Mixing
Find $ int_{S_{c}} underset{bar{}}{F}.d underset{bar{}}{S};$ for truncated cone.
$begingroup$
$$ int_{S_{c}} underset{bar{}}{F}.d underset{bar{}}{S}, $$
where $S_{c}$ is the curved part of the truncated cone given by $x^2 + y^2 = (a-z)^2$, $0≤z≤(1/2)a$ and $underset{bar{}}{F} = (2x + y^2)underset{bar{}}{i} + (y+xz)underset{bar{}}{j} + zunderset{bar{}}{k}$.
Use a normal to $S_{c}$ that points away from the $z$ axis and give answer in the form $ppi a^3/q$.
Hints:
a) Use the divergence theorem
b) The surface of the truncated cone can be split into 2 flat pieces and a curved part.
c) The volume of a cone is 1/3 * base * height
Despite the hints given, I'm still not sure where to start. Would really appreciate some guidance.
integration vector-analysis
edited Jan 17 at 23:45
Dog_69
6441523
asked Jan 17 at 22:34
LiamLiam
135
$endgroup$
add a comment |
$begingroup$
$$ int_{S_{c}} underset{bar{}}{F}.d underset{bar{}}{S}, $$
where $S_{c}$ is the curved part of the truncated cone given by $x^2 + y^2 = (a-z)^2$, $0≤z≤(1/2)a$ and $underset{bar{}}{F} = (2x + y^2)underset{bar{}}{i} + (y+xz)underset{bar{}}{j} + zunderset{bar{}}{k}$.
Use a normal to $S_{c}$ that points away from the $z$ axis and give answer in the form $ppi a^3/q$.
Hints:
a) Use the divergence theorem
b) The surface of the truncated cone can be split into 2 flat pieces and a curved part.
c) The volume of a cone is 1/3 * base * height
Despite the hints given, I'm still not sure where to start. Would really appreciate some guidance.
integration vector-analysis
edited Jan 17 at 23:45
Dog_69
6441523
asked Jan 17 at 22:34
LiamLiam
135
$endgroup$
add a comment |
$begingroup$
$$ int_{S_{c}} underset{bar{}}{F}.d underset{bar{}}{S}, $$
where $S_{c}$ is the curved part of the truncated cone given by $x^2 + y^2 = (a-z)^2$, $0≤z≤(1/2)a$ and $underset{bar{}}{F} = (2x + y^2)underset{bar{}}{i} + (y+xz)underset{bar{}}{j} + zunderset{bar{}}{k}$.
Use a normal to $S_{c}$ that points away from the $z$ axis and give answer in the form $ppi a^3/q$.
Hints:
a) Use the divergence theorem
b) The surface of the truncated cone can be split into 2 flat pieces and a curved part.
c) The volume of a cone is 1/3 * base * height
Despite the hints given, I'm still not sure where to start. Would really appreciate some guidance.
integration vector-analysis
edited Jan 17 at 23:45
Dog_69
6441523
asked Jan 17 at 22:34
LiamLiam
135
$endgroup$
$$ int_{S_{c}} underset{bar{}}{F}.d underset{bar{}}{S}, $$
where $S_{c}$ is the curved part of the truncated cone given by $x^2 + y^2 = (a-z)^2$, $0≤z≤(1/2)a$ and $underset{bar{}}{F} = (2x + y^2)underset{bar{}}{i} + (y+xz)underset{bar{}}{j} + zunderset{bar{}}{k}$.
Use a normal to $S_{c}$ that points away from the $z$ axis and give answer in the form $ppi a^3/q$.
Hints:
a) Use the divergence theorem
b) The surface of the truncated cone can be split into 2 flat pieces and a curved part.
c) The volume of a cone is 1/3 * base * height
Despite the hints given, I'm still not sure where to start. Would really appreciate some guidance.
integration vector-analysis
integration vector-analysis
edited Jan 17 at 23:45
Dog_69
6441523
asked Jan 17 at 22:34
LiamLiam
135
edited Jan 17 at 23:45
Dog_69
6441523
asked Jan 17 at 22:34
LiamLiam
135
edited Jan 17 at 23:45
Dog_69
6441523
edited Jan 17 at 23:45
Dog_69
6441523
edited Jan 17 at 23:45
Dog_69
6441523
6441523
asked Jan 17 at 22:34
LiamLiam
135
asked Jan 17 at 22:34
LiamLiam
135
asked Jan 17 at 22:34
LiamLiam
135
135
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The divergence theorem says $iint F cdot dS = iiint nabla cdot FdV$. We find that $nabla F = 4$. Now find the volume $V$ of the truncated cone, and multiply by $4$. By the divergence theorem this is equivalent to the surface integral on the entire surface of the truncated cone.
