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Calculating the volume of a cone using $int_D f = int_K big[int_h(x)^g(x) f(x,y) dy big] dx$ : $xin mathbbR^n, y in mathbbR$
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There are various ways to calculate the volume of a cone. The one that I am trying is by evaluating the following integral $$textV = int_x^2+y^2 le a^2 int_0^fracha sqrta^2-x^2-y^2 1 dz dK = frac23 pi a^2 h$$ which is twice as the expected one! Where do I do wrong?
PS I got $z$ by using some trigonometric; as $dfracz^2h^2 = dfraca^2-x^2-y^2a^2.$ I guess that here $z$ is not like a 'normal' function of $(x,y)$ different from types like infinite paraboloid but even if so I don't know how it is related.
real-analysis multivariable-calculus volume
edited Aug 2 at 18:37
José Carlos Santos
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asked Aug 2 at 18:26
Edi
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There are various ways to calculate the volume of a cone. The one that I am trying is by evaluating the following integral $$textV = int_x^2+y^2 le a^2 int_0^fracha sqrta^2-x^2-y^2 1 dz dK = frac23 pi a^2 h$$ which is twice as the expected one! Where do I do wrong?
PS I got $z$ by using some trigonometric; as $dfracz^2h^2 = dfraca^2-x^2-y^2a^2.$ I guess that here $z$ is not like a 'normal' function of $(x,y)$ different from types like infinite paraboloid but even if so I don't know how it is related.
real-analysis multivariable-calculus volume
edited Aug 2 at 18:37
José Carlos Santos
112k1696172
asked Aug 2 at 18:26
Edi
1,121728
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
There are various ways to calculate the volume of a cone. The one that I am trying is by evaluating the following integral $$textV = int_x^2+y^2 le a^2 int_0^fracha sqrta^2-x^2-y^2 1 dz dK = frac23 pi a^2 h$$ which is twice as the expected one! Where do I do wrong?
PS I got $z$ by using some trigonometric; as $dfracz^2h^2 = dfraca^2-x^2-y^2a^2.$ I guess that here $z$ is not like a 'normal' function of $(x,y)$ different from types like infinite paraboloid but even if so I don't know how it is related.
real-analysis multivariable-calculus volume
edited Aug 2 at 18:37
José Carlos Santos
112k1696172
asked Aug 2 at 18:26
Edi
1,121728
There are various ways to calculate the volume of a cone. The one that I am trying is by evaluating the following integral $$textV = int_x^2+y^2 le a^2 int_0^fracha sqrta^2-x^2-y^2 1 dz dK = frac23 pi a^2 h$$ which is twice as the expected one! Where do I do wrong?
PS I got $z$ by using some trigonometric; as $dfracz^2h^2 = dfraca^2-x^2-y^2a^2.$ I guess that here $z$ is not like a 'normal' function of $(x,y)$ different from types like infinite paraboloid but even if so I don't know how it is related.
real-analysis multivariable-calculus volume
edited Aug 2 at 18:37
José Carlos Santos
112k1696172
asked Aug 2 at 18:26
Edi
1,121728
edited Aug 2 at 18:37
José Carlos Santos
112k1696172
edited Aug 2 at 18:37
José Carlos Santos
112k1696172
edited Aug 2 at 18:37
José Carlos Santos
112k1696172
112k1696172
asked Aug 2 at 18:26
Edi
1,121728
asked Aug 2 at 18:26
Edi
1,121728
asked Aug 2 at 18:26
Edi
1,121728
1,121728
add a comment |Â
add a comment |Â
4 Answers
4
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Hint: Your dealing with the equation $dfracz^2h^2 = dfraca^2-x^2-y^2a^2$ is wrong, try with
$$left(1-dfraczhright)^2 = dfracx^2+y^2a^2.$$
answered Aug 2 at 18:39
user 108128
18.8k41544
I tried with $left(1-dfraczhright)^2 = dfracx^2+y^2a^2$ and it gives $1/3$. But how $dfracz^2h^2 = dfraca^2-x^2-y^2a^2$ is wrong?! Same triangle same theorem..
