Find the volume bounded by the paraboloid $x^2+y^2=az$, the cylinder $x^2+y^2=2ay$ and the plane $z=0$
$begingroup$
Find the volume bounded by the paraboloid $x^2+y^2=az$, the cylinder $x^2+y^2=2ay$ and the plane $z=0$
My work
Changing to cylindrical coordinates
For Paraboloid
$$x^2+y^2=az;;;implies r^2=az;;;implies z=frac{r^2}{a}$$
For the cylinder
$$x^2+y^2=2ay;;;implies r=2a sin theta$$
Since this volume lies only in first two quadrants $theta$ goes from $0$ to $pi$
Volume=$int_{theta=0}^{pi}int_{r=0}^{2asin theta}int_{z=0}^{r^2/a}r dr dtheta dz$
multiple-integral
$endgroup$
add a comment |
$begingroup$
Find the volume bounded by the paraboloid $x^2+y^2=az$, the cylinder $x^2+y^2=2ay$ and the plane $z=0$
My work
Changing to cylindrical coordinates
For Paraboloid
$$x^2+y^2=az;;;implies r^2=az;;;implies z=frac{r^2}{a}$$
For the cylinder
$$x^2+y^2=2ay;;;implies r=2a sin theta$$
Since this volume lies only in first two quadrants $theta$ goes from $0$ to $pi$
Volume=$int_{theta=0}^{pi}int_{r=0}^{2asin theta}int_{z=0}^{r^2/a}r dr dtheta dz$
multiple-integral
$endgroup$
$begingroup$
Well, $yin[0..2a], xin[-sqrt{2ay-y^2}..sqrt{2ay-y^2}], zin[0..(x^2+y^2)/a]$ but a change of variables might be better.
$endgroup$
– Graham Kemp
Aug 19 '17 at 9:25
add a comment |
$begingroup$
Find the volume bounded by the paraboloid $x^2+y^2=az$, the cylinder $x^2+y^2=2ay$ and the plane $z=0$
My work
Changing to cylindrical coordinates
For Paraboloid
$$x^2+y^2=az;;;implies r^2=az;;;implies z=frac{r^2}{a}$$
For the cylinder
$$x^2+y^2=2ay;;;implies r=2a sin theta$$
Since this volume lies only in first two quadrants $theta$ goes from $0$ to $pi$
Volume=$int_{theta=0}^{pi}int_{r=0}^{2asin theta}int_{z=0}^{r^2/a}r dr dtheta dz$
multiple-integral
$endgroup$
Find the volume bounded by the paraboloid $x^2+y^2=az$, the cylinder $x^2+y^2=2ay$ and the plane $z=0$
My work
Changing to cylindrical coordinates
For Paraboloid
$$x^2+y^2=az;;;implies r^2=az;;;implies z=frac{r^2}{a}$$
For the cylinder
$$x^2+y^2=2ay;;;implies r=2a sin theta$$
Since this volume lies only in first two quadrants $theta$ goes from $0$ to $pi$
Volume=$int_{theta=0}^{pi}int_{r=0}^{2asin theta}int_{z=0}^{r^2/a}r dr dtheta dz$
multiple-integral
multiple-integral
edited Aug 22 '17 at 0:37
user467745
asked Aug 19 '17 at 9:00
user467745user467745
237112
237112
$begingroup$
Well, $yin[0..2a], xin[-sqrt{2ay-y^2}..sqrt{2ay-y^2}], zin[0..(x^2+y^2)/a]$ but a change of variables might be better.
$endgroup$
– Graham Kemp
Aug 19 '17 at 9:25
add a comment |
$begingroup$
Well, $yin[0..2a], xin[-sqrt{2ay-y^2}..sqrt{2ay-y^2}], zin[0..(x^2+y^2)/a]$ but a change of variables might be better.
$endgroup$
– Graham Kemp
Aug 19 '17 at 9:25
$begingroup$
Well, $yin[0..2a], xin[-sqrt{2ay-y^2}..sqrt{2ay-y^2}], zin[0..(x^2+y^2)/a]$ but a change of variables might be better.
$endgroup$
– Graham Kemp
Aug 19 '17 at 9:25
$begingroup$
Well, $yin[0..2a], xin[-sqrt{2ay-y^2}..sqrt{2ay-y^2}], zin[0..(x^2+y^2)/a]$ but a change of variables might be better.