If you want the surface integral for just the curved part of the cone, then calculate $int Fcdot dS$ on $z=0$ (hint: this evaluates to zero, can you see why?) and $int F cdot dS$ on $z=frac{a}{2}$, and subtract this from your $4V$.
Edit for clarity:
Your question asks you to find the flux integral of $F = (2x+y^2)mathbf{i} + (y+xz)mathbf{j} + zmathbf{k}$ on the blue surface shown above. There are two ways to do this.
The first way is to compute this directly through the formula $int mathbf{F}cdot dmathbf{S}$.
This involves finding the normal vector to the blue surface, parameterising the blue surface, and taking the surface integral. Often, this is the more complicated way of doing things.
The second option is to use the divergence theorem.
In this question, the way to do this would be to first calculate the volume enclosed by the three surfaces shown. This question would require a little more work though, since you're only interested in the flux through the blue surface. The divergence theorem would give you the flux through all surfaces enclosing the volume, so you would need to subtract the flux through the green surfaces for your final answer.
Doing this by the divergence theorem, the first step would be to work out $nabla cdot mathbf{F}$. This gives $2+1+1=4$. Since we see that the divergence of $mathbf{F}$ is a constant and not dependent on $x, y$ or $z$, our volume integral becomes $$iiint 4 dV = 4iiint dV$$ So all you need to do is find the volume and multiply by 4 (If the divergence were dependent on $x, y$ or $z$, your integral could be slightly more complicated). Calculating the volume of the truncated cone we have
$$frac{1}{3}(pi a^2a-pifrac{a^2}{4}frac{a}{2})=frac{7pi a^3}{24} $$
So the divergence through the enclosing surfaces is $frac{7pi a^3}{6}.$
The flux integral through the bottom surface where $z=0$ will be given by $$int mathbf{F}cdot mathbf{n} dS.$$ Since $F = (2x+y^2)mathbf{i} + (y+xz)mathbf{j} + 0mathbf{k}$ on $z=0$, and $n=-mathbf{k}$, we can see that $mathbf{F}cdot mathbf{n}=0.$ On the upper surface, where $z=frac{a}{2}$, we have $F = (2x+y^2)mathbf{i} + (y+xz)mathbf{j} + frac{a}{2}mathbf{k}$. Our $mathbf{n}$ remains the same (except for a change in sign, since the normal must point outward from the volume), so $mathbf{F}cdot mathbf{n}=frac{a}{2}.$ The integral over the surface can be calculated using polar co-ordinates as follows
$$iint mathbf{F}cdot mathbf{n} dS = frac{a}{2}int_0^{2pi}int_0^frac{a}{2}rdrdtheta= apiBig[frac{r^2}{2}Big]_0^{frac{a}{2}}=frac{a^3pi}{8}.$$
So your final answer should be $$frac{7pi a^3}{6} - frac{a^3pi}{8} = frac{25a^3pi}{24}.$$
answered Jan 17 at 22:37
cluelessatthiscluelessatthis
402313
$endgroup$
$begingroup$
Thanks for the reply, i really appreciate your help. When calculating the volume of a truncated cone which im assuming is πh(r1²+r2²+r1.r2) ⁄ 3, what would i sub in for r1 and r2? thanks again
$endgroup$
– Liam
Jan 20 at 0:41
$begingroup$
@Liam I'm unsure what your formula here is referring to - to calculate the volume of the truncated cone, calculate the volume of the whole cone, and subtract the cone that's missing, so you would have $frac{1}{3}(pi a^2)a - frac{1}{3}(pi (frac{a}{2})^2) frac{a}{2}$. Do you see how I've reached this?
$endgroup$
– cluelessatthis
Jan 20 at 0:52
$begingroup$
I see. And to calculate$ ∫F⋅dS$ on $z=a/2$, i would have the find the normal with a postive k component which in this case would be $ ∇(x^2+y^2-(a-z)^2)$ = $ (2x,2y,-2(a-z))$. And then $ F.n = ((2x+y^2),(y+xz)+z).((2x,2y,-2(a-z))$. And then $∫F⋅dS$ on $z=a/2 = ∫ F.n dxdx/n.k $ and then use polar coordinates with limits of r to be $a$ and $a/2$. Would this be correct? Thanks again.
