Using spherical coordinates, find the volume.












0












$begingroup$


Find the volume of the solid that lies in the first octant above the cone $z=sqrt{3(x^2+y^2)}$ and inside the sphere $$x^{2}+y^{2}+z^{2}=4z $$ using spherical coordinates:



So here is what I have done, I would honestly just like to make sure if I am doing this properly because normally the "cone" is written just as $z=sqrt{x^2+y^2}$.



$$int_{0}^{frac{pi}{6}} int_{0}^{2pi} int_{0}^{4{cos(phi)}} {rho}^2sin(phi) ,{rm d}{rho},{rm d}{theta}, {rm d}{phi}$$



$$= 2pi int_{0}^{frac{pi}{6}} sin(phi) , {rm d}phi , left[frac{rho^3}{3}right]_0^{4cos(phi)} = 2pi cdot 64 int_0^{frac{pi}{6}} sin(phi)left(frac{cos^3(phi)}{3}right) , {rm d}phi \ = - 2pi cdot 64 cdot frac{1}{3} left[frac{cos^4(phi)}{4}right]_0^{frac{pi}{6}} = - 2pi cdot 64 cdot frac{1}{3} cdot frac{1}{4}left[cos^4left(frac{pi}{6}right) - cos^4(0)right] = left(frac{14pi}{3}right) = frac{14pi}{3}$$






integration multivariable-calculus definite-integrals volume spherical-coordinates







share|cite|improve this question











$endgroup$












  • $begingroup$
    You mean to find the volume enclosed by the finite region of the intersection of the cone $z = sqrt{3(x^2 + y^2)}$ and the ellipsoid-ish body $x^2 + y^2 + z^2 = 4z$. Right?
    $endgroup$
    – amcalde
    Dec 7 '15 at 15:12










  • $begingroup$
    yes except its a sphere
    $endgroup$
    – juliodesa
    Dec 7 '15 at 15:15










  • $begingroup$
    Corrected the RHS of sphere equation.
    $endgroup$
    – Narasimham
    Dec 7 '15 at 15:54










  • $begingroup$
    it was supposed to say 4z @Narasimham
    $endgroup$
    – juliodesa
    Dec 7 '15 at 16:01












  • $begingroup$
    At two places OP states it is a sphere.
    $endgroup$
    – Narasimham
    Dec 7 '15 at 16:50
















0












$begingroup$


Find the volume of the solid that lies in the first octant above the cone $z=sqrt{3(x^2+y^2)}$ and inside the sphere $$x^{2}+y^{2}+z^{2}=4z $$ using spherical coordinates:



So here is what I have done, I would honestly just like to make sure if I am doing this properly because normally the "cone" is written just as $z=sqrt{x^2+y^2}$.



$$int_{0}^{frac{pi}{6}} int_{0}^{2pi} int_{0}^{4{cos(phi)}} {rho}^2sin(phi) ,{rm d}{rho},{rm d}{theta}, {rm d}{phi}$$



$$= 2pi int_{0}^{frac{pi}{6}} sin(phi) , {rm d}phi , left[frac{rho^3}{3}right]_0^{4cos(phi)} = 2pi cdot 64 int_0^{frac{pi}{6}} sin(phi)left(frac{cos^3(phi)}{3}right) , {rm d}phi \ = - 2pi cdot 64 cdot frac{1}{3} left[frac{cos^4(phi)}{4}right]_0^{frac{pi}{6}} = - 2pi cdot 64 cdot frac{1}{3} cdot frac{1}{4}left[cos^4left(frac{pi}{6}right) - cos^4(0)right] = left(frac{14pi}{3}right) = frac{14pi}{3}$$






integration multivariable-calculus definite-integrals volume spherical-coordinates







share|cite|improve this question











$endgroup$












  • $begingroup$
    You mean to find the volume enclosed by the finite region of the intersection of the cone $z = sqrt{3(x^2 + y^2)}$ and the ellipsoid-ish body $x^2 + y^2 + z^2 = 4z$. Right?
    $endgroup$
    – amcalde
    Dec 7 '15 at 15:12










  • $begingroup$
    yes except its a sphere
    $endgroup$
    – juliodesa
    Dec 7 '15 at 15:15










  • $begingroup$
    Corrected the RHS of sphere equation.
    $endgroup$
    – Narasimham
    Dec 7 '15 at 15:54










