$, about the axis of revolution, $y=bx$, as a function of $a$ and $b$, where both variables are positive, real numbers to be
$$V=frac{pi b^5}{30a^3sqrt{1+b^2}}tag{1}$$
I proved this several different ways, but I got stuck at the conclusion of my application of the disk method. The final, integrable equation that I derived was
$$V=int_0^{frac{b}{a}sqrt{1+b^2}}pileft(-frac{1}{b}x-frac{sqrt{1+b^2}}{2ab^2}+frac{sqrt{1+b^2}}{2ab^2}sqrt{4abxsqrt{1+b^2}+1}right)^2dxtag{2}$$
I happen to know that equation 2 does simplify to equation 1 (thank you, Desmos). However, I could not think of a way of performing this simplification without a full-blown expansion of the integrand, separation at the plus/minus signs, integration, returning of limits, and simplification, which (I began but did not finish) took a head-exploding amount of care.
In effect, my question is: What is the most conservative, simplest, best way of evaluating equation 2? All solutions are welcome, but one that uses relatively elementary calculus concepts would be best (I'm a junior in high school taking AP BC Calculus).
Thanks!
calculus algebra-precalculus volume
calculus algebra-precalculus volume
asked Jan 1 at 20:00
Shady PuckShady Puck
1285
asked Jan 1 at 20:00
Shady PuckShady Puck
1285
asked Jan 1 at 20:00
Shady PuckShady Puck
1285
asked Jan 1 at 20:00
Shady PuckShady Puck
1285
asked Jan 1 at 20:00
Shady PuckShady Puck
1285
1285
$begingroup$
In your link to "proved this several ways", $b$ was $1.$
$endgroup$
– coffeemath
Jan 1 at 20:22
1
$begingroup$
@coffeemath Yes — I just meant that I applied the cone, shell, and disk methods described in that question and its answers. I apologize for the confusion.
$endgroup$
– Shady Puck
Jan 1 at 20:46
$begingroup$
Shady Puck-- Just finished getting the rotation volume using Pappus' Theorem [find centroid, get distance $d$ from there to rotation axis, then it's $2pi d A$ with $A$ the area of rotated region]. I got exactly your formula (1). I'm curious how you set up your disc method-- looks like an involved expression to integrate; I assume discs are perpendicular to rotation axis?
$endgroup$
– coffeemath
Jan 2 at 19:45
add a comment |
$begingroup$
In your link to "proved this several ways", $b$ was $1.$
$endgroup$
– coffeemath
Jan 1 at 20:22
1
$begingroup$
@coffeemath Yes — I just meant that I applied the cone, shell, and disk methods described in that question and its answers. I apologize for the confusion.
$endgroup$
– Shady Puck
Jan 1 at 20:46
$begingroup$
Shady Puck-- Just finished getting the rotation volume using Pappus' Theorem [find centroid, get distance $d$ from there to rotation axis, then it's $2pi d A$ with $A$ the area of rotated region]. I got exactly your formula (1). I'm curious how you set up your disc method-- looks like an involved expression to integrate; I assume discs are perpendicular to rotation axis?
$endgroup$
– coffeemath
Jan 2 at 19:45
$begingroup$
In your link to "proved this several ways", $b$ was $1.$
$endgroup$
– coffeemath
Jan 1 at 20:22
$begingroup$
In your link to "proved this several ways", $b$ was $1.$
$endgroup$
– coffeemath
Jan 1 at 20:22
1
1
$begingroup$
@coffeemath Yes — I just meant that I applied the cone, shell, and disk methods described in that question and its answers. I apologize for the confusion.
$endgroup$
– Shady Puck
Jan 1 at 20:46
$begingroup$
@coffeemath Yes — I just meant that I applied the cone, shell, and disk methods described in that question and its answers. I apologize for the confusion.
$endgroup$
– Shady Puck
Jan 1 at 20:46
$begingroup$
Shady Puck-- Just finished getting the rotation volume using Pappus' Theorem [find centroid, get distance $d$ from there to rotation axis, then it's $2pi d A$ with $A$ the area of rotated region]. I got exactly your formula (1). I'm curious how you set up your disc method-- looks like an involved expression to integrate; I assume discs are perpendicular to rotation axis?
$endgroup$
– coffeemath
Jan 2 at 19:45
$begingroup$
Shady Puck-- Just finished getting the rotation volume using Pappus' Theorem [find centroid, get distance $d$ from there to rotation axis, then it's $2pi d A$ with $A$ the area of rotated region]. I got exactly your formula (1). I'm curious how you set up your disc method-- looks like an involved expression to integrate; I assume discs are perpendicular to rotation axis?
$endgroup$
– coffeemath
Jan 2 at 19:45
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Use computer algebra (e.g., Mathematica) which gives a very simple answer:
$$frac{pi b^5}{30 sqrt{b^2+1}}$$
answered Jan 1 at 20:06
David G. StorkDavid G. Stork
11.5k41533
$endgroup$
$begingroup$
Well, yes, that would be very simple. I was asking for a alternate step-by-step solution that did not require the care of a full expansion. BTW, you may want to check your input (as your solution does not match my equation 1).
