Help calculating this volume [closed]
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Being $x,y,z geq 0$ , calculate the volume bounded by $sqrt{x} + sqrt{2y} + sqrt{3z} = 1$
Thanks in advance for the help
integration multivariable-calculus
asked Jan 16 at 18:08
KchernoKcherno
61
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closed as off-topic by mfl, Morgan Rodgers, RRL, Arnaud D., Alexander Gruber♦ Jan 17 at 2:54
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – mfl, Morgan Rodgers, RRL, Arnaud D., Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Being $x,y,z geq 0$ , calculate the volume bounded by $sqrt{x} + sqrt{2y} + sqrt{3z} = 1$
Thanks in advance for the help
integration multivariable-calculus
asked Jan 16 at 18:08
KchernoKcherno
61
$endgroup$
closed as off-topic by mfl, Morgan Rodgers, RRL, Arnaud D., Alexander Gruber♦ Jan 17 at 2:54
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – mfl, Morgan Rodgers, RRL, Arnaud D., Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
2
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What did you tried?
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– DiegoMath
Jan 16 at 18:12
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Did you mean to fund the volume of the region $mathrm{X}$ defined by the relations $x, y,z geq 0$ and $sqrt{x} + sqrt{2y}+sqrt{3z}=1$?
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– Will M.
Jan 16 at 18:13
$begingroup$
Did you tried use some change of variables? What if $x=u^2, 2y=v^2 mbox{and} 3z=w^2$?
$endgroup$
– DiegoMath
Jan 16 at 18:13
$begingroup$
I´ll try to make that change
$endgroup$
– Kcherno
Jan 16 at 18:18
add a comment |
$begingroup$
Being $x,y,z geq 0$ , calculate the volume bounded by $sqrt{x} + sqrt{2y} + sqrt{3z} = 1$
Thanks in advance for the help
integration multivariable-calculus
asked Jan 16 at 18:08
KchernoKcherno
61
$endgroup$
Being $x,y,z geq 0$ , calculate the volume bounded by $sqrt{x} + sqrt{2y} + sqrt{3z} = 1$
Thanks in advance for the help
integration multivariable-calculus
integration multivariable-calculus
asked Jan 16 at 18:08
KchernoKcherno
61
asked Jan 16 at 18:08
KchernoKcherno
61
asked Jan 16 at 18:08
KchernoKcherno
61
asked Jan 16 at 18:08
KchernoKcherno
61
asked Jan 16 at 18:08
KchernoKcherno
61
61
closed as off-topic by mfl, Morgan Rodgers, RRL, Arnaud D., Alexander Gruber♦ Jan 17 at 2:54
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – mfl, Morgan Rodgers, RRL, Arnaud D., Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by mfl, Morgan Rodgers, RRL, Arnaud D., Alexander Gruber♦ Jan 17 at 2:54
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – mfl, Morgan Rodgers, RRL, Arnaud D., Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
2
$begingroup$
What did you tried?
$endgroup$
– DiegoMath
Jan 16 at 18:12
$begingroup$
Did you mean to fund the volume of the region $mathrm{X}$ defined by the relations $x, y,z geq 0$ and $sqrt{x} + sqrt{2y}+sqrt{3z}=1$?
$endgroup$
– Will M.
Jan 16 at 18:13
$begingroup$
Did you tried use some change of variables? What if $x=u^2, 2y=v^2 mbox{and} 3z=w^2$?
$endgroup$
– DiegoMath
Jan 16 at 18:13
$begingroup$
I´ll try to make that change
$endgroup$
– Kcherno
Jan 16 at 18:18
add a comment |
2
$begingroup$
What did you tried?
$endgroup$
– DiegoMath
Jan 16 at 18:12
$begingroup$
Did you mean to fund the volume of the region $mathrm{X}$ defined by the relations $x, y,z geq 0$ and $sqrt{x} + sqrt{2y}+sqrt{3z}=1$?
$endgroup$
– Will M.
Jan 16 at 18:13
$begingroup$
Did you tried use some change of variables? What if $x=u^2, 2y=v^2 mbox{and} 3z=w^2$?
