Find the volume of ice cream cone using cylindrical/spherical coordinates
I'm stuck on what the boundaries are for the volume bounded by the cone $z=-sqrt{(x^2+y^2)}$ and the surface $z=-sqrt{(9-x^2-y^2)}$ $,,$-essentially an upside down ice cream cone
Remember that $r^2=x^2+y^2$
I assumed for cylindrical coordinates the triple integral boundaries would be
$-sqrt{(9-r^2)}le z le -r$
$0le r le 3$
$0le theta le 2pi$
And for spherical coordinates the triple integral boundaries would be
$0le r le 3$
$pi/2le phi le pi$
$0le theta le 2pi$
However upon entering these values into MATLAB, the cylindrical coordinates integral equals to zero, whilst the spherical coordinates integral equals to $18pi$.
So somethings wrong with my boundaries, as both integrals should equal the same volume.
This is my working out for cylindrical coordinate integral so far:
$int_0^{2pi}int_0^3int_{-sqrt{9-r^2}}^{-r} r,, dzdrdtheta$
$int_0^{2pi}int_0^3 [rz]_{-sqrt{9-r^2}}^{-r} ,, dzdrdtheta$
$int_0^{2pi}int_0^3 -r^2+{rsqrt{9-r^2}} ,, dzdrdtheta$
$int_0^{2pi}[(int_0^3 -r^2 ,dr)+(int_0^3 {rsqrt{9-r^2} ,dr)}]dtheta$
Let $u=9-r^2$
$du=-2r,dr$
$int_0^{2pi}[[frac{-r^3}{3}]_0^3,-frac{1}{2}int_0^3 u^{frac{1}{2}} ,du)],dtheta$
$int_0^{2pi}[-frac{27}{3},-frac{1}{2}[frac{2}{3}u^{frac{3}{2}}]_{r=0}^{r=3} ],dtheta$
$int_0^{2pi}[-9,-frac{1}{2}[{frac{2}{3}}(9-r^2)^{frac{3}{2}}]_{0}^{3} ],dtheta$
$int_0^{2pi}[-9,-frac{1}{2}[{frac{2}{3}}(9-3^2)^{frac{3}{2}},-{frac{2}{3}}(9-0^2)^{frac{3}{2}}] ],dtheta$
$int_0^{2pi}[-9,-frac{1}{2}{frac{2}{3}}[(9-9)^{frac{3}{2}},-(9)^{frac{3}{2}}] ],dtheta$
$int_0^{2pi}[-9,-frac{1}{3}[(0)^{frac{3}{2}},-(9)^{frac{3}{2}}] ],dtheta$
$int_0^{2pi}[-9,-frac{1}{3}[-(9)^{frac{3}{2}}] ],dtheta$
$int_0^{2pi}[-9,-frac{1}{3}[-27] ],dtheta$
$int_0^{2pi}[-9,+9],dtheta$
$int_0^{2pi}0,dtheta$
$=0$
So as you can see I can't proceed to the third integral since the second integral equals zero
Any help would be greatly appreciated :)
multivariable-calculus vector-analysis
asked Apr 7 '16 at 8:15
AVelj
3610
add a comment |
I'm stuck on what the boundaries are for the volume bounded by the cone $z=-sqrt{(x^2+y^2)}$ and the surface $z=-sqrt{(9-x^2-y^2)}$ $,,$-essentially an upside down ice cream cone
Remember that $r^2=x^2+y^2$
I assumed for cylindrical coordinates the triple integral boundaries would be
$-sqrt{(9-r^2)}le z le -r$
$0le r le 3$
$0le theta le 2pi$
And for spherical coordinates the triple integral boundaries would be
$0le r le 3$
$pi/2le phi le pi$
$0le theta le 2pi$
However upon entering these values into MATLAB, the cylindrical coordinates integral equals to zero, whilst the spherical coordinates integral equals to $18pi$.