$endgroup$
– Shady Puck
Jan 1 at 20:11
$begingroup$
@ShadyPuck: I re-checked my code indeed implements your Eq. (2). Perhaps you should re-check it.
$endgroup$
– David G. Stork
Jan 1 at 20:14
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%2f3058824%2fsimplest-way-to-evaluate-lengthy-integration-equation-with-succinct-answer%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
$begingroup$
Use computer algebra (e.g., Mathematica) which gives a very simple answer:
$$frac{pi b^5}{30 sqrt{b^2+1}}$$
answered Jan 1 at 20:06
David G. StorkDavid G. Stork
11.5k41533
$endgroup$
$begingroup$
Well, yes, that would be very simple. I was asking for a alternate step-by-step solution that did not require the care of a full expansion. BTW, you may want to check your input (as your solution does not match my equation 1).
$endgroup$
– Shady Puck
Jan 1 at 20:11
$begingroup$
@ShadyPuck: I re-checked my code indeed implements your Eq. (2). Perhaps you should re-check it.
$endgroup$
– David G. Stork
Jan 1 at 20:14
add a comment |
$begingroup$
Use computer algebra (e.g., Mathematica) which gives a very simple answer:
$$frac{pi b^5}{30 sqrt{b^2+1}}$$
answered Jan 1 at 20:06
David G. StorkDavid G. Stork
11.5k41533
$endgroup$
$begingroup$
Well, yes, that would be very simple. I was asking for a alternate step-by-step solution that did not require the care of a full expansion. BTW, you may want to check your input (as your solution does not match my equation 1).
$endgroup$
– Shady Puck
Jan 1 at 20:11
$begingroup$
@ShadyPuck: I re-checked my code indeed implements your Eq. (2). Perhaps you should re-check it.
$endgroup$
– David G. Stork
Jan 1 at 20:14
add a comment |
$begingroup$
Use computer algebra (e.g., Mathematica) which gives a very simple answer:
$$frac{pi b^5}{30 sqrt{b^2+1}}$$
answered Jan 1 at 20:06
David G. StorkDavid G. Stork
11.5k41533
$endgroup$
Use computer algebra (e.g., Mathematica) which gives a very simple answer:
$$frac{pi b^5}{30 sqrt{b^2+1}}$$
answered Jan 1 at 20:06
David G. StorkDavid G. Stork
11.5k41533
answered Jan 1 at 20:06
David G. StorkDavid G. Stork
11.5k41533
answered Jan 1 at 20:06
David G. StorkDavid G. Stork
11.5k41533
answered Jan 1 at 20:06
David G. StorkDavid G. Stork
11.5k41533
11.5k41533
$begingroup$
Well, yes, that would be very simple. I was asking for a alternate step-by-step solution that did not require the care of a full expansion. BTW, you may want to check your input (as your solution does not match my equation 1).
$endgroup$
– Shady Puck
Jan 1 at 20:11
$begingroup$
@ShadyPuck: I re-checked my code indeed implements your Eq. (2). Perhaps you should re-check it.
$endgroup$
– David G. Stork
Jan 1 at 20:14
add a comment |
$begingroup$
Well, yes, that would be very simple. I was asking for a alternate step-by-step solution that did not require the care of a full expansion. BTW, you may want to check your input (as your solution does not match my equation 1).
$endgroup$
– Shady Puck
Jan 1 at 20:11
$begingroup$
@ShadyPuck: I re-checked my code indeed implements your Eq. (2). Perhaps you should re-check it.
$endgroup$
– David G. Stork
Jan 1 at 20:14
$begingroup$
Well, yes, that would be very simple. I was asking for a alternate step-by-step solution that did not require the care of a full expansion. BTW, you may want to check your input (as your solution does not match my equation 1).
$endgroup$
– Shady Puck
Jan 1 at 20:11
$begingroup$
Well, yes, that would be very simple. I was asking for a alternate step-by-step solution that did not require the care of a full expansion. BTW, you may want to check your input (as your solution does not match my equation 1).
$endgroup$
– Shady Puck
Jan 1 at 20:11
$begingroup$
@ShadyPuck: I re-checked my code indeed implements your Eq. (2). Perhaps you should re-check it.
$endgroup$
– David G. Stork
Jan 1 at 20:14
$begingroup$
@ShadyPuck: I re-checked my code indeed implements your Eq. (2). Perhaps you should re-check it.
$endgroup$
– David G. Stork
Jan 1 at 20:14
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%2f3058824%2fsimplest-way-to-evaluate-lengthy-integration-equation-with-succinct-answer%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
$begingroup$
In your link to "proved this several ways", $b$ was $1.$
$endgroup$
– coffeemath
Jan 1 at 20:22
1
$begingroup$
@coffeemath Yes — I just meant that I applied the cone, shell, and disk methods described in that question and its answers. I apologize for the confusion.
$endgroup$
– Shady Puck
Jan 1 at 20:46
$begingroup$
Shady Puck-- Just finished getting the rotation volume using Pappus' Theorem [find centroid, get distance $d$ from there to rotation axis, then it's $2pi d A$ with $A$ the area of rotated region]. I got exactly your formula (1). I'm curious how you set up your disc method-- looks like an involved expression to integrate; I assume discs are perpendicular to rotation axis?
$endgroup$
– coffeemath
Jan 2 at 19:45