Calculate integral where S is the surface of the half ball $x^2 + y^2 + z^2 = 1, space z geq 0,$ and $F = (x...
I am asked to calculate this integral and want to make sure I am doing this set up correctly, So I tried to paramatize this surface by :
$x = rcos(theta), space y = rsin(theta), space z = sqrt{1 - r}$
$0 leq r leq 1 $ and $ 0 leq theta leq 2pi$
then I found :
$Phi_{theta} = -rsin(theta)i + rcos(theta)j + 0k , space Phi_{r} = cos(theta)i + sin(theta)j + frac{1}{2sqrt{1-r}}k$
which would make:
$Phi_{theta} times Phi_{r} = frac{1}{2}bigl(frac{rcos(theta)}{sqrt{1-r}}bigr)i + frac{1}{2}bigl(frac{rsin(theta)}{sqrt{1-r}}bigr)j -rk$
then my surface integral would be :
$ int_0^{2pi} int_0^1 ((rcos(theta) + 3(rsin(theta))^5)i + (rsin(theta) + 10(rcos(theta))(sqrt{1 - r}))j + (sqrt{1 - r} - (rcos(theta)rsin(theta))k )space cdot (frac{1}{2}bigl(frac{rcos(theta)}{sqrt{1-r}}bigr)i + frac{1}{2}bigl(frac{rsin(theta)}{sqrt{1-r}}bigr)j -rk )$
is this set up correct ? I have tried to reduce it after taking the dot product with the trig identity's but I still end up with a fairly complicated expression
multivariable-calculus trigonometry surface-integrals
asked Nov 21 at 23:14
Doug Ray
245112
add a comment |
I am asked to calculate this integral and want to make sure I am doing this set up correctly, So I tried to paramatize this surface by :
$x = rcos(theta), space y = rsin(theta), space z = sqrt{1 - r}$
$0 leq r leq 1 $ and $ 0 leq theta leq 2pi$
then I found :
$Phi_{theta} = -rsin(theta)i + rcos(theta)j + 0k , space Phi_{r} = cos(theta)i + sin(theta)j + frac{1}{2sqrt{1-r}}k$
which would make:
$Phi_{theta} times Phi_{r} = frac{1}{2}bigl(frac{rcos(theta)}{sqrt{1-r}}bigr)i + frac{1}{2}bigl(frac{rsin(theta)}{sqrt{1-r}}bigr)j -rk$
then my surface integral would be :
$ int_0^{2pi} int_0^1 ((rcos(theta) + 3(rsin(theta))^5)i + (rsin(theta) + 10(rcos(theta))(sqrt{1 - r}))j + (sqrt{1 - r} - (rcos(theta)rsin(theta))k )space cdot (frac{1}{2}bigl(frac{rcos(theta)}{sqrt{1-r}}bigr)i + frac{1}{2}bigl(frac{rsin(theta)}{sqrt{1-r}}bigr)j -rk )$
is this set up correct ? I have tried to reduce it after taking the dot product with the trig identity's but I still end up with a fairly complicated expression
multivariable-calculus trigonometry surface-integrals
asked Nov 21 at 23:14
Doug Ray
245112
3
It should be $z = sqrt{1-r^2}$, not $sqrt{1-r}$.