$endgroup$
– Liam
Jan 20 at 13:32
$begingroup$
@Liam actually, what you're doing in this comment is computing the flux integral on the curved surface directly. To compute the surface integral through the top, flat surface of the truncated cone, you would just take $mathbf{F}cdot mathbf{k}$ (Since the flat surface is constant in z, it is parallel to the xy plane, so n is just the unit vector $mathbf{k}$). Then convert to polars, taking limits: $ 0 le r le frac{a}{2}$ and $0 le theta le 2pi $.If you're still a bit confused about what's going on, I can try and expand on my answer to provide more clarity.
$endgroup$
– cluelessatthis
Jan 20 at 14:16
$begingroup$
Thanks, this made it a bit clearer but i am still quite confused so i would really appreciate if you expanded on your answer. Thanks for being so helpful.
$endgroup$
– Liam
Jan 20 at 14:29
|
show 2 more comments
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1 Answer
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$begingroup$
The divergence theorem says $iint F cdot dS = iiint nabla cdot FdV$. We find that $nabla F = 4$. Now find the volume $V$ of the truncated cone, and multiply by $4$. By the divergence theorem this is equivalent to the surface integral on the entire surface of the truncated cone.
If you want the surface integral for just the curved part of the cone, then calculate $int Fcdot dS$ on $z=0$ (hint: this evaluates to zero, can you see why?) and $int F cdot dS$ on $z=frac{a}{2}$, and subtract this from your $4V$.
Edit for clarity:
Your question asks you to find the flux integral of $F = (2x+y^2)mathbf{i} + (y+xz)mathbf{j} + zmathbf{k}$ on the blue surface shown above. There are two ways to do this.
The first way is to compute this directly through the formula $int mathbf{F}cdot dmathbf{S}$.
This involves finding the normal vector to the blue surface, parameterising the blue surface, and taking the surface integral. Often, this is the more complicated way of doing things.
The second option is to use the divergence theorem.
In this question, the way to do this would be to first calculate the volume enclosed by the three surfaces shown. This question would require a little more work though, since you're only interested in the flux through the blue surface. The divergence theorem would give you the flux through all surfaces enclosing the volume, so you would need to subtract the flux through the green surfaces for your final answer.
Doing this by the divergence theorem, the first step would be to work out $nabla cdot mathbf{F}$. This gives $2+1+1=4$. Since we see that the divergence of $mathbf{F}$ is a constant and not dependent on $x, y$ or $z$, our volume integral becomes $$iiint 4 dV = 4iiint dV$$ So all you need to do is find the volume and multiply by 4 (If the divergence were dependent on $x, y$ or $z$, your integral could be slightly more complicated). Calculating the volume of the truncated cone we have
$$frac{1}{3}(pi a^2a-pifrac{a^2}{4}frac{a}{2})=frac{7pi a^3}{24} $$
So the divergence through the enclosing surfaces is $frac{7pi a^3}{6}.$
The flux integral through the bottom surface where $z=0$ will be given by $$int mathbf{F}cdot mathbf{n} dS.$$ Since $F = (2x+y^2)mathbf{i} + (y+xz)mathbf{j} + 0mathbf{k}$ on $z=0$, and $n=-mathbf{k}$, we can see that $mathbf{F}cdot mathbf{n}=0.$ On the upper surface, where $z=frac{a}{2}$, we have $F = (2x+y^2)mathbf{i} + (y+xz)mathbf{j} + frac{a}{2}mathbf{k}$. Our $mathbf{n}$ remains the same (except for a change in sign, since the normal must point outward from the volume), so $mathbf{F}cdot mathbf{n}=frac{a}{2}.$ The integral over the surface can be calculated using polar co-ordinates as follows
$$iint mathbf{F}cdot mathbf{n} dS = frac{a}{2}int_0^{2pi}int_0^frac{a}{2}rdrdtheta= apiBig[frac{r^2}{2}Big]_0^{frac{a}{2}}=frac{a^3pi}{8}.$$
So your final answer should be $$frac{7pi a^3}{6} - frac{a^3pi}{8} = frac{25a^3pi}{24}.$$
answered Jan 17 at 22:37
cluelessatthiscluelessatthis
402313
$endgroup$
$begingroup$
Thanks for the reply, i really appreciate your help. When calculating the volume of a truncated cone which im assuming is πh(r1²+r2²+r1.r2) ⁄ 3, what would i sub in for r1 and r2? thanks again
$endgroup$
– Liam
Jan 20 at 0:41
$begingroup$
@Liam I'm unsure what your formula here is referring to - to calculate the volume of the truncated cone, calculate the volume of the whole cone, and subtract the cone that's missing, so you would have $frac{1}{3}(pi a^2)a - frac{1}{3}(pi (frac{a}{2})^2) frac{a}{2}$. Do you see how I've reached this?