$endgroup$
– Graham Kemp
Aug 19 '17 at 9:25
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Using cylinder coordinate is the best way to solve this problem :
$$left{(rcostheta ,rsintheta +a ,z)mid theta in [0,2pi], rin[0,a], zin left[0,frac{2r^2+2arsin theta+a^2}{a}right]right}.$$
$endgroup$
$begingroup$
Yes, it does look much easier. But what would be the limits in cartesian coords?
$endgroup$
– user467745
Aug 19 '17 at 9:21
$begingroup$
@user467745 : Something as $yin [0,2a]$, $xin [-sqrt{2a-(y-1)^2},sqrt{2a-(y-1)^2}]$ and $zin [0, frac{x^2}{a}+frac{y^2}{a}]$.
$endgroup$
– Surb
Aug 19 '17 at 9:25
$begingroup$
Just to cross check-the answer would be $pi a^3$, right?
$endgroup$
– user467745
Aug 19 '17 at 10:27
$begingroup$
@user467745: I'm not a computer ;-)
$endgroup$
– Surb
Aug 19 '17 at 10:35
$begingroup$
Did I imply that? I am sorry. Thanks for your time
$endgroup$
– user467745
Aug 19 '17 at 10:53
|
show 5 more comments
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%2f2398981%2ffind-the-volume-bounded-by-the-paraboloid-x2y2-az-the-cylinder-x2y2-2a%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
$begingroup$
Using cylinder coordinate is the best way to solve this problem :
$$left{(rcostheta ,rsintheta +a ,z)mid theta in [0,2pi], rin[0,a], zin left[0,frac{2r^2+2arsin theta+a^2}{a}right]right}.$$
$endgroup$
$begingroup$
Yes, it does look much easier. But what would be the limits in cartesian coords?
$endgroup$
– user467745
Aug 19 '17 at 9:21
$begingroup$
@user467745 : Something as $yin [0,2a]$, $xin [-sqrt{2a-(y-1)^2},sqrt{2a-(y-1)^2}]$ and $zin [0, frac{x^2}{a}+frac{y^2}{a}]$.
$endgroup$
– Surb
Aug 19 '17 at 9:25
$begingroup$
Just to cross check-the answer would be $pi a^3$, right?
$endgroup$
– user467745
Aug 19 '17 at 10:27
$begingroup$
@user467745: I'm not a computer ;-)
$endgroup$
– Surb
Aug 19 '17 at 10:35
$begingroup$
Did I imply that? I am sorry. Thanks for your time
$endgroup$
– user467745
Aug 19 '17 at 10:53
|
show 5 more comments
$begingroup$
Using cylinder coordinate is the best way to solve this problem :
$$left{(rcostheta ,rsintheta +a ,z)mid theta in [0,2pi], rin[0,a], zin left[0,frac{2r^2+2arsin theta+a^2}{a}right]right}.$$
$endgroup$
$begingroup$
Yes, it does look much easier. But what would be the limits in cartesian coords?
$endgroup$
– user467745
Aug 19 '17 at 9:21
$begingroup$
@user467745 : Something as $yin [0,2a]$, $xin [-sqrt{2a-(y-1)^2},sqrt{2a-(y-1)^2}]$ and $zin [0, frac{x^2}{a}+frac{y^2}{a}]$.
$endgroup$
– Surb
Aug 19 '17 at 9:25
$begingroup$
Just to cross check-the answer would be $pi a^3$, right?
$endgroup$
– user467745
Aug 19 '17 at 10:27
$begingroup$
@user467745: I'm not a computer ;-)
$endgroup$
– Surb
Aug 19 '17 at 10:35
$begingroup$
Did I imply that? I am sorry. Thanks for your time
$endgroup$
– user467745
Aug 19 '17 at 10:53
|
show 5 more comments
$begingroup$
Using cylinder coordinate is the best way to solve this problem :
$$left{(rcostheta ,rsintheta +a ,z)mid theta in [0,2pi], rin[0,a], zin left[0,frac{2r^2+2arsin theta+a^2}{a}right]right}.$$
$endgroup$
Using cylinder coordinate is the best way to solve this problem :
$$left{(rcostheta ,rsintheta +a ,z)mid theta in [0,2pi], rin[0,a], zin left[0,frac{2r^2+2arsin theta+a^2}{a}right]right}.$$
edited Aug 19 '17 at 15:43
answered Aug 19 '17 at 9:15
SurbSurb
38.4k94476
38.4k94476
$begingroup$
Yes, it does look much easier. But what would be the limits in cartesian coords?