– Edi
Aug 2 at 18:46
1
Let $x=0$ in your equation, then you have a circle (ellipse) equation not a line
– user 108128
Aug 2 at 18:48
Yes again you are right! But I don't understand how $dfracz^2h^2 = dfraca^2-x^2-y^2a^2$ gives a wrong result (i.e. circle) when limited to $x=0$ though it comes from the same trigonometric theorem!
– Edi
Aug 2 at 18:52
Sorry I didn't mean that. Of course it must be a line when limited to any line on plane through the origin; my question is that $z^2/h^2 = (a^2-x^2-y^2)/a^2$ must also give a line when the same limit applies but it fails and why? $z^2/h^2 = (a^2-x^2-y^2)/a^2$ has come from same rule as $(1-dfraczh)^2 = (x^2+y^2)/a^2.$ but it fails when testing a 'boundary' evaluation..
– Edi
Aug 2 at 18:57
1
Sorry. I gave you the true equation and I can't enlightened you dear. I think it's better you try more to find it yourself. This will be worthwhile.
– user 108128
Aug 2 at 19:28
 |Â
show 7 more comments
up vote
1
down vote
That region over which you are integrating is no cone at all. It's the top half of an ellipsoid.
You will get the value that you're after computing the integral$$int_-a^aint_-sqrta^2-x^2^sqrta^2-x^2int_frac hasqrtx^2+y^2^h1,mathrm dz,mathrm dy,mathrm dx.$$Here, my cone is$$left(x,y,z)inmathbbR^3,middle.$$
answered Aug 2 at 18:35
José Carlos Santos
112k1696172
$z^2/h^2 = (a^2-x^2-y^2)/a^2$ and $(1-z/h)^2 = (x^2+y^2)/a^2$ comes from the same theorem of trigonometric on the same object, but I don't understand why the latter gives an ellipsoid; both equations are also squared both use same $z$..
– Edi
Aug 2 at 19:10
$$fracz^2h^2=fraca^2-x^2-y^2a^2iff a^2z^2=(ah)^2-h^2x^2-h^2y^2iff h^2x^2+h^2y^2+a^2z^2=(ah)^2.$$Is this an ellipsoid or not?
– José Carlos Santos
Aug 2 at 19:14
Of course it is but I derived it from a triangle as an intersection a plane through z-axis and the cone. So how it comes an ellipsoid?!
– Edi
Aug 2 at 19:17
The book says "..the cone $x^2+y^2le a^2, 0 le z le h/a^2(a^2-x^2-y^2) $" for hint and I don't know how it deduce $z$?
– Edi
Aug 2 at 19:19
@Edi Then the hint contains a mistake.
– José Carlos Santos
Aug 2 at 19:19
 |Â
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0
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We need 3 constants $ (a,h,z_1):$
Equation of right circular cone with $z_1$ apex shift
$$fracrz-z_1=fracah$$ or squaring,
$$ r^2=x^2+y^2= (z-z_1)^2(a/h)^2.$$
answered Aug 2 at 19:02
Narasimham
20.1k51857
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Because $dfracz^2h^2 = dfraca^2-x^2-y^2a^2$ is wrong. The correct one is $dfraczh = dfraca-sqrtx^2-y^2a$ which gives the correct answer $1/3 pi a^2 h$.
answered Aug 2 at 19:59
Edi
1,121728
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Hint: Your dealing with the equation $dfracz^2h^2 = dfraca^2-x^2-y^2a^2$ is wrong, try with
$$left(1-dfraczhright)^2 = dfracx^2+y^2a^2.$$
answered Aug 2 at 18:39
user 108128
18.8k41544
I tried with $left(1-dfraczhright)^2 = dfracx^2+y^2a^2$ and it gives $1/3$. But how $dfracz^2h^2 = dfraca^2-x^2-y^2a^2$ is wrong?! Same triangle same theorem..
– Edi
Aug 2 at 18:46
1
Let $x=0$ in your equation, then you have a circle (ellipse) equation not a line
– user 108128
Aug 2 at 18:48
Yes again you are right! But I don't understand how $dfracz^2h^2 = dfraca^2-x^2-y^2a^2$ gives a wrong result (i.e. circle) when limited to $x=0$ though it comes from the same trigonometric theorem!