$endgroup$
– cluelessatthis
Jan 20 at 0:52
$begingroup$
I see. And to calculate$ ∫F⋅dS$ on $z=a/2$, i would have the find the normal with a postive k component which in this case would be $ ∇(x^2+y^2-(a-z)^2)$ = $ (2x,2y,-2(a-z))$. And then $ F.n = ((2x+y^2),(y+xz)+z).((2x,2y,-2(a-z))$. And then $∫F⋅dS$ on $z=a/2 = ∫ F.n dxdx/n.k $ and then use polar coordinates with limits of r to be $a$ and $a/2$. Would this be correct? Thanks again.
$endgroup$
– Liam
Jan 20 at 13:32
$begingroup$
@Liam actually, what you're doing in this comment is computing the flux integral on the curved surface directly. To compute the surface integral through the top, flat surface of the truncated cone, you would just take $mathbf{F}cdot mathbf{k}$ (Since the flat surface is constant in z, it is parallel to the xy plane, so n is just the unit vector $mathbf{k}$). Then convert to polars, taking limits: $ 0 le r le frac{a}{2}$ and $0 le theta le 2pi $.If you're still a bit confused about what's going on, I can try and expand on my answer to provide more clarity.
$endgroup$
– cluelessatthis
Jan 20 at 14:16
$begingroup$
Thanks, this made it a bit clearer but i am still quite confused so i would really appreciate if you expanded on your answer. Thanks for being so helpful.
$endgroup$
– Liam
Jan 20 at 14:29
|
show 2 more comments
$begingroup$
The divergence theorem says $iint F cdot dS = iiint nabla cdot FdV$. We find that $nabla F = 4$. Now find the volume $V$ of the truncated cone, and multiply by $4$. By the divergence theorem this is equivalent to the surface integral on the entire surface of the truncated cone.
If you want the surface integral for just the curved part of the cone, then calculate $int Fcdot dS$ on $z=0$ (hint: this evaluates to zero, can you see why?) and $int F cdot dS$ on $z=frac{a}{2}$, and subtract this from your $4V$.
Edit for clarity:
Your question asks you to find the flux integral of $F = (2x+y^2)mathbf{i} + (y+xz)mathbf{j} + zmathbf{k}$ on the blue surface shown above. There are two ways to do this.
The first way is to compute this directly through the formula $int mathbf{F}cdot dmathbf{S}$.
This involves finding the normal vector to the blue surface, parameterising the blue surface, and taking the surface integral. Often, this is the more complicated way of doing things.
The second option is to use the divergence theorem.
In this question, the way to do this would be to first calculate the volume enclosed by the three surfaces shown. This question would require a little more work though, since you're only interested in the flux through the blue surface. The divergence theorem would give you the flux through all surfaces enclosing the volume, so you would need to subtract the flux through the green surfaces for your final answer.
Doing this by the divergence theorem, the first step would be to work out $nabla cdot mathbf{F}$. This gives $2+1+1=4$. Since we see that the divergence of $mathbf{F}$ is a constant and not dependent on $x, y$ or $z$, our volume integral becomes $$iiint 4 dV = 4iiint dV$$ So all you need to do is find the volume and multiply by 4 (If the divergence were dependent on $x, y$ or $z$, your integral could be slightly more complicated). Calculating the volume of the truncated cone we have
$$frac{1}{3}(pi a^2a-pifrac{a^2}{4}frac{a}{2})=frac{7pi a^3}{24} $$
So the divergence through the enclosing surfaces is $frac{7pi a^3}{6}.$
The flux integral through the bottom surface where $z=0$ will be given by $$int mathbf{F}cdot mathbf{n} dS.$$ Since $F = (2x+y^2)mathbf{i} + (y+xz)mathbf{j} + 0mathbf{k}$ on $z=0$, and $n=-mathbf{k}$, we can see that $mathbf{F}cdot mathbf{n}=0.$ On the upper surface, where $z=frac{a}{2}$, we have $F = (2x+y^2)mathbf{i} + (y+xz)mathbf{j} + frac{a}{2}mathbf{k}$. Our $mathbf{n}$ remains the same (except for a change in sign, since the normal must point outward from the volume), so $mathbf{F}cdot mathbf{n}=frac{a}{2}.$ The integral over the surface can be calculated using polar co-ordinates as follows
$$iint mathbf{F}cdot mathbf{n} dS = frac{a}{2}int_0^{2pi}int_0^frac{a}{2}rdrdtheta= apiBig[frac{r^2}{2}Big]_0^{frac{a}{2}}=frac{a^3pi}{8}.$$
So your final answer should be $$frac{7pi a^3}{6} - frac{a^3pi}{8} = frac{25a^3pi}{24}.$$
answered Jan 17 at 22:37
cluelessatthiscluelessatthis
402313
$endgroup$
$begingroup$
Thanks for the reply, i really appreciate your help. When calculating the volume of a truncated cone which im assuming is πh(r1²+r2²+r1.r2) ⁄ 3, what would i sub in for r1 and r2? thanks again
$endgroup$
– Liam
Jan 20 at 0:41
$begingroup$
@Liam I'm unsure what your formula here is referring to - to calculate the volume of the truncated cone, calculate the volume of the whole cone, and subtract the cone that's missing, so you would have $frac{1}{3}(pi a^2)a - frac{1}{3}(pi (frac{a}{2})^2) frac{a}{2}$. Do you see how I've reached this?
