$ int_{S_{c}} underset{bar{}}{F}.d underset{bar{}}{S};$ where $S_{c}$ is the curved part of the truncated...












0












$begingroup$


$ int_{S_{c}} underset{bar{}}{F}.d underset{bar{}}{S};$ where $S_{c}$ is the curved part of the truncated cone $x^2 + y^2 = (a-z)^2 , 0≤z≤(1/2)a$



Where $underset{bar{}}{F}$ = $(2x + y^2)underset{bar{}}{i} + (y+xz)underset{bar{}}{j} + zunderset{bar{}}{k}$. Use a normal to $S_{c}$ that points away from the z axis.



My first thought is to split the surface area $S_{c}$ of the truncated cone into 3 parts and use the divergence theorem. $S_{1}$ = the top, $S_{2}$ = the bottom, $S_{3}$= the sides(middle part).
$S_{c}$ is the union $S_{1}$∪$S_{2}$∪$S_{3}$.



For the divergence theorem you need a double integral but the question gives a single integral. How can i overcome this?






vector-fields divergence







share|cite|improve this question









$endgroup$












  • $begingroup$
    “The curved part” says to me that you’re not meant to integrate over the top or bottom.
    $endgroup$
    – amd
    Jan 6 at 22:02










  • $begingroup$
    is the 'curved part' supposed to be the middle part (the sides)?
    $endgroup$
    – Neels
    Jan 6 at 22:20
















0












$begingroup$


$ int_{S_{c}} underset{bar{}}{F}.d underset{bar{}}{S};$ where $S_{c}$ is the curved part of the truncated cone $x^2 + y^2 = (a-z)^2 , 0≤z≤(1/2)a$



Where $underset{bar{}}{F}$ = $(2x + y^2)underset{bar{}}{i} + (y+xz)underset{bar{}}{j} + zunderset{bar{}}{k}$. Use a normal to $S_{c}$ that points away from the z axis.



My first thought is to split the surface area $S_{c}$ of the truncated cone into 3 parts and use the divergence theorem. $S_{1}$ = the top, $S_{2}$ = the bottom, $S_{3}$= the sides(middle part).
$S_{c}$ is the union $S_{1}$∪$S_{2}$∪$S_{3}$.



For the divergence theorem you need a double integral but the question gives a single integral. How can i overcome this?






vector-fields divergence







share|cite|improve this question









$endgroup$












  • $begingroup$
    “The curved part” says to me that you’re not meant to integrate over the top or bottom.
    $endgroup$
    – amd
    Jan 6 at 22:02










  • $begingroup$
    is the 'curved part' supposed to be the middle part (the sides)?
    $endgroup$
    – Neels
    Jan 6 at 22:20














0












0








0





$begingroup$


$ int_{S_{c}} underset{bar{}}{F}.d underset{bar{}}{S};$ where $S_{c}$ is the curved part of the truncated cone $x^2 + y^2 = (a-z)^2 , 0≤z≤(1/2)a$



Where $underset{bar{}}{F}$ = $(2x + y^2)underset{bar{}}{i} + (y+xz)underset{bar{}}{j} + zunderset{bar{}}{k}$. Use a normal to $S_{c}$ that points away from the z axis.



My first thought is to split the surface area $S_{c}$ of the truncated cone into 3 parts and use the divergence theorem. $S_{1}$ = the top, $S_{2}$ = the bottom, $S_{3}$= the sides(middle part).
$S_{c}$ is the union $S_{1}$∪$S_{2}$∪$S_{3}$.



For the divergence theorem you need a double integral but the question gives a single integral. How can i overcome this?






vector-fields divergence







share|cite|improve this question









$endgroup$




$ int_{S_{c}} underset{bar{}}{F}.d underset{bar{}}{S};$ where $S_{c}$ is the curved part of the truncated cone $x^2 + y^2 = (a-z)^2 , 0≤z≤(1/2)a$



Where $underset{bar{}}{F}$ = $(2x + y^2)underset{bar{}}{i} + (y+xz)underset{bar{}}{j} + zunderset{bar{}}{k}$. Use a normal to $S_{c}$ that points away from the z axis.



My first thought is to split the surface area $S_{c}$ of the truncated cone into 3 parts and use the divergence theorem. $S_{1}$ = the top, $S_{2}$ = the bottom, $S_{3}$= the sides(middle part).
$S_{c}$ is the union $S_{1}$∪$S_{2}$∪$S_{3}$.



For the divergence theorem you need a double integral but the question gives a single integral. How can i overcome this?






vector-fields divergence




vector-fields divergence






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 6 at 21:45









NeelsNeels

367




367












  • $begingroup$
    “The curved part” says to me that you’re not meant to integrate over the top or bottom.
    $endgroup$
    – amd
    Jan 6 at 22:02










  • $begingroup$
    is the 'curved part' supposed to be the middle part (the sides)?
    $endgroup$
    – Neels
    Jan 6 at 22:20


















  • $begingroup$
    “The curved part” says to me that you’re not meant to integrate over the top or bottom.
    $endgroup$
    – amd
    Jan 6 at 22:02










  • $begingroup$
    is the 'curved part' supposed to be the middle part (the sides)?
    $endgroup$
    – Neels
    Jan 6 at 22:20
















$begingroup$
“The curved part” says to me that you’re not meant to integrate over the top or bottom.
$endgroup$
– amd
Jan 6 at 22:02




$begingroup$
“The curved part” says to me that you’re not meant to integrate over the top or bottom.
$endgroup$
– amd
Jan 6 at 22:02












$begingroup$
is the 'curved part' supposed to be the middle part (the sides)?
$endgroup$
– Neels
Jan 6 at 22:20




$begingroup$
is the 'curved part' supposed to be the middle part (the sides)?
$endgroup$
– Neels
Jan 6 at 22:20










1 Answer
1






active

oldest

votes


















0












$begingroup$

First, choose a surface internal parameters:
$$ P = { (u,v) in ([0..2pi],[0..(1/2)a]) } $$



Then describe the surface based on those parameters:
$$ S(a)(u,v) = { (x,y,z)in ((a-v)*cos(u),(a-v)*sin(u),v) }$$



Calculating surface area:
$$ A = pi r_0 s_0 - pi r_1 s_1 $$
where $r_0$ and $s_0$ are the radius and slant height of the cone, and the $r_1$ and $s_1$ are the radius and slant height of the (invisible) cone removed from the larger cone.