– Edi
Aug 2 at 18:52
Sorry I didn't mean that. Of course it must be a line when limited to any line on plane through the origin; my question is that $z^2/h^2 = (a^2-x^2-y^2)/a^2$ must also give a line when the same limit applies but it fails and why? $z^2/h^2 = (a^2-x^2-y^2)/a^2$ has come from same rule as $(1-dfraczh)^2 = (x^2+y^2)/a^2.$ but it fails when testing a 'boundary' evaluation..
– Edi
Aug 2 at 18:57
1
Sorry. I gave you the true equation and I can't enlightened you dear. I think it's better you try more to find it yourself. This will be worthwhile.
– user 108128
Aug 2 at 19:28
 |Â
show 7 more comments
up vote
1
down vote
Hint: Your dealing with the equation $dfracz^2h^2 = dfraca^2-x^2-y^2a^2$ is wrong, try with
$$left(1-dfraczhright)^2 = dfracx^2+y^2a^2.$$
answered Aug 2 at 18:39
user 108128
18.8k41544
I tried with $left(1-dfraczhright)^2 = dfracx^2+y^2a^2$ and it gives $1/3$. But how $dfracz^2h^2 = dfraca^2-x^2-y^2a^2$ is wrong?! Same triangle same theorem..
– Edi
Aug 2 at 18:46
1
Let $x=0$ in your equation, then you have a circle (ellipse) equation not a line
– user 108128
Aug 2 at 18:48
Yes again you are right! But I don't understand how $dfracz^2h^2 = dfraca^2-x^2-y^2a^2$ gives a wrong result (i.e. circle) when limited to $x=0$ though it comes from the same trigonometric theorem!
– Edi
Aug 2 at 18:52
Sorry I didn't mean that. Of course it must be a line when limited to any line on plane through the origin; my question is that $z^2/h^2 = (a^2-x^2-y^2)/a^2$ must also give a line when the same limit applies but it fails and why? $z^2/h^2 = (a^2-x^2-y^2)/a^2$ has come from same rule as $(1-dfraczh)^2 = (x^2+y^2)/a^2.$ but it fails when testing a 'boundary' evaluation..
– Edi
Aug 2 at 18:57
1
Sorry. I gave you the true equation and I can't enlightened you dear. I think it's better you try more to find it yourself. This will be worthwhile.
– user 108128
Aug 2 at 19:28
 |Â
show 7 more comments
up vote
1
down vote
up vote
1
down vote
Hint: Your dealing with the equation $dfracz^2h^2 = dfraca^2-x^2-y^2a^2$ is wrong, try with
$$left(1-dfraczhright)^2 = dfracx^2+y^2a^2.$$
answered Aug 2 at 18:39
user 108128
18.8k41544
Hint: Your dealing with the equation $dfracz^2h^2 = dfraca^2-x^2-y^2a^2$ is wrong, try with
$$left(1-dfraczhright)^2 = dfracx^2+y^2a^2.$$
answered Aug 2 at 18:39
user 108128
18.8k41544
answered Aug 2 at 18:39
user 108128
18.8k41544
answered Aug 2 at 18:39
user 108128
18.8k41544
answered Aug 2 at 18:39
user 108128
18.8k41544
18.8k41544
I tried with $left(1-dfraczhright)^2 = dfracx^2+y^2a^2$ and it gives $1/3$. But how $dfracz^2h^2 = dfraca^2-x^2-y^2a^2$ is wrong?! Same triangle same theorem..
– Edi
Aug 2 at 18:46
1
Let $x=0$ in your equation, then you have a circle (ellipse) equation not a line
– user 108128
Aug 2 at 18:48
Yes again you are right! But I don't understand how $dfracz^2h^2 = dfraca^2-x^2-y^2a^2$ gives a wrong result (i.e. circle) when limited to $x=0$ though it comes from the same trigonometric theorem!
– Edi
Aug 2 at 18:52
Sorry I didn't mean that. Of course it must be a line when limited to any line on plane through the origin; my question is that $z^2/h^2 = (a^2-x^2-y^2)/a^2$ must also give a line when the same limit applies but it fails and why? $z^2/h^2 = (a^2-x^2-y^2)/a^2$ has come from same rule as $(1-dfraczh)^2 = (x^2+y^2)/a^2.$ but it fails when testing a 'boundary' evaluation..