$endgroup$
– cluelessatthis
Jan 20 at 0:52
$begingroup$
I see. And to calculate$ ∫F⋅dS$ on $z=a/2$, i would have the find the normal with a postive k component which in this case would be $ ∇(x^2+y^2-(a-z)^2)$ = $ (2x,2y,-2(a-z))$. And then $ F.n = ((2x+y^2),(y+xz)+z).((2x,2y,-2(a-z))$. And then $∫F⋅dS$ on $z=a/2 = ∫ F.n dxdx/n.k $ and then use polar coordinates with limits of r to be $a$ and $a/2$. Would this be correct? Thanks again.
$endgroup$
– Liam
Jan 20 at 13:32
$begingroup$
@Liam actually, what you're doing in this comment is computing the flux integral on the curved surface directly. To compute the surface integral through the top, flat surface of the truncated cone, you would just take $mathbf{F}cdot mathbf{k}$ (Since the flat surface is constant in z, it is parallel to the xy plane, so n is just the unit vector $mathbf{k}$). Then convert to polars, taking limits: $ 0 le r le frac{a}{2}$ and $0 le theta le 2pi $.If you're still a bit confused about what's going on, I can try and expand on my answer to provide more clarity.
$endgroup$
– cluelessatthis
Jan 20 at 14:16
$begingroup$
Thanks, this made it a bit clearer but i am still quite confused so i would really appreciate if you expanded on your answer. Thanks for being so helpful.
$endgroup$
– Liam
Jan 20 at 14:29
|
show 2 more comments
$begingroup$
The divergence theorem says $iint F cdot dS = iiint nabla cdot FdV$. We find that $nabla F = 4$. Now find the volume $V$ of the truncated cone, and multiply by $4$. By the divergence theorem this is equivalent to the surface integral on the entire surface of the truncated cone.
If you want the surface integral for just the curved part of the cone, then calculate $int Fcdot dS$ on $z=0$ (hint: this evaluates to zero, can you see why?) and $int F cdot dS$ on $z=frac{a}{2}$, and subtract this from your $4V$.
Edit for clarity:
Your question asks you to find the flux integral of $F = (2x+y^2)mathbf{i} + (y+xz)mathbf{j} + zmathbf{k}$ on the blue surface shown above. There are two ways to do this.
The first way is to compute this directly through the formula $int mathbf{F}cdot dmathbf{S}$.
This involves finding the normal vector to the blue surface, parameterising the blue surface, and taking the surface integral. Often, this is the more complicated way of doing things.
The second option is to use the divergence theorem.
In this question, the way to do this would be to first calculate the volume enclosed by the three surfaces shown. This question would require a little more work though, since you're only interested in the flux through the blue surface. The divergence theorem would give you the flux through all surfaces enclosing the volume, so you would need to subtract the flux through the green surfaces for your final answer.
