Mixing
Find the volume enclosed by $left({x^2over a^2}+{y^2over b^2}+{z^2over c^2}right)^2={xover h}$
$begingroup$
Find the volume enclosed by $left({x^2over a^2}+{y^2over b^2}+{z^2over c^2}right)^2={xover h}$, where a, b, c, h are real constants.
My attempt:
I use polar coordinates transformation and get
$$x=a r sinphi costheta$$
$$y=b r sinphi sintheta$$
$$z=c r cosphi$$
The the original equation becomes $r^3={aover h}sinphi costheta$. so I get $rin(0,1)$, $phiin(0,pi)$, $thetain(0,pi/2)$ and $(3pi/2,2pi)$.
and the volume will be: $newcommand{d}{;mathrm{d}}$
$$V=int_0^{pi/2}int_0^piint_0^1 abcr^2sinphi d{r} dphi dtheta+int_{3pi/2}^{2pi}int_0^piint_0^1 abcr^2 sinphi d{r} dphi dtheta$$
I'm not confident with my answer at all. Can anyone show me how to find the volume of this?
calculus
edited Jan 23 at 9:33
Anakhand
259114
asked Jan 23 at 9:02
Yibei HeYibei He
3139
$endgroup$
add a comment |
$begingroup$
Find the volume enclosed by $left({x^2over a^2}+{y^2over b^2}+{z^2over c^2}right)^2={xover h}$, where a, b, c, h are real constants.
My attempt:
I use polar coordinates transformation and get
$$x=a r sinphi costheta$$
$$y=b r sinphi sintheta$$
$$z=c r cosphi$$
The the original equation becomes $r^3={aover h}sinphi costheta$. so I get $rin(0,1)$, $phiin(0,pi)$, $thetain(0,pi/2)$ and $(3pi/2,2pi)$.
and the volume will be: $newcommand{d}{;mathrm{d}}$
$$V=int_0^{pi/2}int_0^piint_0^1 abcr^2sinphi d{r} dphi dtheta+int_{3pi/2}^{2pi}int_0^piint_0^1 abcr^2 sinphi d{r} dphi dtheta$$
I'm not confident with my answer at all. Can anyone show me how to find the volume of this?
calculus
edited Jan 23 at 9:33
Anakhand
259114
asked Jan 23 at 9:02
Yibei HeYibei He
3139
$endgroup$
add a comment |
$begingroup$
Find the volume enclosed by $left({x^2over a^2}+{y^2over b^2}+{z^2over c^2}right)^2={xover h}$, where a, b, c, h are real constants.
My attempt:
I use polar coordinates transformation and get
$$x=a r sinphi costheta$$
$$y=b r sinphi sintheta$$
$$z=c r cosphi$$
The the original equation becomes $r^3={aover h}sinphi costheta$. so I get $rin(0,1)$, $phiin(0,pi)$, $thetain(0,pi/2)$ and $(3pi/2,2pi)$.
and the volume will be: $newcommand{d}{;mathrm{d}}$
$$V=int_0^{pi/2}int_0^piint_0^1 abcr^2sinphi d{r} dphi dtheta+int_{3pi/2}^{2pi}int_0^piint_0^1 abcr^2 sinphi d{r} dphi dtheta$$
I'm not confident with my answer at all. Can anyone show me how to find the volume of this?
calculus
edited Jan 23 at 9:33
Anakhand
259114
asked Jan 23 at 9:02
Yibei HeYibei He
3139
$endgroup$
Find the volume enclosed by $left({x^2over a^2}+{y^2over b^2}+{z^2over c^2}right)^2={xover h}$, where a, b, c, h are real constants.
My attempt:
I use polar coordinates transformation and get
$$x=a r sinphi costheta$$
$$y=b r sinphi sintheta$$
$$z=c r cosphi$$
The the original equation becomes $r^3={aover h}sinphi costheta$. so I get $rin(0,1)$, $phiin(0,pi)$, $thetain(0,pi/2)$ and $(3pi/2,2pi)$.
and the volume will be: $newcommand{d}{;mathrm{d}}$
$$V=int_0^{pi/2}int_0^piint_0^1 abcr^2sinphi d{r} dphi dtheta+int_{3pi/2}^{2pi}int_0^piint_0^1 abcr^2 sinphi d{r} dphi dtheta$$
I'm not confident with my answer at all. Can anyone show me how to find the volume of this?
calculus
calculus
edited Jan 23 at 9:33
Anakhand
259114
asked Jan 23 at 9:02
Yibei HeYibei He
3139
edited Jan 23 at 9:33
Anakhand
259114
asked Jan 23 at 9:02
Yibei HeYibei He
3139
edited Jan 23 at 9:33
Anakhand
259114
edited Jan 23 at 9:33
Anakhand
259114
edited Jan 23 at 9:33
Anakhand
259114
259114
asked Jan 23 at 9:02
Yibei HeYibei He
3139
asked Jan 23 at 9:02
Yibei HeYibei He
3139
asked Jan 23 at 9:02
Yibei HeYibei He
3139
3139
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Since your volume is enclosed by the surface $r^3=frac{a}{h}sin{phi}cos{theta}$, the upper limit for the integral over r should be $r=sqrt[3]{frac{a}{h}sin{phi}cos{theta}}$. The volume should be
$$V=int_0^{pi/2} int_0^{pi}int_0^{sqrt[3]{frac{a}{h}sin{phi}cos{theta}}} mathrm{d}r mathrm{d}phi mathrm{d} theta abcr^2 sin phi \ + int_{frac{3pi}{2}}^{2pi} int_0^{pi}int_0^{sqrt[3]{frac{a}{h}sin{phi}cos{theta}}} mathrm{d}r mathrm{d}phi mathrm{d} theta abcr^2 sin phi$$
answered Jan 23 at 9:33
Angela RichardsonAngela Richardson
5,40411734
$endgroup$
$begingroup$
should I discuss whether $agt h$ since $rin (0,1)$?