Then the integral is simply average of all samples picked from the surface multiplied by the surface area:
$$ int_{S_c}{Fcdot dS} = A timesfrac{sum_{n=1}^{k}F(S(a)(u_n,v_n))}{k}, k>10000 $$, where $(u_n,v_n)$ are randomly picked surface point from $P$.



Note: edited the cone surface area formula to more correct one.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    what are s0 and s1 suppose to be?
    $endgroup$
    – Neels
    Jan 6 at 22:56











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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









0












$begingroup$

First, choose a surface internal parameters:
$$ P = { (u,v) in ([0..2pi],[0..(1/2)a]) } $$



Then describe the surface based on those parameters:
$$ S(a)(u,v) = { (x,y,z)in ((a-v)*cos(u),(a-v)*sin(u),v) }$$



Calculating surface area:
$$ A = pi r_0 s_0 - pi r_1 s_1 $$
where $r_0$ and $s_0$ are the radius and slant height of the cone, and the $r_1$ and $s_1$ are the radius and slant height of the (invisible) cone removed from the larger cone.



Then the integral is simply average of all samples picked from the surface multiplied by the surface area:
$$ int_{S_c}{Fcdot dS} = A timesfrac{sum_{n=1}^{k}F(S(a)(u_n,v_n))}{k}, k>10000 $$, where $(u_n,v_n)$ are randomly picked surface point from $P$.



Note: edited the cone surface area formula to more correct one.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    what are s0 and s1 suppose to be?
    $endgroup$
    – Neels
    Jan 6 at 22:56
















0












$begingroup$

First, choose a surface internal parameters:
$$ P = { (u,v) in ([0..2pi],[0..(1/2)a]) } $$



Then describe the surface based on those parameters:
$$ S(a)(u,v) = { (x,y,z)in ((a-v)*cos(u),(a-v)*sin(u),v) }$$



Calculating surface area:
$$ A = pi r_0 s_0 - pi r_1 s_1 $$
where $r_0$ and $s_0$ are the radius and slant height of the cone, and the $r_1$ and $s_1$ are the radius and slant height of the (invisible) cone removed from the larger cone.



Then the integral is simply average of all samples picked from the surface multiplied by the surface area:
$$ int_{S_c}{Fcdot dS} = A timesfrac{sum_{n=1}^{k}F(S(a)(u_n,v_n))}{k}, k>10000 $$, where $(u_n,v_n)$ are randomly picked surface point from $P$.



Note: edited the cone surface area formula to more correct one.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    what are s0 and s1 suppose to be?
    $endgroup$
    – Neels
    Jan 6 at 22:56














0












0








0





$begingroup$

First, choose a surface internal parameters:
$$ P = { (u,v) in ([0..2pi],[0..(1/2)a]) } $$



Then describe the surface based on those parameters:
$$ S(a)(u,v) = { (x,y,z)in ((a-v)*cos(u),(a-v)*sin(u),v) }$$



Calculating surface area:
$$ A = pi r_0 s_0 - pi r_1 s_1 $$
where $r_0$ and $s_0$ are the radius and slant height of the cone, and the $r_1$ and $s_1$ are the radius and slant height of the (invisible) cone removed from the larger cone.



Then the integral is simply average of all samples picked from the surface multiplied by the surface area:
$$ int_{S_c}{Fcdot dS} = A timesfrac{sum_{n=1}^{k}F(S(a)(u_n,v_n))}{k}, k>10000 $$, where $(u_n,v_n)$ are randomly picked surface point from $P$.



Note: edited the cone surface area formula to more correct one.






share|cite|improve this answer











$endgroup$



First, choose a surface internal parameters:
$$ P = { (u,v) in ([0..2pi],[0..(1/2)a]) } $$



Then describe the surface based on those parameters:
$$ S(a)(u,v) = { (x,y,z)in ((a-v)*cos(u),(a-v)*sin(u),v) }$$



Calculating surface area:
$$ A = pi r_0 s_0 - pi r_1 s_1 $$
where $r_0$ and $s_0$ are the radius and slant height of the cone, and the $r_1$ and $s_1$ are the radius and slant height of the (invisible) cone removed from the larger cone.



Then the integral is simply average of all samples picked from the surface multiplied by the surface area:
$$ int_{S_c}{Fcdot dS} = A timesfrac{sum_{n=1}^{k}F(S(a)(u_n,v_n))}{k}, k>10000 $$, where $(u_n,v_n)$ are randomly picked surface point from $P$.



Note: edited the cone surface area formula to more correct one.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 7 at 0:13

























answered Jan 6 at 22:48









tp1tp1

31648




31648












  • $begingroup$
    what are s0 and s1 suppose to be?
    $endgroup$
    – Neels
    Jan 6 at 22:56


















  • $begingroup$
    what are s0 and s1 suppose to be?
    $endgroup$
    – Neels
    Jan 6 at 22:56
















$begingroup$
what are s0 and s1 suppose to be?
$endgroup$
– Neels
Jan 6 at 22:56




$begingroup$
what are s0 and s1 suppose to be?
$endgroup$
– Neels
Jan 6 at 22:56


















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