Doing this by the divergence theorem, the first step would be to work out $nabla cdot mathbf{F}$. This gives $2+1+1=4$. Since we see that the divergence of $mathbf{F}$ is a constant and not dependent on $x, y$ or $z$, our volume integral becomes $$iiint 4 dV = 4iiint dV$$ So all you need to do is find the volume and multiply by 4 (If the divergence were dependent on $x, y$ or $z$, your integral could be slightly more complicated). Calculating the volume of the truncated cone we have
$$frac{1}{3}(pi a^2a-pifrac{a^2}{4}frac{a}{2})=frac{7pi a^3}{24} $$
So the divergence through the enclosing surfaces is $frac{7pi a^3}{6}.$
The flux integral through the bottom surface where $z=0$ will be given by $$int mathbf{F}cdot mathbf{n} dS.$$ Since $F = (2x+y^2)mathbf{i} + (y+xz)mathbf{j} + 0mathbf{k}$ on $z=0$, and $n=-mathbf{k}$, we can see that $mathbf{F}cdot mathbf{n}=0.$ On the upper surface, where $z=frac{a}{2}$, we have $F = (2x+y^2)mathbf{i} + (y+xz)mathbf{j} + frac{a}{2}mathbf{k}$. Our $mathbf{n}$ remains the same (except for a change in sign, since the normal must point outward from the volume), so $mathbf{F}cdot mathbf{n}=frac{a}{2}.$ The integral over the surface can be calculated using polar co-ordinates as follows
$$iint mathbf{F}cdot mathbf{n} dS = frac{a}{2}int_0^{2pi}int_0^frac{a}{2}rdrdtheta= apiBig[frac{r^2}{2}Big]_0^{frac{a}{2}}=frac{a^3pi}{8}.$$
So your final answer should be $$frac{7pi a^3}{6} - frac{a^3pi}{8} = frac{25a^3pi}{24}.$$
answered Jan 17 at 22:37
cluelessatthiscluelessatthis
402313
$endgroup$
The divergence theorem says $iint F cdot dS = iiint nabla cdot FdV$. We find that $nabla F = 4$. Now find the volume $V$ of the truncated cone, and multiply by $4$. By the divergence theorem this is equivalent to the surface integral on the entire surface of the truncated cone.
If you want the surface integral for just the curved part of the cone, then calculate $int Fcdot dS$ on $z=0$ (hint: this evaluates to zero, can you see why?) and $int F cdot dS$ on $z=frac{a}{2}$, and subtract this from your $4V$.
Edit for clarity:
Your question asks you to find the flux integral of $F = (2x+y^2)mathbf{i} + (y+xz)mathbf{j} + zmathbf{k}$ on the blue surface shown above. There are two ways to do this.
The first way is to compute this directly through the formula $int mathbf{F}cdot dmathbf{S}$.
This involves finding the normal vector to the blue surface, parameterising the blue surface, and taking the surface integral. Often, this is the more complicated way of doing things.
The second option is to use the divergence theorem.
In this question, the way to do this would be to first calculate the volume enclosed by the three surfaces shown. This question would require a little more work though, since you're only interested in the flux through the blue surface. The divergence theorem would give you the flux through all surfaces enclosing the volume, so you would need to subtract the flux through the green surfaces for your final answer.
Doing this by the divergence theorem, the first step would be to work out $nabla cdot mathbf{F}$. This gives $2+1+1=4$. Since we see that the divergence of $mathbf{F}$ is a constant and not dependent on $x, y$ or $z$, our volume integral becomes $$iiint 4 dV = 4iiint dV$$ So all you need to do is find the volume and multiply by 4 (If the divergence were dependent on $x, y$ or $z$, your integral could be slightly more complicated). Calculating the volume of the truncated cone we have
$$frac{1}{3}(pi a^2a-pifrac{a^2}{4}frac{a}{2})=frac{7pi a^3}{24} $$
So the divergence through the enclosing surfaces is $frac{7pi a^3}{6}.$
The flux integral through the bottom surface where $z=0$ will be given by $$int mathbf{F}cdot mathbf{n} dS.$$ Since $F = (2x+y^2)mathbf{i} + (y+xz)mathbf{j} + 0mathbf{k}$ on $z=0$, and $n=-mathbf{k}$, we can see that $mathbf{F}cdot mathbf{n}=0.$ On the upper surface, where $z=frac{a}{2}$, we have $F = (2x+y^2)mathbf{i} + (y+xz)mathbf{j} + frac{a}{2}mathbf{k}$. Our $mathbf{n}$ remains the same (except for a change in sign, since the normal must point outward from the volume), so $mathbf{F}cdot mathbf{n}=frac{a}{2}.$ The integral over the surface can be calculated using polar co-ordinates as follows
$$iint mathbf{F}cdot mathbf{n} dS = frac{a}{2}int_0^{2pi}int_0^frac{a}{2}rdrdtheta= apiBig[frac{r^2}{2}Big]_0^{frac{a}{2}}=frac{a^3pi}{8}.$$
So your final answer should be $$frac{7pi a^3}{6} - frac{a^3pi}{8} = frac{25a^3pi}{24}.$$
answered Jan 17 at 22:37
cluelessatthiscluelessatthis
402313
edited Jan 21 at 14:21
answered Jan 17 at 22:37
cluelessatthiscluelessatthis
402313
answered Jan 17 at 22:37
cluelessatthiscluelessatthis
402313
answered Jan 17 at 22:37
cluelessatthiscluelessatthis
402313
402313
$begingroup$
Thanks for the reply, i really appreciate your help. When calculating the volume of a truncated cone which im assuming is πh(r1²+r2²+r1.r2) ⁄ 3, what would i sub in for r1 and r2? thanks again
$endgroup$
– Liam
Jan 20 at 0:41
$begingroup$
@Liam I'm unsure what your formula here is referring to - to calculate the volume of the truncated cone, calculate the volume of the whole cone, and subtract the cone that's missing, so you would have $frac{1}{3}(pi a^2)a - frac{1}{3}(pi (frac{a}{2})^2) frac{a}{2}$. Do you see how I've reached this?