  • $begingroup$
    it was supposed to say 4z @Narasimham
    $endgroup$
    – juliodesa
    Dec 7 '15 at 16:01












  • $begingroup$
    At two places OP states it is a sphere.
    $endgroup$
    – Narasimham
    Dec 7 '15 at 16:50














0












0








0





$begingroup$


Find the volume of the solid that lies in the first octant above the cone $z=sqrt{3(x^2+y^2)}$ and inside the sphere $$x^{2}+y^{2}+z^{2}=4z $$ using spherical coordinates:



So here is what I have done, I would honestly just like to make sure if I am doing this properly because normally the "cone" is written just as $z=sqrt{x^2+y^2}$.



$$int_{0}^{frac{pi}{6}} int_{0}^{2pi} int_{0}^{4{cos(phi)}} {rho}^2sin(phi) ,{rm d}{rho},{rm d}{theta}, {rm d}{phi}$$



$$= 2pi int_{0}^{frac{pi}{6}} sin(phi) , {rm d}phi , left[frac{rho^3}{3}right]_0^{4cos(phi)} = 2pi cdot 64 int_0^{frac{pi}{6}} sin(phi)left(frac{cos^3(phi)}{3}right) , {rm d}phi \ = - 2pi cdot 64 cdot frac{1}{3} left[frac{cos^4(phi)}{4}right]_0^{frac{pi}{6}} = - 2pi cdot 64 cdot frac{1}{3} cdot frac{1}{4}left[cos^4left(frac{pi}{6}right) - cos^4(0)right] = left(frac{14pi}{3}right) = frac{14pi}{3}$$






integration multivariable-calculus definite-integrals volume spherical-coordinates







share|cite|improve this question











$endgroup$




Find the volume of the solid that lies in the first octant above the cone $z=sqrt{3(x^2+y^2)}$ and inside the sphere $$x^{2}+y^{2}+z^{2}=4z $$ using spherical coordinates:



So here is what I have done, I would honestly just like to make sure if I am doing this properly because normally the "cone" is written just as $z=sqrt{x^2+y^2}$.



$$int_{0}^{frac{pi}{6}} int_{0}^{2pi} int_{0}^{4{cos(phi)}} {rho}^2sin(phi) ,{rm d}{rho},{rm d}{theta}, {rm d}{phi}$$



$$= 2pi int_{0}^{frac{pi}{6}} sin(phi) , {rm d}phi , left[frac{rho^3}{3}right]_0^{4cos(phi)} = 2pi cdot 64 int_0^{frac{pi}{6}} sin(phi)left(frac{cos^3(phi)}{3}right) , {rm d}phi \ = - 2pi cdot 64 cdot frac{1}{3} left[frac{cos^4(phi)}{4}right]_0^{frac{pi}{6}} = - 2pi cdot 64 cdot frac{1}{3} cdot frac{1}{4}left[cos^4left(frac{pi}{6}right) - cos^4(0)right] = left(frac{14pi}{3}right) = frac{14pi}{3}$$






integration multivariable-calculus definite-integrals volume spherical-coordinates




integration multivariable-calculus definite-integrals volume spherical-coordinates






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 7 '15 at 16:02







juliodesa

















asked Dec 7 '15 at 15:07









juliodesajuliodesa

5817




5817












  • $begingroup$
    You mean to find the volume enclosed by the finite region of the intersection of the cone $z = sqrt{3(x^2 + y^2)}$ and the ellipsoid-ish body $x^2 + y^2 + z^2 = 4z$. Right?
    $endgroup$
    – amcalde
    Dec 7 '15 at 15:12










  • $begingroup$
    yes except its a sphere
    $endgroup$
    – juliodesa
    Dec 7 '15 at 15:15










  • $begingroup$
    Corrected the RHS of sphere equation.
    $endgroup$
    – Narasimham
    Dec 7 '15 at 15:54










  • $begingroup$
    it was supposed to say 4z @Narasimham
    $endgroup$
    – juliodesa
    Dec 7 '15 at 16:01












  • $begingroup$
    At two places OP states it is a sphere.
    $endgroup$
    – Narasimham
    Dec 7 '15 at 16:50


















  • $begingroup$
    You mean to find the volume enclosed by the finite region of the intersection of the cone $z = sqrt{3(x^2 + y^2)}$ and the ellipsoid-ish body $x^2 + y^2 + z^2 = 4z$. Right?
    $endgroup$
    – amcalde
    Dec 7 '15 at 15:12










