Verifying Gauss's divergence theorem on a upside down truncated cone
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I have a surface with the geometric equation $$(z+1)^2=x^2+y^2$$ where $1leq zleq 0$ which give a sort of upside down truncated cone like this
I want to verify Gauss's divergence theorem for this volume with $mathbf{F}=(x,y,0)$.
So I first went about parameterizing the surfaces for the curved part we have
$$(u,v) rightarrow (ucos(v),usin(v),u-1) ,quad 1leq u leq 2, quad 0leq v leq 2pi $$ the other surfaces are just this same parameterization with z component $z=0$ and $0leq u leq 1$ for the bottom circle and z component $z=1$ with $0leq u leq 2$ for the top circle.
When I calculate the surface integrals for the each surface flux integral for the top and bottom circles I get that they are just $0$, since for the top circle we get $n=(0,0,1)$ as the outward normal and for the bottom circle the outward normal is $n=(0,0,-1)$ and we only really need the normal vector not the unit normal since the area element $dS=lvert n rvert dudv$. So $F cdot n = 0$ and so the surface integrals for thee top and bottom circle are also $0$.
When I calculate the surface flux integral for the curved surface I find that the normal $n=(-ucos(v),-usin(v),u)$ by finding the tangent vector with respect to u and the tangent vector with respect to v and taking their cross product. So $F.n=-u^2$, so I evaluate my surface integral $$int_{v=0}^{v=2pi}int_{u=1}^{u=2}-u^2dS$$ get that it is $frac{-14pi}{3}$ so the divergence theorem says that
$$intintint_V nabla cdot F dV = intint_S F cdot hat{n} dS$$
$nabla cdot F = 2$ so our volume integral is just 2 time the volume of V using the formula for area of a truncated cone I get that this is $frac{10pi}{3}$ which isn't right so i know i made a mistake somewhere I've went through my working like 100 times I really cant tell where i went wrong though.
calculus integration vector-analysis vector-fields
asked Dec 25 '18 at 16:00
QuestloveQuestlove
18711
$endgroup$
add a comment |
$begingroup$
I have a surface with the geometric equation $$(z+1)^2=x^2+y^2$$ where $1leq zleq 0$ which give a sort of upside down truncated cone like this
I want to verify Gauss's divergence theorem for this volume with $mathbf{F}=(x,y,0)$.
So I first went about parameterizing the surfaces for the curved part we have
$$(u,v) rightarrow (ucos(v),usin(v),u-1) ,quad 1leq u leq 2, quad 0leq v leq 2pi $$ the other surfaces are just this same parameterization with z component $z=0$ and $0leq u leq 1$ for the bottom circle and z component $z=1$ with $0leq u leq 2$ for the top circle.
When I calculate the surface integrals for the each surface flux integral for the top and bottom circles I get that they are just $0$, since for the top circle we get $n=(0,0,1)$ as the outward normal and for the bottom circle the outward normal is $n=(0,0,-1)$ and we only really need the normal vector not the unit normal since the area element $dS=lvert n rvert dudv$. So $F cdot n = 0$ and so the surface integrals for thee top and bottom circle are also $0$.
When I calculate the surface flux integral for the curved surface I find that the normal $n=(-ucos(v),-usin(v),u)$ by finding the tangent vector with respect to u and the tangent vector with respect to v and taking their cross product. So $F.n=-u^2$, so I evaluate my surface integral $$int_{v=0}^{v=2pi}int_{u=1}^{u=2}-u^2dS$$ get that it is $frac{-14pi}{3}$ so the divergence theorem says that
$$intintint_V nabla cdot F dV = intint_S F cdot hat{n} dS$$
$nabla cdot F = 2$ so our volume integral is just 2 time the volume of V using the formula for area of a truncated cone I get that this is $frac{10pi}{3}$ which isn't right so i know i made a mistake somewhere I've went through my working like 100 times I really cant tell where i went wrong though.
calculus integration vector-analysis vector-fields
asked Dec 25 '18 at 16:00
QuestloveQuestlove
18711
$endgroup$
add a comment |
$begingroup$
I have a surface with the geometric equation $$(z+1)^2=x^2+y^2$$ where $1leq zleq 0$ which give a sort of upside down truncated cone like this
I want to verify Gauss's divergence theorem for this volume with $mathbf{F}=(x,y,0)$.
So I first went about parameterizing the surfaces for the curved part we have
$$(u,v) rightarrow (ucos(v),usin(v),u-1) ,quad 1leq u leq 2, quad 0leq v leq 2pi $$ the other surfaces are just this same parameterization with z component $z=0$ and $0leq u leq 1$ for the bottom circle and z component $z=1$ with $0leq u leq 2$ for the top circle.
When I calculate the surface integrals for the each surface flux integral for the top and bottom circles I get that they are just $0$, since for the top circle we get $n=(0,0,1)$ as the outward normal and for the bottom circle the outward normal is $n=(0,0,-1)$ and we only really need the normal vector not the unit normal since the area element $dS=lvert n rvert dudv$. So $F cdot n = 0$ and so the surface integrals for thee top and bottom circle are also $0$.