So somethings wrong with my boundaries, as both integrals should equal the same volume.
This is my working out for cylindrical coordinate integral so far:
$int_0^{2pi}int_0^3int_{-sqrt{9-r^2}}^{-r} r,, dzdrdtheta$
$int_0^{2pi}int_0^3 [rz]_{-sqrt{9-r^2}}^{-r} ,, dzdrdtheta$
$int_0^{2pi}int_0^3 -r^2+{rsqrt{9-r^2}} ,, dzdrdtheta$
$int_0^{2pi}[(int_0^3 -r^2 ,dr)+(int_0^3 {rsqrt{9-r^2} ,dr)}]dtheta$
Let $u=9-r^2$
$du=-2r,dr$
$int_0^{2pi}[[frac{-r^3}{3}]_0^3,-frac{1}{2}int_0^3 u^{frac{1}{2}} ,du)],dtheta$
$int_0^{2pi}[-frac{27}{3},-frac{1}{2}[frac{2}{3}u^{frac{3}{2}}]_{r=0}^{r=3} ],dtheta$
$int_0^{2pi}[-9,-frac{1}{2}[{frac{2}{3}}(9-r^2)^{frac{3}{2}}]_{0}^{3} ],dtheta$
$int_0^{2pi}[-9,-frac{1}{2}[{frac{2}{3}}(9-3^2)^{frac{3}{2}},-{frac{2}{3}}(9-0^2)^{frac{3}{2}}] ],dtheta$
$int_0^{2pi}[-9,-frac{1}{2}{frac{2}{3}}[(9-9)^{frac{3}{2}},-(9)^{frac{3}{2}}] ],dtheta$
$int_0^{2pi}[-9,-frac{1}{3}[(0)^{frac{3}{2}},-(9)^{frac{3}{2}}] ],dtheta$
$int_0^{2pi}[-9,-frac{1}{3}[-(9)^{frac{3}{2}}] ],dtheta$
$int_0^{2pi}[-9,-frac{1}{3}[-27] ],dtheta$
$int_0^{2pi}[-9,+9],dtheta$
$int_0^{2pi}0,dtheta$
$=0$
So as you can see I can't proceed to the third integral since the second integral equals zero
Any help would be greatly appreciated :)
multivariable-calculus vector-analysis
asked Apr 7 '16 at 8:15
AVelj
3610
Hi again. I don't think $18pi$ is correct either. Consider a sphere with radius of $3$. The volume would be $frac43pitimes3^3=36pi$. Your answer is half of this while your shape is clearly less than half a sphere. The reason for your spherical answer giving this is incorrect limits on $phi$. You should use $frac{3pi}{4}leqphileqpi$ as your cone has an angle of $frac{pi}{4}$. (I'll try to give you an actual answer once I get back to my computer.)
– Ian Miller
Apr 7 '16 at 8:23
add a comment |
I'm stuck on what the boundaries are for the volume bounded by the cone $z=-sqrt{(x^2+y^2)}$ and the surface $z=-sqrt{(9-x^2-y^2)}$ $,,$-essentially an upside down ice cream cone
Remember that $r^2=x^2+y^2$
I assumed for cylindrical coordinates the triple integral boundaries would be
$-sqrt{(9-r^2)}le z le -r$
$0le r le 3$
$0le theta le 2pi$
And for spherical coordinates the triple integral boundaries would be
$0le r le 3$
$pi/2le phi le pi$
$0le theta le 2pi$
However upon entering these values into MATLAB, the cylindrical coordinates integral equals to zero, whilst the spherical coordinates integral equals to $18pi$.
So somethings wrong with my boundaries, as both integrals should equal the same volume.