  • $begingroup$
    yes except its a sphere
    $endgroup$
    – juliodesa
    Dec 7 '15 at 15:15










  • $begingroup$
    Corrected the RHS of sphere equation.
    $endgroup$
    – Narasimham
    Dec 7 '15 at 15:54










  • $begingroup$
    it was supposed to say 4z @Narasimham
    $endgroup$
    – juliodesa
    Dec 7 '15 at 16:01












  • $begingroup$
    At two places OP states it is a sphere.
    $endgroup$
    – Narasimham
    Dec 7 '15 at 16:50
















$begingroup$
You mean to find the volume enclosed by the finite region of the intersection of the cone $z = sqrt{3(x^2 + y^2)}$ and the ellipsoid-ish body $x^2 + y^2 + z^2 = 4z$. Right?
$endgroup$
– amcalde
Dec 7 '15 at 15:12




$begingroup$
You mean to find the volume enclosed by the finite region of the intersection of the cone $z = sqrt{3(x^2 + y^2)}$ and the ellipsoid-ish body $x^2 + y^2 + z^2 = 4z$. Right?
$endgroup$
– amcalde
Dec 7 '15 at 15:12












$begingroup$
yes except its a sphere
$endgroup$
– juliodesa
Dec 7 '15 at 15:15




$begingroup$
yes except its a sphere
$endgroup$
– juliodesa
Dec 7 '15 at 15:15












$begingroup$
Corrected the RHS of sphere equation.
$endgroup$
– Narasimham
Dec 7 '15 at 15:54




$begingroup$
Corrected the RHS of sphere equation.
$endgroup$
– Narasimham
Dec 7 '15 at 15:54












$begingroup$
it was supposed to say 4z @Narasimham
$endgroup$
– juliodesa
Dec 7 '15 at 16:01






$begingroup$
it was supposed to say 4z @Narasimham
$endgroup$
– juliodesa
Dec 7 '15 at 16:01














$begingroup$
At two places OP states it is a sphere.
$endgroup$
– Narasimham
Dec 7 '15 at 16:50




$begingroup$
At two places OP states it is a sphere.
$endgroup$
– Narasimham
Dec 7 '15 at 16:50










1 Answer
1






active

oldest

votes


















0












$begingroup$

Your problem has azimuthal symmetry so I can immediately write:
$$I = int!!!int!!!int_Omega dxdydz =2pi int_{theta=0}^{theta_1}int_0^{r(theta)}r^2 sin theta dtheta= frac{2pi}{3} int_{theta=0}^{theta_1}(r(theta))^3 sin theta dtheta$$
The trick is to calculate what $theta_1$ is (where do the surfaces intersect?) And also the function $r(theta)$ which we can do via direct substitution.



$$x^2+y^2+z^2 = 4z$$
becomes
$$r^2=4rcos theta$$
So we have
$$r(theta) = 4cos theta$$
The angle is calculated:
$$theta_1=arccosfrac{z}{r} = arccosfrac{sqrt{3(x^2+y^2)}}{sqrt{x^2+y^2+z^2}}=arccosfrac{sqrt{3}}{2}$$
So
$$I = 2pi int_{theta=0}^{pi/6} frac{(4 cos theta )^3}{3} sin theta dtheta = -frac{128pi}{3} int_{theta=0}^{pi/6} (cos theta )^3 d(cos theta)$$



$$I=-frac{32pi}{3} [(cos(pi/6))^4 - (cos 0 )^4] = frac{32pi}{3} frac{7}{16} = 14pi/3$$



So I get the same thing. Hopefully this is what you were looking for. Let me know if you have any other questions about my answer.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    yes!!!! Thank you very much. @amcalde
    $endgroup$
    – juliodesa
    Dec 7 '15 at 18:43












Your Answer








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
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1564267%2fusing-spherical-coordinates-find-the-volume%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









0












$begingroup$

Your problem has azimuthal symmetry so I can immediately write:
$$I = int!!!int!!!int_Omega dxdydz =2pi int_{theta=0}^{theta_1}int_0^{r(theta)}r^2 sin theta dtheta= frac{2pi}{3} int_{theta=0}^{theta_1}(r(theta))^3 sin theta dtheta$$
The trick is to calculate what $theta_1$ is (where do the surfaces intersect?) And also the function $r(theta)$ which we can do via direct substitution.