$endgroup$
– cluelessatthis
Jan 20 at 0:52
$begingroup$
I see. And to calculate$ ∫F⋅dS$ on $z=a/2$, i would have the find the normal with a postive k component which in this case would be $ ∇(x^2+y^2-(a-z)^2)$ = $ (2x,2y,-2(a-z))$. And then $ F.n = ((2x+y^2),(y+xz)+z).((2x,2y,-2(a-z))$. And then $∫F⋅dS$ on $z=a/2 = ∫ F.n dxdx/n.k $ and then use polar coordinates with limits of r to be $a$ and $a/2$. Would this be correct? Thanks again.
$endgroup$
– Liam
Jan 20 at 13:32
$begingroup$
@Liam actually, what you're doing in this comment is computing the flux integral on the curved surface directly. To compute the surface integral through the top, flat surface of the truncated cone, you would just take $mathbf{F}cdot mathbf{k}$ (Since the flat surface is constant in z, it is parallel to the xy plane, so n is just the unit vector $mathbf{k}$). Then convert to polars, taking limits: $ 0 le r le frac{a}{2}$ and $0 le theta le 2pi $.If you're still a bit confused about what's going on, I can try and expand on my answer to provide more clarity.
$endgroup$
– cluelessatthis
Jan 20 at 14:16
$begingroup$
Thanks, this made it a bit clearer but i am still quite confused so i would really appreciate if you expanded on your answer. Thanks for being so helpful.
$endgroup$
– Liam
Jan 20 at 14:29
|
show 2 more comments
$begingroup$
Thanks for the reply, i really appreciate your help. When calculating the volume of a truncated cone which im assuming is πh(r1²+r2²+r1.r2) ⁄ 3, what would i sub in for r1 and r2? thanks again
$endgroup$
– Liam
Jan 20 at 0:41
$begingroup$
@Liam I'm unsure what your formula here is referring to - to calculate the volume of the truncated cone, calculate the volume of the whole cone, and subtract the cone that's missing, so you would have $frac{1}{3}(pi a^2)a - frac{1}{3}(pi (frac{a}{2})^2) frac{a}{2}$. Do you see how I've reached this?
$endgroup$
– cluelessatthis
Jan 20 at 0:52
$begingroup$
I see. And to calculate$ ∫F⋅dS$ on $z=a/2$, i would have the find the normal with a postive k component which in this case would be $ ∇(x^2+y^2-(a-z)^2)$ = $ (2x,2y,-2(a-z))$. And then $ F.n = ((2x+y^2),(y+xz)+z).((2x,2y,-2(a-z))$. And then $∫F⋅dS$ on $z=a/2 = ∫ F.n dxdx/n.k $ and then use polar coordinates with limits of r to be $a$ and $a/2$. Would this be correct? Thanks again.
$endgroup$
– Liam
Jan 20 at 13:32
$begingroup$
@Liam actually, what you're doing in this comment is computing the flux integral on the curved surface directly. To compute the surface integral through the top, flat surface of the truncated cone, you would just take $mathbf{F}cdot mathbf{k}$ (Since the flat surface is constant in z, it is parallel to the xy plane, so n is just the unit vector $mathbf{k}$). Then convert to polars, taking limits: $ 0 le r le frac{a}{2}$ and $0 le theta le 2pi $.If you're still a bit confused about what's going on, I can try and expand on my answer to provide more clarity.