$endgroup$
– DiegoMath
Jan 16 at 18:13
$begingroup$
I´ll try to make that change
$endgroup$
– Kcherno
Jan 16 at 18:18
2
2
$begingroup$
What did you tried?
$endgroup$
– DiegoMath
Jan 16 at 18:12
$begingroup$
What did you tried?
$endgroup$
– DiegoMath
Jan 16 at 18:12
$begingroup$
Did you mean to fund the volume of the region $mathrm{X}$ defined by the relations $x, y,z geq 0$ and $sqrt{x} + sqrt{2y}+sqrt{3z}=1$?
$endgroup$
– Will M.
Jan 16 at 18:13
$begingroup$
Did you mean to fund the volume of the region $mathrm{X}$ defined by the relations $x, y,z geq 0$ and $sqrt{x} + sqrt{2y}+sqrt{3z}=1$?
$endgroup$
– Will M.
Jan 16 at 18:13
$begingroup$
Did you tried use some change of variables? What if $x=u^2, 2y=v^2 mbox{and} 3z=w^2$?
$endgroup$
– DiegoMath
Jan 16 at 18:13
$begingroup$
Did you tried use some change of variables? What if $x=u^2, 2y=v^2 mbox{and} 3z=w^2$?
$endgroup$
– DiegoMath
Jan 16 at 18:13
$begingroup$
I´ll try to make that change
$endgroup$
– Kcherno
Jan 16 at 18:18
$begingroup$
I´ll try to make that change
$endgroup$
– Kcherno
Jan 16 at 18:18
add a comment |
1 Answer
1
active
oldest
votes
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As suggested, use $x=u^2, 2y=v^2 mbox{and} 3z=w^2$. Then your volume is
$$
V = int int int frac43 u v w ; du ;dv ;dw
$$
with $u,v,w > 0$ and $u + v + w le 1$. The boundaries can be implemented by
$$
V = int_0^{1} int_0^{1-u} int_0^{1-u-v} frac43 u v w ; du ;dv ;dw
$$
so
$$
V = frac46 int_0^{1} int_0^{1-u} u v (1-u-v)^2 ; du ;dv
$$
$$
V = frac46 frac{1}{12} int_0^{1} u (u-1)^4 ; du = frac46 frac{1}{12} frac{1}{30} = frac{1}{540} simeq 0.00185
$$
answered Jan 16 at 21:50
AndreasAndreas
8,4161137
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$begingroup$
Wow, a downvote on this one. I'm looking forward to the alternative solution to come up from this enthusiast coworker.
$endgroup$
– Andreas
Jan 17 at 8:22
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
As suggested, use $x=u^2, 2y=v^2 mbox{and} 3z=w^2$. Then your volume is
$$
V = int int int frac43 u v w ; du ;dv ;dw
$$
with $u,v,w > 0$ and $u + v + w le 1$. The boundaries can be implemented by
$$
V = int_0^{1} int_0^{1-u} int_0^{1-u-v} frac43 u v w ; du ;dv ;dw
$$
so
$$
V = frac46 int_0^{1} int_0^{1-u} u v (1-u-v)^2 ; du ;dv
$$
$$
V = frac46 frac{1}{12} int_0^{1} u (u-1)^4 ; du = frac46 frac{1}{12} frac{1}{30} = frac{1}{540} simeq 0.00185
$$
answered Jan 16 at 21:50
AndreasAndreas
8,4161137
$endgroup$
$begingroup$
Wow, a downvote on this one. I'm looking forward to the alternative solution to come up from this enthusiast coworker.