$$x^2+y^2+z^2 = 4z$$
becomes
$$r^2=4rcos theta$$
So we have
$$r(theta) = 4cos theta$$
The angle is calculated:
$$theta_1=arccosfrac{z}{r} = arccosfrac{sqrt{3(x^2+y^2)}}{sqrt{x^2+y^2+z^2}}=arccosfrac{sqrt{3}}{2}$$
So
$$I = 2pi int_{theta=0}^{pi/6} frac{(4 cos theta )^3}{3} sin theta dtheta = -frac{128pi}{3} int_{theta=0}^{pi/6} (cos theta )^3 d(cos theta)$$



$$I=-frac{32pi}{3} [(cos(pi/6))^4 - (cos 0 )^4] = frac{32pi}{3} frac{7}{16} = 14pi/3$$



So I get the same thing. Hopefully this is what you were looking for. Let me know if you have any other questions about my answer.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    yes!!!! Thank you very much. @amcalde
    $endgroup$
    – juliodesa
    Dec 7 '15 at 18:43
















0












$begingroup$

Your problem has azimuthal symmetry so I can immediately write:
$$I = int!!!int!!!int_Omega dxdydz =2pi int_{theta=0}^{theta_1}int_0^{r(theta)}r^2 sin theta dtheta= frac{2pi}{3} int_{theta=0}^{theta_1}(r(theta))^3 sin theta dtheta$$
The trick is to calculate what $theta_1$ is (where do the surfaces intersect?) And also the function $r(theta)$ which we can do via direct substitution.



$$x^2+y^2+z^2 = 4z$$
becomes
$$r^2=4rcos theta$$
So we have
$$r(theta) = 4cos theta$$
The angle is calculated:
$$theta_1=arccosfrac{z}{r} = arccosfrac{sqrt{3(x^2+y^2)}}{sqrt{x^2+y^2+z^2}}=arccosfrac{sqrt{3}}{2}$$
So
$$I = 2pi int_{theta=0}^{pi/6} frac{(4 cos theta )^3}{3} sin theta dtheta = -frac{128pi}{3} int_{theta=0}^{pi/6} (cos theta )^3 d(cos theta)$$



$$I=-frac{32pi}{3} [(cos(pi/6))^4 - (cos 0 )^4] = frac{32pi}{3} frac{7}{16} = 14pi/3$$



So I get the same thing. Hopefully this is what you were looking for. Let me know if you have any other questions about my answer.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    yes!!!! Thank you very much. @amcalde
    $endgroup$
    – juliodesa
    Dec 7 '15 at 18:43














0












0








0





$begingroup$

Your problem has azimuthal symmetry so I can immediately write:
$$I = int!!!int!!!int_Omega dxdydz =2pi int_{theta=0}^{theta_1}int_0^{r(theta)}r^2 sin theta dtheta= frac{2pi}{3} int_{theta=0}^{theta_1}(r(theta))^3 sin theta dtheta$$
The trick is to calculate what $theta_1$ is (where do the surfaces intersect?) And also the function $r(theta)$ which we can do via direct substitution.



$$x^2+y^2+z^2 = 4z$$
becomes
$$r^2=4rcos theta$$
So we have
$$r(theta) = 4cos theta$$
The angle is calculated:
$$theta_1=arccosfrac{z}{r} = arccosfrac{sqrt{3(x^2+y^2)}}{sqrt{x^2+y^2+z^2}}=arccosfrac{sqrt{3}}{2}$$
So
$$I = 2pi int_{theta=0}^{pi/6} frac{(4 cos theta )^3}{3} sin theta dtheta = -frac{128pi}{3} int_{theta=0}^{pi/6} (cos theta )^3 d(cos theta)$$



$$I=-frac{32pi}{3} [(cos(pi/6))^4 - (cos 0 )^4] = frac{32pi}{3} frac{7}{16} = 14pi/3$$



So I get the same thing. Hopefully this is what you were looking for. Let me know if you have any other questions about my answer.