$endgroup$
– cluelessatthis
Jan 20 at 14:16
$begingroup$
Thanks, this made it a bit clearer but i am still quite confused so i would really appreciate if you expanded on your answer. Thanks for being so helpful.
$endgroup$
– Liam
Jan 20 at 14:29
$begingroup$
Thanks for the reply, i really appreciate your help. When calculating the volume of a truncated cone which im assuming is πh(r1²+r2²+r1.r2) ⁄ 3, what would i sub in for r1 and r2? thanks again
$endgroup$
– Liam
Jan 20 at 0:41
$begingroup$
Thanks for the reply, i really appreciate your help. When calculating the volume of a truncated cone which im assuming is πh(r1²+r2²+r1.r2) ⁄ 3, what would i sub in for r1 and r2? thanks again
$endgroup$
– Liam
Jan 20 at 0:41
$begingroup$
@Liam I'm unsure what your formula here is referring to - to calculate the volume of the truncated cone, calculate the volume of the whole cone, and subtract the cone that's missing, so you would have $frac{1}{3}(pi a^2)a - frac{1}{3}(pi (frac{a}{2})^2) frac{a}{2}$. Do you see how I've reached this?
$endgroup$
– cluelessatthis
Jan 20 at 0:52
$begingroup$
@Liam I'm unsure what your formula here is referring to - to calculate the volume of the truncated cone, calculate the volume of the whole cone, and subtract the cone that's missing, so you would have $frac{1}{3}(pi a^2)a - frac{1}{3}(pi (frac{a}{2})^2) frac{a}{2}$. Do you see how I've reached this?
$endgroup$
– cluelessatthis
Jan 20 at 0:52
$begingroup$
I see. And to calculate$ ∫F⋅dS$ on $z=a/2$, i would have the find the normal with a postive k component which in this case would be $ ∇(x^2+y^2-(a-z)^2)$ = $ (2x,2y,-2(a-z))$. And then $ F.n = ((2x+y^2),(y+xz)+z).((2x,2y,-2(a-z))$. And then $∫F⋅dS$ on $z=a/2 = ∫ F.n dxdx/n.k $ and then use polar coordinates with limits of r to be $a$ and $a/2$. Would this be correct? Thanks again.
$endgroup$
– Liam
Jan 20 at 13:32
$begingroup$
I see. And to calculate$ ∫F⋅dS$ on $z=a/2$, i would have the find the normal with a postive k component which in this case would be $ ∇(x^2+y^2-(a-z)^2)$ = $ (2x,2y,-2(a-z))$. And then $ F.n = ((2x+y^2),(y+xz)+z).((2x,2y,-2(a-z))$. And then $∫F⋅dS$ on $z=a/2 = ∫ F.n dxdx/n.k $ and then use polar coordinates with limits of r to be $a$ and $a/2$. Would this be correct? Thanks again.
$endgroup$
– Liam
Jan 20 at 13:32
$begingroup$
@Liam actually, what you're doing in this comment is computing the flux integral on the curved surface directly. To compute the surface integral through the top, flat surface of the truncated cone, you would just take $mathbf{F}cdot mathbf{k}$ (Since the flat surface is constant in z, it is parallel to the xy plane, so n is just the unit vector $mathbf{k}$). Then convert to polars, taking limits: $ 0 le r le frac{a}{2}$ and $0 le theta le 2pi $.If you're still a bit confused about what's going on, I can try and expand on my answer to provide more clarity.
$endgroup$
– cluelessatthis
Jan 20 at 14:16
$begingroup$
@Liam actually, what you're doing in this comment is computing the flux integral on the curved surface directly. To compute the surface integral through the top, flat surface of the truncated cone, you would just take $mathbf{F}cdot mathbf{k}$ (Since the flat surface is constant in z, it is parallel to the xy plane, so n is just the unit vector $mathbf{k}$). Then convert to polars, taking limits: $ 0 le r le frac{a}{2}$ and $0 le theta le 2pi $.If you're still a bit confused about what's going on, I can try and expand on my answer to provide more clarity.
$endgroup$
– cluelessatthis
Jan 20 at 14:16
$begingroup$
Thanks, this made it a bit clearer but i am still quite confused so i would really appreciate if you expanded on your answer. Thanks for being so helpful.
$endgroup$
– Liam
Jan 20 at 14:29
$begingroup$
Thanks, this made it a bit clearer but i am still quite confused so i would really appreciate if you expanded on your answer. Thanks for being so helpful.
$endgroup$
– Liam
Jan 20 at 14:29
|
show 2 more comments
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