– Edi
Aug 2 at 18:57
1
Sorry. I gave you the true equation and I can't enlightened you dear. I think it's better you try more to find it yourself. This will be worthwhile.
– user 108128
Aug 2 at 19:28
 |Â
show 7 more comments
I tried with $left(1-dfraczhright)^2 = dfracx^2+y^2a^2$ and it gives $1/3$. But how $dfracz^2h^2 = dfraca^2-x^2-y^2a^2$ is wrong?! Same triangle same theorem..
– Edi
Aug 2 at 18:46
1
Let $x=0$ in your equation, then you have a circle (ellipse) equation not a line
– user 108128
Aug 2 at 18:48
Yes again you are right! But I don't understand how $dfracz^2h^2 = dfraca^2-x^2-y^2a^2$ gives a wrong result (i.e. circle) when limited to $x=0$ though it comes from the same trigonometric theorem!
– Edi
Aug 2 at 18:52
Sorry I didn't mean that. Of course it must be a line when limited to any line on plane through the origin; my question is that $z^2/h^2 = (a^2-x^2-y^2)/a^2$ must also give a line when the same limit applies but it fails and why? $z^2/h^2 = (a^2-x^2-y^2)/a^2$ has come from same rule as $(1-dfraczh)^2 = (x^2+y^2)/a^2.$ but it fails when testing a 'boundary' evaluation..
– Edi
Aug 2 at 18:57
1
Sorry. I gave you the true equation and I can't enlightened you dear. I think it's better you try more to find it yourself. This will be worthwhile.
– user 108128
Aug 2 at 19:28
I tried with $left(1-dfraczhright)^2 = dfracx^2+y^2a^2$ and it gives $1/3$. But how $dfracz^2h^2 = dfraca^2-x^2-y^2a^2$ is wrong?! Same triangle same theorem..
– Edi
Aug 2 at 18:46
I tried with $left(1-dfraczhright)^2 = dfracx^2+y^2a^2$ and it gives $1/3$. But how $dfracz^2h^2 = dfraca^2-x^2-y^2a^2$ is wrong?! Same triangle same theorem..
– Edi
Aug 2 at 18:46
1
1
Let $x=0$ in your equation, then you have a circle (ellipse) equation not a line
– user 108128
Aug 2 at 18:48
Let $x=0$ in your equation, then you have a circle (ellipse) equation not a line
– user 108128
Aug 2 at 18:48
Yes again you are right! But I don't understand how $dfracz^2h^2 = dfraca^2-x^2-y^2a^2$ gives a wrong result (i.e. circle) when limited to $x=0$ though it comes from the same trigonometric theorem!
– Edi
Aug 2 at 18:52
Yes again you are right! But I don't understand how $dfracz^2h^2 = dfraca^2-x^2-y^2a^2$ gives a wrong result (i.e. circle) when limited to $x=0$ though it comes from the same trigonometric theorem!
– Edi
Aug 2 at 18:52
Sorry I didn't mean that. Of course it must be a line when limited to any line on plane through the origin; my question is that $z^2/h^2 = (a^2-x^2-y^2)/a^2$ must also give a line when the same limit applies but it fails and why? $z^2/h^2 = (a^2-x^2-y^2)/a^2$ has come from same rule as $(1-dfraczh)^2 = (x^2+y^2)/a^2.$ but it fails when testing a 'boundary' evaluation..
– Edi
Aug 2 at 18:57
Sorry I didn't mean that. Of course it must be a line when limited to any line on plane through the origin; my question is that $z^2/h^2 = (a^2-x^2-y^2)/a^2$ must also give a line when the same limit applies but it fails and why? $z^2/h^2 = (a^2-x^2-y^2)/a^2$ has come from same rule as $(1-dfraczh)^2 = (x^2+y^2)/a^2.$ but it fails when testing a 'boundary' evaluation..
