Using spherical coordinates, find the volume.
$begingroup$
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
asked Dec 7 '15 at 15:07
juliodesajuliodesa
5817
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show 2 more comments
$begingroup$
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
asked Dec 7 '15 at 15:07
juliodesajuliodesa
5817
$endgroup$
$begingroup$
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
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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
|
show 2 more comments
$begingroup$
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
asked Dec 7 '15 at 15:07
juliodesajuliodesa
5817
$endgroup$
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
asked Dec 7 '15 at 15:07
juliodesajuliodesa
5817
asked Dec 7 '15 at 15:07
juliodesajuliodesa
5817
edited Dec 7 '15 at 16:02
juliodesa
asked Dec 7 '15 at 15:07
juliodesajuliodesa
5817
asked Dec 7 '15 at 15:07
juliodesajuliodesa
5817
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
|
show 2 more comments
$begingroup$
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
|
show 2 more comments
1 Answer
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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.
answered Dec 7 '15 at 18:39
amcaldeamcalde
3,851832
$endgroup$
$begingroup$
yes!!!! Thank you very much. @amcalde
$endgroup$
– juliodesa
Dec 7 '15 at 18:43
add a comment |
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1 Answer
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1 Answer
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$begingroup$
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.
answered Dec 7 '15 at 18:39
amcaldeamcalde
3,851832
$endgroup$
$begingroup$
yes!!!! Thank you very much. @amcalde
$endgroup$
– juliodesa
Dec 7 '15 at 18:43
add a comment |
$begingroup$
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.
answered Dec 7 '15 at 18:39
amcaldeamcalde
3,851832
$endgroup$
$begingroup$
yes!!!! Thank you very much. @amcalde
$endgroup$
– juliodesa
Dec 7 '15 at 18:43
add a comment |
$begingroup$
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.
answered Dec 7 '15 at 18:39
amcaldeamcalde
3,851832
$endgroup$
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.
answered Dec 7 '15 at 18:39
amcaldeamcalde
3,851832
edited Dec 7 '15 at 18:44
answered Dec 7 '15 at 18:39
amcaldeamcalde
3,851832
answered Dec 7 '15 at 18:39
amcaldeamcalde
3,851832
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
add a comment |
$begingroup$
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
add a comment |
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$begingroup$
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