This is my working out for cylindrical coordinate integral so far:
$int_0^{2pi}int_0^3int_{-sqrt{9-r^2}}^{-r} r,, dzdrdtheta$
$int_0^{2pi}int_0^3 [rz]_{-sqrt{9-r^2}}^{-r} ,, dzdrdtheta$
$int_0^{2pi}int_0^3 -r^2+{rsqrt{9-r^2}} ,, dzdrdtheta$
$int_0^{2pi}[(int_0^3 -r^2 ,dr)+(int_0^3 {rsqrt{9-r^2} ,dr)}]dtheta$
Let $u=9-r^2$
$du=-2r,dr$
$int_0^{2pi}[[frac{-r^3}{3}]_0^3,-frac{1}{2}int_0^3 u^{frac{1}{2}} ,du)],dtheta$
$int_0^{2pi}[-frac{27}{3},-frac{1}{2}[frac{2}{3}u^{frac{3}{2}}]_{r=0}^{r=3} ],dtheta$
$int_0^{2pi}[-9,-frac{1}{2}[{frac{2}{3}}(9-r^2)^{frac{3}{2}}]_{0}^{3} ],dtheta$
$int_0^{2pi}[-9,-frac{1}{2}[{frac{2}{3}}(9-3^2)^{frac{3}{2}},-{frac{2}{3}}(9-0^2)^{frac{3}{2}}] ],dtheta$
$int_0^{2pi}[-9,-frac{1}{2}{frac{2}{3}}[(9-9)^{frac{3}{2}},-(9)^{frac{3}{2}}] ],dtheta$
$int_0^{2pi}[-9,-frac{1}{3}[(0)^{frac{3}{2}},-(9)^{frac{3}{2}}] ],dtheta$
$int_0^{2pi}[-9,-frac{1}{3}[-(9)^{frac{3}{2}}] ],dtheta$
$int_0^{2pi}[-9,-frac{1}{3}[-27] ],dtheta$
$int_0^{2pi}[-9,+9],dtheta$
$int_0^{2pi}0,dtheta$
$=0$
So as you can see I can't proceed to the third integral since the second integral equals zero
Any help would be greatly appreciated :)
multivariable-calculus vector-analysis
asked Apr 7 '16 at 8:15
AVelj
3610
I'm stuck on what the boundaries are for the volume bounded by the cone $z=-sqrt{(x^2+y^2)}$ and the surface $z=-sqrt{(9-x^2-y^2)}$ $,,$-essentially an upside down ice cream cone
Remember that $r^2=x^2+y^2$
I assumed for cylindrical coordinates the triple integral boundaries would be
$-sqrt{(9-r^2)}le z le -r$
$0le r le 3$
$0le theta le 2pi$
And for spherical coordinates the triple integral boundaries would be
$0le r le 3$
$pi/2le phi le pi$
$0le theta le 2pi$
However upon entering these values into MATLAB, the cylindrical coordinates integral equals to zero, whilst the spherical coordinates integral equals to $18pi$.
So somethings wrong with my boundaries, as both integrals should equal the same volume.