– Edi
Aug 2 at 18:57
1
1
Sorry. I gave you the true equation and I can't enlightened you dear. I think it's better you try more to find it yourself. This will be worthwhile.
– user 108128
Aug 2 at 19:28
Sorry. I gave you the true equation and I can't enlightened you dear. I think it's better you try more to find it yourself. This will be worthwhile.
– user 108128
Aug 2 at 19:28
 |Â
show 7 more comments
up vote
1
down vote
That region over which you are integrating is no cone at all. It's the top half of an ellipsoid.
You will get the value that you're after computing the integral$$int_-a^aint_-sqrta^2-x^2^sqrta^2-x^2int_frac hasqrtx^2+y^2^h1,mathrm dz,mathrm dy,mathrm dx.$$Here, my cone is$$left(x,y,z)inmathbbR^3,middle.$$
answered Aug 2 at 18:35
José Carlos Santos
112k1696172
$z^2/h^2 = (a^2-x^2-y^2)/a^2$ and $(1-z/h)^2 = (x^2+y^2)/a^2$ comes from the same theorem of trigonometric on the same object, but I don't understand why the latter gives an ellipsoid; both equations are also squared both use same $z$..
– Edi
Aug 2 at 19:10
$$fracz^2h^2=fraca^2-x^2-y^2a^2iff a^2z^2=(ah)^2-h^2x^2-h^2y^2iff h^2x^2+h^2y^2+a^2z^2=(ah)^2.$$Is this an ellipsoid or not?
– José Carlos Santos
Aug 2 at 19:14
Of course it is but I derived it from a triangle as an intersection a plane through z-axis and the cone. So how it comes an ellipsoid?!
– Edi
Aug 2 at 19:17
The book says "..the cone $x^2+y^2le a^2, 0 le z le h/a^2(a^2-x^2-y^2) $" for hint and I don't know how it deduce $z$?
– Edi
Aug 2 at 19:19
@Edi Then the hint contains a mistake.
– José Carlos Santos
Aug 2 at 19:19
 |Â
show 5 more comments
up vote
1
down vote
That region over which you are integrating is no cone at all. It's the top half of an ellipsoid.
You will get the value that you're after computing the integral$$int_-a^aint_-sqrta^2-x^2^sqrta^2-x^2int_frac hasqrtx^2+y^2^h1,mathrm dz,mathrm dy,mathrm dx.$$Here, my cone is$$left(x,y,z)inmathbbR^3,middle.$$
answered Aug 2 at 18:35
José Carlos Santos
112k1696172
$z^2/h^2 = (a^2-x^2-y^2)/a^2$ and $(1-z/h)^2 = (x^2+y^2)/a^2$ comes from the same theorem of trigonometric on the same object, but I don't understand why the latter gives an ellipsoid; both equations are also squared both use same $z$..
– Edi
Aug 2 at 19:10
$$fracz^2h^2=fraca^2-x^2-y^2a^2iff a^2z^2=(ah)^2-h^2x^2-h^2y^2iff h^2x^2+h^2y^2+a^2z^2=(ah)^2.$$Is this an ellipsoid or not?
– José Carlos Santos
Aug 2 at 19:14
Of course it is but I derived it from a triangle as an intersection a plane through z-axis and the cone. So how it comes an ellipsoid?!
– Edi
Aug 2 at 19:17
The book says "..the cone $x^2+y^2le a^2, 0 le z le h/a^2(a^2-x^2-y^2) $" for hint and I don't know how it deduce $z$?
– Edi
Aug 2 at 19:19
@Edi Then the hint contains a mistake.
– José Carlos Santos
Aug 2 at 19:19
 |Â
show 5 more comments
up vote
1
down vote
up vote
1
down vote
That region over which you are integrating is no cone at all. It's the top half of an ellipsoid.
You will get the value that you're after computing the integral$$int_-a^aint_-sqrta^2-x^2^sqrta^2-x^2int_frac hasqrtx^2+y^2^h1,mathrm dz,mathrm dy,mathrm dx.$$Here, my cone is$$left(x,y,z)inmathbbR^3,middle.$$
answered Aug 2 at 18:35
José Carlos Santos
112k1696172
That region over which you are integrating is no cone at all. It's the top half of an ellipsoid.