– Nick
Nov 21 at 23:17
add a comment |
I am asked to calculate this integral and want to make sure I am doing this set up correctly, So I tried to paramatize this surface by :
$x = rcos(theta), space y = rsin(theta), space z = sqrt{1 - r}$
$0 leq r leq 1 $ and $ 0 leq theta leq 2pi$
then I found :
$Phi_{theta} = -rsin(theta)i + rcos(theta)j + 0k , space Phi_{r} = cos(theta)i + sin(theta)j + frac{1}{2sqrt{1-r}}k$
which would make:
$Phi_{theta} times Phi_{r} = frac{1}{2}bigl(frac{rcos(theta)}{sqrt{1-r}}bigr)i + frac{1}{2}bigl(frac{rsin(theta)}{sqrt{1-r}}bigr)j -rk$
then my surface integral would be :
$ int_0^{2pi} int_0^1 ((rcos(theta) + 3(rsin(theta))^5)i + (rsin(theta) + 10(rcos(theta))(sqrt{1 - r}))j + (sqrt{1 - r} - (rcos(theta)rsin(theta))k )space cdot (frac{1}{2}bigl(frac{rcos(theta)}{sqrt{1-r}}bigr)i + frac{1}{2}bigl(frac{rsin(theta)}{sqrt{1-r}}bigr)j -rk )$
is this set up correct ? I have tried to reduce it after taking the dot product with the trig identity's but I still end up with a fairly complicated expression
multivariable-calculus trigonometry surface-integrals
asked Nov 21 at 23:14
Doug Ray
245112
I am asked to calculate this integral and want to make sure I am doing this set up correctly, So I tried to paramatize this surface by :
$x = rcos(theta), space y = rsin(theta), space z = sqrt{1 - r}$
$0 leq r leq 1 $ and $ 0 leq theta leq 2pi$
then I found :
$Phi_{theta} = -rsin(theta)i + rcos(theta)j + 0k , space Phi_{r} = cos(theta)i + sin(theta)j + frac{1}{2sqrt{1-r}}k$
which would make:
$Phi_{theta} times Phi_{r} = frac{1}{2}bigl(frac{rcos(theta)}{sqrt{1-r}}bigr)i + frac{1}{2}bigl(frac{rsin(theta)}{sqrt{1-r}}bigr)j -rk$
then my surface integral would be :
$ int_0^{2pi} int_0^1 ((rcos(theta) + 3(rsin(theta))^5)i + (rsin(theta) + 10(rcos(theta))(sqrt{1 - r}))j + (sqrt{1 - r} - (rcos(theta)rsin(theta))k )space cdot (frac{1}{2}bigl(frac{rcos(theta)}{sqrt{1-r}}bigr)i + frac{1}{2}bigl(frac{rsin(theta)}{sqrt{1-r}}bigr)j -rk )$
is this set up correct ? I have tried to reduce it after taking the dot product with the trig identity's but I still end up with a fairly complicated expression
multivariable-calculus trigonometry surface-integrals
multivariable-calculus trigonometry surface-integrals
asked Nov 21 at 23:14
Doug Ray
245112
asked Nov 21 at 23:14
Doug Ray
245112
edited Nov 24 at 23:07
asked Nov 21 at 23:14
Doug Ray
245112
asked Nov 21 at 23:14
Doug Ray
245112
asked Nov 21 at 23:14
Doug Ray
245112
245112
3
It should be $z = sqrt{1-r^2}$, not $sqrt{1-r}$.
– Nick
Nov 21 at 23:17
add a comment |
3
It should be $z = sqrt{1-r^2}$, not $sqrt{1-r}$.
– Nick
Nov 21 at 23:17
3
3
It should be $z = sqrt{1-r^2}$, not $sqrt{1-r}$.
– Nick
Nov 21 at 23:17
It should be $z = sqrt{1-r^2}$, not $sqrt{1-r}$.
– Nick
Nov 21 at 23:17
add a comment |
1 Answer
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The problem is ambiguous. If the surface of the half ball includes the flat surface at $z=0$ the problem simplifies to a very simple integral using the divergence theorem. If not, the problem simplifies too but needs to calculate the flux through that flat surface.
1.- For the flat surface included ($B$ is the half ball, $partial B$ its surface and $mathbf F=F_xmathbf i+F_ymathbf j+F_zmathbf k$), the divergence theorem gives:
$$phi=int_{partial B}mathbf F·mathbb dmathbf S=int_Bnabla·mathbf F,mathbb dV$$
$$nabla·mathbf F=dfrac{partial F_x}{partial x}+dfrac{partial F_y}{partial y}+dfrac{partial F_z}{partial z}=1+1+1=3$$
In spherical coordinates $(r,theta,phi)$
$$phi=int_B 3mathbb d V=int_0^1int_0^{pi/2}int_o^{2pi}3r^2sintheta,mathbb dphi,mathbb dtheta,mathbb d r=2pi$$
2.- If the flat surface, $D={x^2+y^2leq 1;z=0}$, is not included we need to subtract the flux through it:
For the surface element with its vector pointing outwards is $-mathbf k$:
$$mathbb dmathbf S=-mathbb dxmathbb dy,mathbf k$$
For the flux through the surface element ($z=0$):
$$mathbf F·mathbb dmathbf S=((x + 3y^5)mathbf i + (y + 10xz)mathbf j + (z - xy)mathbf k)·(-mathbb dxmathbb dy,mathbf k)=xy,mathbb dxmathbb dy$$
$$phi_D=int_{-1}^1int_{-sqrt{1-x^2}}^{sqrt{1-x^2}}xy,mathbb dxmathbb dy=0$$
The flux is zero. It seems that in any case the answer is $2pi$
answered Nov 22 at 9:31
Rafa Budría
5,4951825
add a comment |
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1 Answer
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The problem is ambiguous. If the surface of the half ball includes the flat surface at $z=0$ the problem simplifies to a very simple integral using the divergence theorem. If not, the problem simplifies too but needs to calculate the flux through that flat surface.