This is my working out for cylindrical coordinate integral so far:
$int_0^{2pi}int_0^3int_{-sqrt{9-r^2}}^{-r} r,, dzdrdtheta$
$int_0^{2pi}int_0^3 [rz]_{-sqrt{9-r^2}}^{-r} ,, dzdrdtheta$
$int_0^{2pi}int_0^3 -r^2+{rsqrt{9-r^2}} ,, dzdrdtheta$
$int_0^{2pi}[(int_0^3 -r^2 ,dr)+(int_0^3 {rsqrt{9-r^2} ,dr)}]dtheta$
Let $u=9-r^2$
$du=-2r,dr$
$int_0^{2pi}[[frac{-r^3}{3}]_0^3,-frac{1}{2}int_0^3 u^{frac{1}{2}} ,du)],dtheta$
$int_0^{2pi}[-frac{27}{3},-frac{1}{2}[frac{2}{3}u^{frac{3}{2}}]_{r=0}^{r=3} ],dtheta$
$int_0^{2pi}[-9,-frac{1}{2}[{frac{2}{3}}(9-r^2)^{frac{3}{2}}]_{0}^{3} ],dtheta$
$int_0^{2pi}[-9,-frac{1}{2}[{frac{2}{3}}(9-3^2)^{frac{3}{2}},-{frac{2}{3}}(9-0^2)^{frac{3}{2}}] ],dtheta$
$int_0^{2pi}[-9,-frac{1}{2}{frac{2}{3}}[(9-9)^{frac{3}{2}},-(9)^{frac{3}{2}}] ],dtheta$
$int_0^{2pi}[-9,-frac{1}{3}[(0)^{frac{3}{2}},-(9)^{frac{3}{2}}] ],dtheta$
$int_0^{2pi}[-9,-frac{1}{3}[-(9)^{frac{3}{2}}] ],dtheta$
$int_0^{2pi}[-9,-frac{1}{3}[-27] ],dtheta$
$int_0^{2pi}[-9,+9],dtheta$
$int_0^{2pi}0,dtheta$
$=0$
So as you can see I can't proceed to the third integral since the second integral equals zero
Any help would be greatly appreciated :)
multivariable-calculus vector-analysis
multivariable-calculus vector-analysis
asked Apr 7 '16 at 8:15
AVelj
3610
asked Apr 7 '16 at 8:15
AVelj
3610
asked Apr 7 '16 at 8:15
AVelj
3610
asked Apr 7 '16 at 8:15
AVelj
3610
asked Apr 7 '16 at 8:15
AVelj
3610
3610
Hi again. I don't think $18pi$ is correct either. Consider a sphere with radius of $3$. The volume would be $frac43pitimes3^3=36pi$. Your answer is half of this while your shape is clearly less than half a sphere. The reason for your spherical answer giving this is incorrect limits on $phi$. You should use $frac{3pi}{4}leqphileqpi$ as your cone has an angle of $frac{pi}{4}$. (I'll try to give you an actual answer once I get back to my computer.)
– Ian Miller
Apr 7 '16 at 8:23
add a comment |
Hi again. I don't think $18pi$ is correct either. Consider a sphere with radius of $3$. The volume would be $frac43pitimes3^3=36pi$. Your answer is half of this while your shape is clearly less than half a sphere. The reason for your spherical answer giving this is incorrect limits on $phi$. You should use $frac{3pi}{4}leqphileqpi$ as your cone has an angle of $frac{pi}{4}$. (I'll try to give you an actual answer once I get back to my computer.)
– Ian Miller
Apr 7 '16 at 8:23
Hi again. I don't think $18pi$ is correct either. Consider a sphere with radius of $3$. The volume would be $frac43pitimes3^3=36pi$. Your answer is half of this while your shape is clearly less than half a sphere. The reason for your spherical answer giving this is incorrect limits on $phi$. You should use $frac{3pi}{4}leqphileqpi$ as your cone has an angle of $frac{pi}{4}$. (I'll try to give you an actual answer once I get back to my computer.)
– Ian Miller
Apr 7 '16 at 8:23
Hi again. I don't think $18pi$ is correct either. Consider a sphere with radius of $3$. The volume would be $frac43pitimes3^3=36pi$. Your answer is half of this while your shape is clearly less than half a sphere. The reason for your spherical answer giving this is incorrect limits on $phi$. You should use $frac{3pi}{4}leqphileqpi$ as your cone has an angle of $frac{pi}{4}$. (I'll try to give you an actual answer once I get back to my computer.)
– Ian Miller
Apr 7 '16 at 8:23
add a comment |
2 Answers
2
active
oldest
votes
By cylindrical coordinates the set up of the integral should be
$$int_0^{2pi}int_{-3}^{-frac3{sqrt2}}int_{0}^{sqrt{9-z^2}} r,, dzdrdtheta+int_0^{2pi}int_{-frac3{sqrt2}}^0int_{0}^{-z} r,, dzdrdtheta$$
answered Feb 3 at 3:20
gimusi
1
Wait, the polar coordinates are not for double integrals? I think it should be cylindrical coordinates, altough the first two coordinates are the same for the third integral we need a third variable.