You will get the value that you're after computing the integral$$int_-a^aint_-sqrta^2-x^2^sqrta^2-x^2int_frac hasqrtx^2+y^2^h1,mathrm dz,mathrm dy,mathrm dx.$$Here, my cone is$$left(x,y,z)inmathbbR^3,middle.$$
answered Aug 2 at 18:35
José Carlos Santos
112k1696172
edited Aug 2 at 19:02
answered Aug 2 at 18:35
José Carlos Santos
112k1696172
answered Aug 2 at 18:35
José Carlos Santos
112k1696172
answered Aug 2 at 18:35
José Carlos Santos
112k1696172
112k1696172
$z^2/h^2 = (a^2-x^2-y^2)/a^2$ and $(1-z/h)^2 = (x^2+y^2)/a^2$ comes from the same theorem of trigonometric on the same object, but I don't understand why the latter gives an ellipsoid; both equations are also squared both use same $z$..
– Edi
Aug 2 at 19:10
$$fracz^2h^2=fraca^2-x^2-y^2a^2iff a^2z^2=(ah)^2-h^2x^2-h^2y^2iff h^2x^2+h^2y^2+a^2z^2=(ah)^2.$$Is this an ellipsoid or not?
– José Carlos Santos
Aug 2 at 19:14
Of course it is but I derived it from a triangle as an intersection a plane through z-axis and the cone. So how it comes an ellipsoid?!
– Edi
Aug 2 at 19:17
The book says "..the cone $x^2+y^2le a^2, 0 le z le h/a^2(a^2-x^2-y^2) $" for hint and I don't know how it deduce $z$?
– Edi
Aug 2 at 19:19
@Edi Then the hint contains a mistake.
– José Carlos Santos
Aug 2 at 19:19
 |Â
show 5 more comments
$z^2/h^2 = (a^2-x^2-y^2)/a^2$ and $(1-z/h)^2 = (x^2+y^2)/a^2$ comes from the same theorem of trigonometric on the same object, but I don't understand why the latter gives an ellipsoid; both equations are also squared both use same $z$..
– Edi
Aug 2 at 19:10
$$fracz^2h^2=fraca^2-x^2-y^2a^2iff a^2z^2=(ah)^2-h^2x^2-h^2y^2iff h^2x^2+h^2y^2+a^2z^2=(ah)^2.$$Is this an ellipsoid or not?
– José Carlos Santos
Aug 2 at 19:14
Of course it is but I derived it from a triangle as an intersection a plane through z-axis and the cone. So how it comes an ellipsoid?!
– Edi
Aug 2 at 19:17
The book says "..the cone $x^2+y^2le a^2, 0 le z le h/a^2(a^2-x^2-y^2) $" for hint and I don't know how it deduce $z$?
– Edi
Aug 2 at 19:19
@Edi Then the hint contains a mistake.
– José Carlos Santos
Aug 2 at 19:19
$z^2/h^2 = (a^2-x^2-y^2)/a^2$ and $(1-z/h)^2 = (x^2+y^2)/a^2$ comes from the same theorem of trigonometric on the same object, but I don't understand why the latter gives an ellipsoid; both equations are also squared both use same $z$..
– Edi
Aug 2 at 19:10
$z^2/h^2 = (a^2-x^2-y^2)/a^2$ and $(1-z/h)^2 = (x^2+y^2)/a^2$ comes from the same theorem of trigonometric on the same object, but I don't understand why the latter gives an ellipsoid; both equations are also squared both use same $z$..
– Edi
Aug 2 at 19:10
$$fracz^2h^2=fraca^2-x^2-y^2a^2iff a^2z^2=(ah)^2-h^2x^2-h^2y^2iff h^2x^2+h^2y^2+a^2z^2=(ah)^2.$$Is this an ellipsoid or not?
– José Carlos Santos
Aug 2 at 19:14
$$fracz^2h^2=fraca^2-x^2-y^2a^2iff a^2z^2=(ah)^2-h^2x^2-h^2y^2iff h^2x^2+h^2y^2+a^2z^2=(ah)^2.$$Is this an ellipsoid or not?