$endgroup$
– user467745
Aug 19 '17 at 9:21
$begingroup$
@user467745 : Something as $yin [0,2a]$, $xin [-sqrt{2a-(y-1)^2},sqrt{2a-(y-1)^2}]$ and $zin [0, frac{x^2}{a}+frac{y^2}{a}]$.
$endgroup$
– Surb
Aug 19 '17 at 9:25
$begingroup$
Just to cross check-the answer would be $pi a^3$, right?
$endgroup$
– user467745
Aug 19 '17 at 10:27
$begingroup$
@user467745: I'm not a computer ;-)
$endgroup$
– Surb
Aug 19 '17 at 10:35
$begingroup$
Did I imply that? I am sorry. Thanks for your time
$endgroup$
– user467745
Aug 19 '17 at 10:53
|
show 5 more comments
$begingroup$
Yes, it does look much easier. But what would be the limits in cartesian coords?
$endgroup$
– user467745
Aug 19 '17 at 9:21
$begingroup$
@user467745 : Something as $yin [0,2a]$, $xin [-sqrt{2a-(y-1)^2},sqrt{2a-(y-1)^2}]$ and $zin [0, frac{x^2}{a}+frac{y^2}{a}]$.
$endgroup$
– Surb
Aug 19 '17 at 9:25
$begingroup$
Just to cross check-the answer would be $pi a^3$, right?
$endgroup$
– user467745
Aug 19 '17 at 10:27
$begingroup$
@user467745: I'm not a computer ;-)
$endgroup$
– Surb
Aug 19 '17 at 10:35
$begingroup$
Did I imply that? I am sorry. Thanks for your time
$endgroup$
– user467745
Aug 19 '17 at 10:53
$begingroup$
Yes, it does look much easier. But what would be the limits in cartesian coords?
$endgroup$
– user467745
Aug 19 '17 at 9:21
$begingroup$
Yes, it does look much easier. But what would be the limits in cartesian coords?
$endgroup$
– user467745
Aug 19 '17 at 9:21
$begingroup$
@user467745 : Something as $yin [0,2a]$, $xin [-sqrt{2a-(y-1)^2},sqrt{2a-(y-1)^2}]$ and $zin [0, frac{x^2}{a}+frac{y^2}{a}]$.
$endgroup$
– Surb
Aug 19 '17 at 9:25
$begingroup$
@user467745 : Something as $yin [0,2a]$, $xin [-sqrt{2a-(y-1)^2},sqrt{2a-(y-1)^2}]$ and $zin [0, frac{x^2}{a}+frac{y^2}{a}]$.
$endgroup$
– Surb
Aug 19 '17 at 9:25
$begingroup$
Just to cross check-the answer would be $pi a^3$, right?
$endgroup$
– user467745
Aug 19 '17 at 10:27
$begingroup$
Just to cross check-the answer would be $pi a^3$, right?
$endgroup$
– user467745
Aug 19 '17 at 10:27
$begingroup$
@user467745: I'm not a computer ;-)
$endgroup$
– Surb
Aug 19 '17 at 10:35
$begingroup$
@user467745: I'm not a computer ;-)
$endgroup$
– Surb
Aug 19 '17 at 10:35
$begingroup$
Did I imply that? I am sorry. Thanks for your time
$endgroup$
– user467745
Aug 19 '17 at 10:53
$begingroup$
Did I imply that? I am sorry. Thanks for your time
$endgroup$
– user467745
Aug 19 '17 at 10:53
|
show 5 more comments
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%2f2398981%2ffind-the-volume-bounded-by-the-paraboloid-x2y2-az-the-cylinder-x2y2-2a%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
$begingroup$
Well, $yin[0..2a], xin[-sqrt{2ay-y^2}..sqrt{2ay-y^2}], zin[0..(x^2+y^2)/a]$ but a change of variables might be better.
$endgroup$
– Graham Kemp
Aug 19 '17 at 9:25