– manooooh
Nov 26 at 3:08
1
@manooooh Yes of course, I’ve used the uncorrect term here! I fix that, thanks.
– gimusi
Nov 26 at 6:06
add a comment |
The surfaces intersect when
$$-r=-sqrt{9-r^2}$$
which is $r^2=frac92$, not $r^2=9$ as you have in your cylindrical attempt.
For spherical coordinates $phi$ should be from $frac34pi$ to $pi$. I believe the rest is correct, as long as you are writing $r^2=x^2+y^2+z^2$ in the spherical case. (It is common to call this $rho^2$ in order to distinguish it from $r^2=x^2+y^2$.)
answered Apr 7 '16 at 8:24
David
67.7k663126
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%2f1731695%2ffind-the-volume-of-ice-cream-cone-using-cylindrical-spherical-coordinates%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
By cylindrical coordinates the set up of the integral should be
$$int_0^{2pi}int_{-3}^{-frac3{sqrt2}}int_{0}^{sqrt{9-z^2}} r,, dzdrdtheta+int_0^{2pi}int_{-frac3{sqrt2}}^0int_{0}^{-z} r,, dzdrdtheta$$
answered Feb 3 at 3:20
gimusi
1
Wait, the polar coordinates are not for double integrals? I think it should be cylindrical coordinates, altough the first two coordinates are the same for the third integral we need a third variable.
– manooooh
Nov 26 at 3:08
1
@manooooh Yes of course, I’ve used the uncorrect term here! I fix that, thanks.
– gimusi
Nov 26 at 6:06
add a comment |
By cylindrical coordinates the set up of the integral should be
$$int_0^{2pi}int_{-3}^{-frac3{sqrt2}}int_{0}^{sqrt{9-z^2}} r,, dzdrdtheta+int_0^{2pi}int_{-frac3{sqrt2}}^0int_{0}^{-z} r,, dzdrdtheta$$
answered Feb 3 at 3:20
gimusi
1
Wait, the polar coordinates are not for double integrals? I think it should be cylindrical coordinates, altough the first two coordinates are the same for the third integral we need a third variable.
– manooooh
Nov 26 at 3:08
1
@manooooh Yes of course, I’ve used the uncorrect term here! I fix that, thanks.
– gimusi
Nov 26 at 6:06
add a comment |
By cylindrical coordinates the set up of the integral should be
$$int_0^{2pi}int_{-3}^{-frac3{sqrt2}}int_{0}^{sqrt{9-z^2}} r,, dzdrdtheta+int_0^{2pi}int_{-frac3{sqrt2}}^0int_{0}^{-z} r,, dzdrdtheta$$
answered Feb 3 at 3:20
gimusi
1
By cylindrical coordinates the set up of the integral should be
$$int_0^{2pi}int_{-3}^{-frac3{sqrt2}}int_{0}^{sqrt{9-z^2}} r,, dzdrdtheta+int_0^{2pi}int_{-frac3{sqrt2}}^0int_{0}^{-z} r,, dzdrdtheta$$
answered Feb 3 at 3:20
gimusi
1
edited Nov 26 at 6:07
answered Feb 3 at 3:20
gimusi
1
answered Feb 3 at 3:20
gimusi
1
answered Feb 3 at 3:20
gimusi
1
1
Wait, the polar coordinates are not for double integrals? I think it should be cylindrical coordinates, altough the first two coordinates are the same for the third integral we need a third variable.
– manooooh
Nov 26 at 3:08
1
@manooooh Yes of course, I’ve used the uncorrect term here! I fix that, thanks.
– gimusi
Nov 26 at 6:06
add a comment |
Wait, the polar coordinates are not for double integrals? I think it should be cylindrical coordinates, altough the first two coordinates are the same for the third integral we need a third variable.