share|cite|improve this answer











$endgroup$



Your problem has azimuthal symmetry so I can immediately write:
$$I = int!!!int!!!int_Omega dxdydz =2pi int_{theta=0}^{theta_1}int_0^{r(theta)}r^2 sin theta dtheta= frac{2pi}{3} int_{theta=0}^{theta_1}(r(theta))^3 sin theta dtheta$$
The trick is to calculate what $theta_1$ is (where do the surfaces intersect?) And also the function $r(theta)$ which we can do via direct substitution.



$$x^2+y^2+z^2 = 4z$$
becomes
$$r^2=4rcos theta$$
So we have
$$r(theta) = 4cos theta$$
The angle is calculated:
$$theta_1=arccosfrac{z}{r} = arccosfrac{sqrt{3(x^2+y^2)}}{sqrt{x^2+y^2+z^2}}=arccosfrac{sqrt{3}}{2}$$
So
$$I = 2pi int_{theta=0}^{pi/6} frac{(4 cos theta )^3}{3} sin theta dtheta = -frac{128pi}{3} int_{theta=0}^{pi/6} (cos theta )^3 d(cos theta)$$



$$I=-frac{32pi}{3} [(cos(pi/6))^4 - (cos 0 )^4] = frac{32pi}{3} frac{7}{16} = 14pi/3$$



So I get the same thing. Hopefully this is what you were looking for. Let me know if you have any other questions about my answer.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 7 '15 at 18:44

























answered Dec 7 '15 at 18:39









amcaldeamcalde

3,851832




3,851832












  • $begingroup$
    yes!!!! Thank you very much. @amcalde
    $endgroup$
    – juliodesa
    Dec 7 '15 at 18:43


















  • $begingroup$
    yes!!!! Thank you very much. @amcalde
    $endgroup$
    – juliodesa
    Dec 7 '15 at 18:43
















$begingroup$
yes!!!! Thank you very much. @amcalde
$endgroup$
– juliodesa
Dec 7 '15 at 18:43




$begingroup$
yes!!!! Thank you very much. @amcalde
$endgroup$
– juliodesa
Dec 7 '15 at 18:43


















draft saved

draft discarded




















































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.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1564267%2fusing-spherical-coordinates-find-the-volume%23new-answer', 'question_page');
}
);

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







-

Popular posts from this blog

How to change which sound is reproduced for terminal bell?

Image
3 Recently upgrade to 18.10 from 18.04, and using the terminal is becoming annoying: the new alert sound is a drip ( /usr/share/sounds/gnome/default/alerts/drip.ogg ), and the previous one is still located at ( /usr/share/sounds/ubuntu/stereo/bell.ogg ). My issue is that at the gnome-terminal, when autocompleting or pressing left BOTH sounds are reproduced, at the same time. I want to just have ubuntu/stereo/bell.ogg active, not both . At the moment I have been only able to reduce the volume of the alerts (but still both are played), or turn the alerts down (but I dislike both "solutions"). gnome-terminal 18.10 teminal-bell share | improve this question
Read more

Can I use Tabulator js library in my java Spring + Thymeleaf project?

Image
0 I want to add Tabulator js to my java project. I added tabulator.min.js, tabulator.min.css to resources folder. After that added to my html page these dependencies as recommending in readme.md. <link rel="stylesheet" th:href="@{/css/tabulator.min.css}"> <script type="text/javascript" th:src="@{/js/tabulator.min.js}"></script> Now my tables view all elements, that I have in the html page, but not correct. In the documentation of Tabulator is example how to create header and table, var table = new Tabulator("#example-table", { columns:[ {title:"Name", field:"name", sorter:"string", width:200, editor:true}, {title:"Age", field:"age", sorter:"
Read more

Title Spacing in Bjornstrup Chapter, Removing Chapter Number From Contents

Image
up vote 5 down vote favorite 2 I would like to make it so that the vertical space below the title of each chapter is the same when using the Bjornstrup package (i.e. so that the descenders do not affect the spacing - as shown by the red arrow in the picture below). I think I'm fairly happy with the spacing around "A chapter", so I'd really like titles without descenders (e.g. "Contents") to have the same vertical spacing as "A chapter" below the title. I would also like to remove the large "0" in the contents title. Any help appreciated, as always. documentclass[11pt,a4paper]{book} % Packages[![enter image description here][1]][1] usepackage[left=3.5cm, right=3.5cm, top=4.3cm, bottom=4.25cm]{geometry} usepackage[utf8]{inputenc} usepackage{libertine} usepackage[libertine]{newtx
Read more