Find the volume of ice cream cone using cylindrical/spherical coordinates
$begingroup$
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
AVeljAVelj
3610
$endgroup$
add a comment |
$begingroup$
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
AVeljAVelj
3610
$endgroup$
$begingroup$
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.)
$endgroup$
– Ian Miller
Apr 7 '16 at 8:23
add a comment |
$begingroup$
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
AVeljAVelj
3610
$endgroup$
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
AVeljAVelj
3610
asked Apr 7 '16 at 8:15
AVeljAVelj
3610
asked Apr 7 '16 at 8:15
AVeljAVelj
3610
asked Apr 7 '16 at 8:15
AVeljAVelj
3610
asked Apr 7 '16 at 8:15
AVeljAVelj
3610
3610
$begingroup$
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.)
$endgroup$
– Ian Miller
Apr 7 '16 at 8:23
add a comment |
$begingroup$
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.)
$endgroup$
– Ian Miller
Apr 7 '16 at 8:23
$begingroup$
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.)
$endgroup$
– Ian Miller
Apr 7 '16 at 8:23
$begingroup$
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.)
$endgroup$
– Ian Miller
Apr 7 '16 at 8:23
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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 '18 at 3:20
gimusigimusi
92.9k94494
$endgroup$
$begingroup$
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.
$endgroup$
– manooooh
Nov 26 '18 at 3:08
1
$begingroup$
@manooooh Yes of course, I’ve used the uncorrect term here! I fix that, thanks.
$endgroup$
– gimusi
Nov 26 '18 at 6:06
add a comment |
$begingroup$
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
DavidDavid
68.1k664126
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
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 '18 at 3:20
gimusigimusi
92.9k94494
$endgroup$
$begingroup$
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.
$endgroup$
– manooooh
Nov 26 '18 at 3:08
1
$begingroup$
@manooooh Yes of course, I’ve used the uncorrect term here! I fix that, thanks.
$endgroup$
– gimusi
Nov 26 '18 at 6:06
add a comment |
$begingroup$
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 '18 at 3:20
gimusigimusi
92.9k94494
$endgroup$
$begingroup$
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.
$endgroup$
– manooooh
Nov 26 '18 at 3:08
1
$begingroup$
@manooooh Yes of course, I’ve used the uncorrect term here! I fix that, thanks.
$endgroup$
– gimusi
Nov 26 '18 at 6:06
add a comment |
$begingroup$
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 '18 at 3:20
gimusigimusi
92.9k94494
$endgroup$
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 '18 at 3:20
gimusigimusi
92.9k94494
edited Nov 26 '18 at 6:07
answered Feb 3 '18 at 3:20
gimusigimusi
92.9k94494
answered Feb 3 '18 at 3:20
gimusigimusi
92.9k94494
answered Feb 3 '18 at 3:20
gimusigimusi
92.9k94494
92.9k94494
$begingroup$
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.
$endgroup$
– manooooh
Nov 26 '18 at 3:08
1
$begingroup$
@manooooh Yes of course, I’ve used the uncorrect term here! I fix that, thanks.
$endgroup$
– gimusi
Nov 26 '18 at 6:06
add a comment |
$begingroup$
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.
$endgroup$
– manooooh
Nov 26 '18 at 3:08
1
$begingroup$
@manooooh Yes of course, I’ve used the uncorrect term here! I fix that, thanks.
$endgroup$
– gimusi
Nov 26 '18 at 6:06
$begingroup$
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.
$endgroup$
– manooooh
Nov 26 '18 at 3:08
$begingroup$
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.
$endgroup$
– manooooh
Nov 26 '18 at 3:08
1
1
$begingroup$
@manooooh Yes of course, I’ve used the uncorrect term here! I fix that, thanks.
$endgroup$
– gimusi
Nov 26 '18 at 6:06
$begingroup$
@manooooh Yes of course, I’ve used the uncorrect term here! I fix that, thanks.
$endgroup$
– gimusi
Nov 26 '18 at 6:06
add a comment |
$begingroup$
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
DavidDavid
68.1k664126
$endgroup$
add a comment |
$begingroup$
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
DavidDavid
68.1k664126
$endgroup$
add a comment |
$begingroup$
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
DavidDavid
68.1k664126
$endgroup$
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
DavidDavid
68.1k664126
answered Apr 7 '16 at 8:24
DavidDavid
68.1k664126
answered Apr 7 '16 at 8:24
DavidDavid
68.1k664126
answered Apr 7 '16 at 8:24
DavidDavid
68.1k664126
68.1k664126
add a comment |
add a comment |
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$begingroup$
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.)
$endgroup$
– Ian Miller
Apr 7 '16 at 8:23