– manooooh
Nov 26 at 3:08
1
@manooooh Yes of course, I’ve used the uncorrect term here! I fix that, thanks.
– gimusi
Nov 26 at 6:06
Wait, the polar coordinates are not for double integrals? I think it should be cylindrical coordinates, altough the first two coordinates are the same for the third integral we need a third variable.
– manooooh
Nov 26 at 3:08
Wait, the polar coordinates are not for double integrals? I think it should be cylindrical coordinates, altough the first two coordinates are the same for the third integral we need a third variable.
– manooooh
Nov 26 at 3:08
1
1
@manooooh Yes of course, I’ve used the uncorrect term here! I fix that, thanks.
– gimusi
Nov 26 at 6:06
@manooooh Yes of course, I’ve used the uncorrect term here! I fix that, thanks.
– gimusi
Nov 26 at 6:06
add a comment |
The surfaces intersect when
$$-r=-sqrt{9-r^2}$$
which is $r^2=frac92$, not $r^2=9$ as you have in your cylindrical attempt.
For spherical coordinates $phi$ should be from $frac34pi$ to $pi$. I believe the rest is correct, as long as you are writing $r^2=x^2+y^2+z^2$ in the spherical case. (It is common to call this $rho^2$ in order to distinguish it from $r^2=x^2+y^2$.)
answered Apr 7 '16 at 8:24
David
67.7k663126
add a comment |
The surfaces intersect when
$$-r=-sqrt{9-r^2}$$
which is $r^2=frac92$, not $r^2=9$ as you have in your cylindrical attempt.
For spherical coordinates $phi$ should be from $frac34pi$ to $pi$. I believe the rest is correct, as long as you are writing $r^2=x^2+y^2+z^2$ in the spherical case. (It is common to call this $rho^2$ in order to distinguish it from $r^2=x^2+y^2$.)
answered Apr 7 '16 at 8:24
David
67.7k663126
add a comment |
The surfaces intersect when
$$-r=-sqrt{9-r^2}$$
which is $r^2=frac92$, not $r^2=9$ as you have in your cylindrical attempt.
For spherical coordinates $phi$ should be from $frac34pi$ to $pi$. I believe the rest is correct, as long as you are writing $r^2=x^2+y^2+z^2$ in the spherical case. (It is common to call this $rho^2$ in order to distinguish it from $r^2=x^2+y^2$.)
answered Apr 7 '16 at 8:24
David
67.7k663126
The surfaces intersect when
$$-r=-sqrt{9-r^2}$$
which is $r^2=frac92$, not $r^2=9$ as you have in your cylindrical attempt.
For spherical coordinates $phi$ should be from $frac34pi$ to $pi$. I believe the rest is correct, as long as you are writing $r^2=x^2+y^2+z^2$ in the spherical case. (It is common to call this $rho^2$ in order to distinguish it from $r^2=x^2+y^2$.)
answered Apr 7 '16 at 8:24
David
67.7k663126
answered Apr 7 '16 at 8:24
David
67.7k663126
answered Apr 7 '16 at 8:24
David
67.7k663126
answered Apr 7 '16 at 8:24
David
67.7k663126
67.7k663126
add a comment |
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.
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%2f1731695%2ffind-the-volume-of-ice-cream-cone-using-cylindrical-spherical-coordinates%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
Hi again. I don't think $18pi$ is correct either. Consider a sphere with radius of $3$. The volume would be $frac43pitimes3^3=36pi$. Your answer is half of this while your shape is clearly less than half a sphere. The reason for your spherical answer giving this is incorrect limits on $phi$. You should use $frac{3pi}{4}leqphileqpi$ as your cone has an angle of $frac{pi}{4}$. (I'll try to give you an actual answer once I get back to my computer.)
– Ian Miller
Apr 7 '16 at 8:23