– José Carlos Santos
Aug 2 at 19:14
Of course it is but I derived it from a triangle as an intersection a plane through z-axis and the cone. So how it comes an ellipsoid?!
– Edi
Aug 2 at 19:17
Of course it is but I derived it from a triangle as an intersection a plane through z-axis and the cone. So how it comes an ellipsoid?!
– Edi
Aug 2 at 19:17
The book says "..the cone $x^2+y^2le a^2, 0 le z le h/a^2(a^2-x^2-y^2) $" for hint and I don't know how it deduce $z$?
– Edi
Aug 2 at 19:19
The book says "..the cone $x^2+y^2le a^2, 0 le z le h/a^2(a^2-x^2-y^2) $" for hint and I don't know how it deduce $z$?
– Edi
Aug 2 at 19:19
@Edi Then the hint contains a mistake.
– José Carlos Santos
Aug 2 at 19:19
@Edi Then the hint contains a mistake.
– José Carlos Santos
Aug 2 at 19:19
 |Â
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up vote
0
down vote
We need 3 constants $ (a,h,z_1):$
Equation of right circular cone with $z_1$ apex shift
$$fracrz-z_1=fracah$$ or squaring,
$$ r^2=x^2+y^2= (z-z_1)^2(a/h)^2.$$
answered Aug 2 at 19:02
Narasimham
20.1k51857
add a comment |Â
up vote
0
down vote
We need 3 constants $ (a,h,z_1):$
Equation of right circular cone with $z_1$ apex shift
$$fracrz-z_1=fracah$$ or squaring,
$$ r^2=x^2+y^2= (z-z_1)^2(a/h)^2.$$
answered Aug 2 at 19:02
Narasimham
20.1k51857
add a comment |Â
up vote
0
down vote
up vote
0
down vote
We need 3 constants $ (a,h,z_1):$
Equation of right circular cone with $z_1$ apex shift
$$fracrz-z_1=fracah$$ or squaring,
$$ r^2=x^2+y^2= (z-z_1)^2(a/h)^2.$$
answered Aug 2 at 19:02
Narasimham
20.1k51857
We need 3 constants $ (a,h,z_1):$
Equation of right circular cone with $z_1$ apex shift
$$fracrz-z_1=fracah$$ or squaring,
$$ r^2=x^2+y^2= (z-z_1)^2(a/h)^2.$$
answered Aug 2 at 19:02
Narasimham
20.1k51857
answered Aug 2 at 19:02
Narasimham
20.1k51857
answered Aug 2 at 19:02
Narasimham
20.1k51857
answered Aug 2 at 19:02
Narasimham
20.1k51857
20.1k51857
add a comment |Â
add a comment |Â
up vote
0
down vote
accepted
Because $dfracz^2h^2 = dfraca^2-x^2-y^2a^2$ is wrong. The correct one is $dfraczh = dfraca-sqrtx^2-y^2a$ which gives the correct answer $1/3 pi a^2 h$.
answered Aug 2 at 19:59
Edi
1,121728
add a comment |Â
up vote
0
down vote
accepted
Because $dfracz^2h^2 = dfraca^2-x^2-y^2a^2$ is wrong. The correct one is $dfraczh = dfraca-sqrtx^2-y^2a$ which gives the correct answer $1/3 pi a^2 h$.
answered Aug 2 at 19:59
Edi
1,121728
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Because $dfracz^2h^2 = dfraca^2-x^2-y^2a^2$ is wrong. The correct one is $dfraczh = dfraca-sqrtx^2-y^2a$ which gives the correct answer $1/3 pi a^2 h$.
answered Aug 2 at 19:59
Edi
1,121728
Because $dfracz^2h^2 = dfraca^2-x^2-y^2a^2$ is wrong. The correct one is $dfraczh = dfraca-sqrtx^2-y^2a$ which gives the correct answer $1/3 pi a^2 h$.
answered Aug 2 at 19:59
Edi
1,121728
answered Aug 2 at 19:59
Edi
1,121728
answered Aug 2 at 19:59
Edi
1,121728
answered Aug 2 at 19:59
Edi
1,121728
1,121728
add a comment |Â
add a comment |Â
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