When I calculate the surface flux integral for the curved surface I find that the normal $n=(-ucos(v),-usin(v),u)$ by finding the tangent vector with respect to u and the tangent vector with respect to v and taking their cross product. So $F.n=-u^2$, so I evaluate my surface integral $$int_{v=0}^{v=2pi}int_{u=1}^{u=2}-u^2dS$$ get that it is $frac{-14pi}{3}$ so the divergence theorem says that
$$intintint_V nabla cdot F dV = intint_S F cdot hat{n} dS$$
$nabla cdot F = 2$ so our volume integral is just 2 time the volume of V using the formula for area of a truncated cone I get that this is $frac{10pi}{3}$ which isn't right so i know i made a mistake somewhere I've went through my working like 100 times I really cant tell where i went wrong though.
calculus integration vector-analysis vector-fields
asked Dec 25 '18 at 16:00
QuestloveQuestlove
18711
$endgroup$
I have a surface with the geometric equation $$(z+1)^2=x^2+y^2$$ where $1leq zleq 0$ which give a sort of upside down truncated cone like this
I want to verify Gauss's divergence theorem for this volume with $mathbf{F}=(x,y,0)$.
So I first went about parameterizing the surfaces for the curved part we have
$$(u,v) rightarrow (ucos(v),usin(v),u-1) ,quad 1leq u leq 2, quad 0leq v leq 2pi $$ the other surfaces are just this same parameterization with z component $z=0$ and $0leq u leq 1$ for the bottom circle and z component $z=1$ with $0leq u leq 2$ for the top circle.
When I calculate the surface integrals for the each surface flux integral for the top and bottom circles I get that they are just $0$, since for the top circle we get $n=(0,0,1)$ as the outward normal and for the bottom circle the outward normal is $n=(0,0,-1)$ and we only really need the normal vector not the unit normal since the area element $dS=lvert n rvert dudv$. So $F cdot n = 0$ and so the surface integrals for thee top and bottom circle are also $0$.
When I calculate the surface flux integral for the curved surface I find that the normal $n=(-ucos(v),-usin(v),u)$ by finding the tangent vector with respect to u and the tangent vector with respect to v and taking their cross product. So $F.n=-u^2$, so I evaluate my surface integral $$int_{v=0}^{v=2pi}int_{u=1}^{u=2}-u^2dS$$ get that it is $frac{-14pi}{3}$ so the divergence theorem says that
$$intintint_V nabla cdot F dV = intint_S F cdot hat{n} dS$$
$nabla cdot F = 2$ so our volume integral is just 2 time the volume of V using the formula for area of a truncated cone I get that this is $frac{10pi}{3}$ which isn't right so i know i made a mistake somewhere I've went through my working like 100 times I really cant tell where i went wrong though.
calculus integration vector-analysis vector-fields
calculus integration vector-analysis vector-fields
asked Dec 25 '18 at 16:00
QuestloveQuestlove
18711
asked Dec 25 '18 at 16:00
QuestloveQuestlove
18711
asked Dec 25 '18 at 16:00
QuestloveQuestlove
18711
asked Dec 25 '18 at 16:00
QuestloveQuestlove
18711
asked Dec 25 '18 at 16:00
QuestloveQuestlove
18711
18711
add a comment |
add a comment |
1 Answer
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Problem is in the last step, the volume of the cone is
$$
V = frac{pi}{3}left(R_1^2 h_1 - R_2^2 h_2right)
$$
where $R_1 = 2 = h_1$ (the large cone) and $R_2 = 1 = h_2$ (removed tip), so the result is
$$
V = frac{7pi}{3}
$$
answered Dec 25 '18 at 16:27
caveraccaverac
14.8k31130
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1 Answer
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1 Answer
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active
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oldest
votes
$begingroup$
Problem is in the last step, the volume of the cone is
$$
V = frac{pi}{3}left(R_1^2 h_1 - R_2^2 h_2right)
$$
where $R_1 = 2 = h_1$ (the large cone) and $R_2 = 1 = h_2$ (removed tip), so the result is
$$
V = frac{7pi}{3}
$$
answered Dec 25 '18 at 16:27
caveraccaverac
14.8k31130
$endgroup$
add a comment |
$begingroup$
Problem is in the last step, the volume of the cone is
$$
V = frac{pi}{3}left(R_1^2 h_1 - R_2^2 h_2right)
$$
where $R_1 = 2 = h_1$ (the large cone) and $R_2 = 1 = h_2$ (removed tip), so the result is
$$
V = frac{7pi}{3}
$$
answered Dec 25 '18 at 16:27
caveraccaverac
14.8k31130
$endgroup$
add a comment |
$begingroup$
Problem is in the last step, the volume of the cone is
$$
V = frac{pi}{3}left(R_1^2 h_1 - R_2^2 h_2right)
$$
where $R_1 = 2 = h_1$ (the large cone) and $R_2 = 1 = h_2$ (removed tip), so the result is
$$
V = frac{7pi}{3}
$$
answered Dec 25 '18 at 16:27
caveraccaverac
14.8k31130
$endgroup$
Problem is in the last step, the volume of the cone is
$$
V = frac{pi}{3}left(R_1^2 h_1 - R_2^2 h_2right)
$$
where $R_1 = 2 = h_1$ (the large cone) and $R_2 = 1 = h_2$ (removed tip), so the result is
$$
V = frac{7pi}{3}
$$
answered Dec 25 '18 at 16:27
caveraccaverac
14.8k31130
answered Dec 25 '18 at 16:27
caveraccaverac
14.8k31130
answered Dec 25 '18 at 16:27
caveraccaverac
14.8k31130
answered Dec 25 '18 at 16:27
caveraccaverac
14.8k31130
14.8k31130
add a comment |
add a comment |
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