1.- For the flat surface included ($B$ is the half ball, $partial B$ its surface and $mathbf F=F_xmathbf i+F_ymathbf j+F_zmathbf k$), the divergence theorem gives:
$$phi=int_{partial B}mathbf F·mathbb dmathbf S=int_Bnabla·mathbf F,mathbb dV$$
$$nabla·mathbf F=dfrac{partial F_x}{partial x}+dfrac{partial F_y}{partial y}+dfrac{partial F_z}{partial z}=1+1+1=3$$
In spherical coordinates $(r,theta,phi)$
$$phi=int_B 3mathbb d V=int_0^1int_0^{pi/2}int_o^{2pi}3r^2sintheta,mathbb dphi,mathbb dtheta,mathbb d r=2pi$$
2.- If the flat surface, $D={x^2+y^2leq 1;z=0}$, is not included we need to subtract the flux through it:
For the surface element with its vector pointing outwards is $-mathbf k$:
$$mathbb dmathbf S=-mathbb dxmathbb dy,mathbf k$$
For the flux through the surface element ($z=0$):
$$mathbf F·mathbb dmathbf S=((x + 3y^5)mathbf i + (y + 10xz)mathbf j + (z - xy)mathbf k)·(-mathbb dxmathbb dy,mathbf k)=xy,mathbb dxmathbb dy$$
$$phi_D=int_{-1}^1int_{-sqrt{1-x^2}}^{sqrt{1-x^2}}xy,mathbb dxmathbb dy=0$$
The flux is zero. It seems that in any case the answer is $2pi$
answered Nov 22 at 9:31
Rafa Budría
5,4951825
add a comment |
The problem is ambiguous. If the surface of the half ball includes the flat surface at $z=0$ the problem simplifies to a very simple integral using the divergence theorem. If not, the problem simplifies too but needs to calculate the flux through that flat surface.
1.- For the flat surface included ($B$ is the half ball, $partial B$ its surface and $mathbf F=F_xmathbf i+F_ymathbf j+F_zmathbf k$), the divergence theorem gives:
$$phi=int_{partial B}mathbf F·mathbb dmathbf S=int_Bnabla·mathbf F,mathbb dV$$
$$nabla·mathbf F=dfrac{partial F_x}{partial x}+dfrac{partial F_y}{partial y}+dfrac{partial F_z}{partial z}=1+1+1=3$$
In spherical coordinates $(r,theta,phi)$
$$phi=int_B 3mathbb d V=int_0^1int_0^{pi/2}int_o^{2pi}3r^2sintheta,mathbb dphi,mathbb dtheta,mathbb d r=2pi$$
2.- If the flat surface, $D={x^2+y^2leq 1;z=0}$, is not included we need to subtract the flux through it:
For the surface element with its vector pointing outwards is $-mathbf k$:
$$mathbb dmathbf S=-mathbb dxmathbb dy,mathbf k$$
For the flux through the surface element ($z=0$):
$$mathbf F·mathbb dmathbf S=((x + 3y^5)mathbf i + (y + 10xz)mathbf j + (z - xy)mathbf k)·(-mathbb dxmathbb dy,mathbf k)=xy,mathbb dxmathbb dy$$
$$phi_D=int_{-1}^1int_{-sqrt{1-x^2}}^{sqrt{1-x^2}}xy,mathbb dxmathbb dy=0$$
The flux is zero. It seems that in any case the answer is $2pi$
answered Nov 22 at 9:31
Rafa Budría
5,4951825
add a comment |
The problem is ambiguous. If the surface of the half ball includes the flat surface at $z=0$ the problem simplifies to a very simple integral using the divergence theorem. If not, the problem simplifies too but needs to calculate the flux through that flat surface.