$endgroup$
– Yibei He
Jan 26 at 7:07
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%2f3084244%2ffind-the-volume-enclosed-by-leftx2-over-a2y2-over-b2z2-over-c2%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
$begingroup$
Since your volume is enclosed by the surface $r^3=frac{a}{h}sin{phi}cos{theta}$, the upper limit for the integral over r should be $r=sqrt[3]{frac{a}{h}sin{phi}cos{theta}}$. The volume should be
$$V=int_0^{pi/2} int_0^{pi}int_0^{sqrt[3]{frac{a}{h}sin{phi}cos{theta}}} mathrm{d}r mathrm{d}phi mathrm{d} theta abcr^2 sin phi \ + int_{frac{3pi}{2}}^{2pi} int_0^{pi}int_0^{sqrt[3]{frac{a}{h}sin{phi}cos{theta}}} mathrm{d}r mathrm{d}phi mathrm{d} theta abcr^2 sin phi$$
answered Jan 23 at 9:33
Angela RichardsonAngela Richardson
5,40411734
$endgroup$
$begingroup$
should I discuss whether $agt h$ since $rin (0,1)$?
$endgroup$
– Yibei He
Jan 26 at 7:07
add a comment |
$begingroup$
Since your volume is enclosed by the surface $r^3=frac{a}{h}sin{phi}cos{theta}$, the upper limit for the integral over r should be $r=sqrt[3]{frac{a}{h}sin{phi}cos{theta}}$. The volume should be
$$V=int_0^{pi/2} int_0^{pi}int_0^{sqrt[3]{frac{a}{h}sin{phi}cos{theta}}} mathrm{d}r mathrm{d}phi mathrm{d} theta abcr^2 sin phi \ + int_{frac{3pi}{2}}^{2pi} int_0^{pi}int_0^{sqrt[3]{frac{a}{h}sin{phi}cos{theta}}} mathrm{d}r mathrm{d}phi mathrm{d} theta abcr^2 sin phi$$
answered Jan 23 at 9:33
Angela RichardsonAngela Richardson
5,40411734
$endgroup$
$begingroup$
should I discuss whether $agt h$ since $rin (0,1)$?
$endgroup$
– Yibei He
Jan 26 at 7:07
add a comment |
$begingroup$
Since your volume is enclosed by the surface $r^3=frac{a}{h}sin{phi}cos{theta}$, the upper limit for the integral over r should be $r=sqrt[3]{frac{a}{h}sin{phi}cos{theta}}$. The volume should be
$$V=int_0^{pi/2} int_0^{pi}int_0^{sqrt[3]{frac{a}{h}sin{phi}cos{theta}}} mathrm{d}r mathrm{d}phi mathrm{d} theta abcr^2 sin phi \ + int_{frac{3pi}{2}}^{2pi} int_0^{pi}int_0^{sqrt[3]{frac{a}{h}sin{phi}cos{theta}}} mathrm{d}r mathrm{d}phi mathrm{d} theta abcr^2 sin phi$$
answered Jan 23 at 9:33
Angela RichardsonAngela Richardson
5,40411734
$endgroup$
Since your volume is enclosed by the surface $r^3=frac{a}{h}sin{phi}cos{theta}$, the upper limit for the integral over r should be $r=sqrt[3]{frac{a}{h}sin{phi}cos{theta}}$. The volume should be
$$V=int_0^{pi/2} int_0^{pi}int_0^{sqrt[3]{frac{a}{h}sin{phi}cos{theta}}} mathrm{d}r mathrm{d}phi mathrm{d} theta abcr^2 sin phi \ + int_{frac{3pi}{2}}^{2pi} int_0^{pi}int_0^{sqrt[3]{frac{a}{h}sin{phi}cos{theta}}} mathrm{d}r mathrm{d}phi mathrm{d} theta abcr^2 sin phi$$
answered Jan 23 at 9:33
Angela RichardsonAngela Richardson
5,40411734
edited Jan 27 at 6:18
answered Jan 23 at 9:33
Angela RichardsonAngela Richardson
5,40411734
answered Jan 23 at 9:33
Angela RichardsonAngela Richardson
5,40411734
answered Jan 23 at 9:33
Angela RichardsonAngela Richardson
5,40411734
5,40411734
$begingroup$
should I discuss whether $agt h$ since $rin (0,1)$?
$endgroup$
– Yibei He
Jan 26 at 7:07
add a comment |
$begingroup$
should I discuss whether $agt h$ since $rin (0,1)$?
$endgroup$
– Yibei He
Jan 26 at 7:07
$begingroup$
should I discuss whether $agt h$ since $rin (0,1)$?
$endgroup$
– Yibei He
Jan 26 at 7:07
$begingroup$
should I discuss whether $agt h$ since $rin (0,1)$?
$endgroup$
– Yibei He
Jan 26 at 7:07
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%2f3084244%2ffind-the-volume-enclosed-by-leftx2-over-a2y2-over-b2z2-over-c2%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