$endgroup$
– Andreas
Jan 17 at 8:22
add a comment |
$begingroup$
As suggested, use $x=u^2, 2y=v^2 mbox{and} 3z=w^2$. Then your volume is
$$
V = int int int frac43 u v w ; du ;dv ;dw
$$
with $u,v,w > 0$ and $u + v + w le 1$. The boundaries can be implemented by
$$
V = int_0^{1} int_0^{1-u} int_0^{1-u-v} frac43 u v w ; du ;dv ;dw
$$
so
$$
V = frac46 int_0^{1} int_0^{1-u} u v (1-u-v)^2 ; du ;dv
$$
$$
V = frac46 frac{1}{12} int_0^{1} u (u-1)^4 ; du = frac46 frac{1}{12} frac{1}{30} = frac{1}{540} simeq 0.00185
$$
answered Jan 16 at 21:50
AndreasAndreas
8,4161137
$endgroup$
$begingroup$
Wow, a downvote on this one. I'm looking forward to the alternative solution to come up from this enthusiast coworker.
$endgroup$
– Andreas
Jan 17 at 8:22
add a comment |
$begingroup$
As suggested, use $x=u^2, 2y=v^2 mbox{and} 3z=w^2$. Then your volume is
$$
V = int int int frac43 u v w ; du ;dv ;dw
$$
with $u,v,w > 0$ and $u + v + w le 1$. The boundaries can be implemented by
$$
V = int_0^{1} int_0^{1-u} int_0^{1-u-v} frac43 u v w ; du ;dv ;dw
$$
so
$$
V = frac46 int_0^{1} int_0^{1-u} u v (1-u-v)^2 ; du ;dv
$$
$$
V = frac46 frac{1}{12} int_0^{1} u (u-1)^4 ; du = frac46 frac{1}{12} frac{1}{30} = frac{1}{540} simeq 0.00185
$$
answered Jan 16 at 21:50
AndreasAndreas
8,4161137
$endgroup$
As suggested, use $x=u^2, 2y=v^2 mbox{and} 3z=w^2$. Then your volume is
$$
V = int int int frac43 u v w ; du ;dv ;dw
$$
with $u,v,w > 0$ and $u + v + w le 1$. The boundaries can be implemented by
$$
V = int_0^{1} int_0^{1-u} int_0^{1-u-v} frac43 u v w ; du ;dv ;dw
$$
so
$$
V = frac46 int_0^{1} int_0^{1-u} u v (1-u-v)^2 ; du ;dv
$$
$$
V = frac46 frac{1}{12} int_0^{1} u (u-1)^4 ; du = frac46 frac{1}{12} frac{1}{30} = frac{1}{540} simeq 0.00185
$$
answered Jan 16 at 21:50
AndreasAndreas
8,4161137
answered Jan 16 at 21:50
AndreasAndreas
8,4161137
answered Jan 16 at 21:50
AndreasAndreas
8,4161137
answered Jan 16 at 21:50
AndreasAndreas
8,4161137
8,4161137
$begingroup$
Wow, a downvote on this one. I'm looking forward to the alternative solution to come up from this enthusiast coworker.
$endgroup$
– Andreas
Jan 17 at 8:22
add a comment |
$begingroup$
Wow, a downvote on this one. I'm looking forward to the alternative solution to come up from this enthusiast coworker.
$endgroup$
– Andreas
Jan 17 at 8:22
$begingroup$
Wow, a downvote on this one. I'm looking forward to the alternative solution to come up from this enthusiast coworker.
$endgroup$
– Andreas
Jan 17 at 8:22
$begingroup$
Wow, a downvote on this one. I'm looking forward to the alternative solution to come up from this enthusiast coworker.
$endgroup$
– Andreas
Jan 17 at 8:22
add a comment |
2
$begingroup$
What did you tried?
$endgroup$
– DiegoMath
Jan 16 at 18:12
$begingroup$
Did you mean to fund the volume of the region $mathrm{X}$ defined by the relations $x, y,z geq 0$ and $sqrt{x} + sqrt{2y}+sqrt{3z}=1$?
$endgroup$
– Will M.
Jan 16 at 18:13
$begingroup$
Did you tried use some change of variables? What if $x=u^2, 2y=v^2 mbox{and} 3z=w^2$?
$endgroup$
– DiegoMath
Jan 16 at 18:13
$begingroup$
I´ll try to make that change
$endgroup$
– Kcherno
Jan 16 at 18:18