1.- For the flat surface included ($B$ is the half ball, $partial B$ its surface and $mathbf F=F_xmathbf i+F_ymathbf j+F_zmathbf k$), the divergence theorem gives:
$$phi=int_{partial B}mathbf F·mathbb dmathbf S=int_Bnabla·mathbf F,mathbb dV$$
$$nabla·mathbf F=dfrac{partial F_x}{partial x}+dfrac{partial F_y}{partial y}+dfrac{partial F_z}{partial z}=1+1+1=3$$
In spherical coordinates $(r,theta,phi)$
$$phi=int_B 3mathbb d V=int_0^1int_0^{pi/2}int_o^{2pi}3r^2sintheta,mathbb dphi,mathbb dtheta,mathbb d r=2pi$$
2.- If the flat surface, $D={x^2+y^2leq 1;z=0}$, is not included we need to subtract the flux through it:
For the surface element with its vector pointing outwards is $-mathbf k$:
$$mathbb dmathbf S=-mathbb dxmathbb dy,mathbf k$$
For the flux through the surface element ($z=0$):
$$mathbf F·mathbb dmathbf S=((x + 3y^5)mathbf i + (y + 10xz)mathbf j + (z - xy)mathbf k)·(-mathbb dxmathbb dy,mathbf k)=xy,mathbb dxmathbb dy$$
$$phi_D=int_{-1}^1int_{-sqrt{1-x^2}}^{sqrt{1-x^2}}xy,mathbb dxmathbb dy=0$$
The flux is zero. It seems that in any case the answer is $2pi$
answered Nov 22 at 9:31
Rafa Budría
5,4951825
The problem is ambiguous. If the surface of the half ball includes the flat surface at $z=0$ the problem simplifies to a very simple integral using the divergence theorem. If not, the problem simplifies too but needs to calculate the flux through that flat surface.
1.- For the flat surface included ($B$ is the half ball, $partial B$ its surface and $mathbf F=F_xmathbf i+F_ymathbf j+F_zmathbf k$), the divergence theorem gives:
$$phi=int_{partial B}mathbf F·mathbb dmathbf S=int_Bnabla·mathbf F,mathbb dV$$
$$nabla·mathbf F=dfrac{partial F_x}{partial x}+dfrac{partial F_y}{partial y}+dfrac{partial F_z}{partial z}=1+1+1=3$$
In spherical coordinates $(r,theta,phi)$
$$phi=int_B 3mathbb d V=int_0^1int_0^{pi/2}int_o^{2pi}3r^2sintheta,mathbb dphi,mathbb dtheta,mathbb d r=2pi$$
2.- If the flat surface, $D={x^2+y^2leq 1;z=0}$, is not included we need to subtract the flux through it:
For the surface element with its vector pointing outwards is $-mathbf k$:
$$mathbb dmathbf S=-mathbb dxmathbb dy,mathbf k$$
For the flux through the surface element ($z=0$):
$$mathbf F·mathbb dmathbf S=((x + 3y^5)mathbf i + (y + 10xz)mathbf j + (z - xy)mathbf k)·(-mathbb dxmathbb dy,mathbf k)=xy,mathbb dxmathbb dy$$
$$phi_D=int_{-1}^1int_{-sqrt{1-x^2}}^{sqrt{1-x^2}}xy,mathbb dxmathbb dy=0$$
The flux is zero. It seems that in any case the answer is $2pi$
answered Nov 22 at 9:31
Rafa Budría
5,4951825
edited Nov 22 at 13:26
answered Nov 22 at 9:31
Rafa Budría
5,4951825
answered Nov 22 at 9:31
Rafa Budría
5,4951825
answered Nov 22 at 9:31
Rafa Budría
5,4951825
5,4951825
add a comment |
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3
It should be $z = sqrt{1-r^2}$, not $sqrt{1-r}$.
– Nick
Nov 21 at 23:17