Mixing
Find the volume of the region bounded by the paraboloids $z = 2x^2 + y^2$ and $z = 12 - x^2 - 2y^2$
$begingroup$
Find the volume of the region bounded by the paraboloids $z = 2x^2 + y^2$ and $z = 12 - x^2 - 2y^2$
The answer says $24pi$ but I just never got it :) Thanks!
multivariable-calculus
edited Feb 3 at 9:02
Jneven
953322
asked Feb 3 at 8:45
Ploen VoraprukpisutPloen Voraprukpisut
1
$endgroup$
add a comment |
$begingroup$
Find the volume of the region bounded by the paraboloids $z = 2x^2 + y^2$ and $z = 12 - x^2 - 2y^2$
The answer says $24pi$ but I just never got it :) Thanks!
multivariable-calculus
edited Feb 3 at 9:02
Jneven
953322
asked Feb 3 at 8:45
Ploen VoraprukpisutPloen Voraprukpisut
1
$endgroup$
add a comment |
$begingroup$
Find the volume of the region bounded by the paraboloids $z = 2x^2 + y^2$ and $z = 12 - x^2 - 2y^2$
The answer says $24pi$ but I just never got it :) Thanks!
multivariable-calculus
edited Feb 3 at 9:02
Jneven
953322
asked Feb 3 at 8:45
Ploen VoraprukpisutPloen Voraprukpisut
1
$endgroup$
Find the volume of the region bounded by the paraboloids $z = 2x^2 + y^2$ and $z = 12 - x^2 - 2y^2$
The answer says $24pi$ but I just never got it :) Thanks!
multivariable-calculus
multivariable-calculus
edited Feb 3 at 9:02
Jneven
953322
asked Feb 3 at 8:45
Ploen VoraprukpisutPloen Voraprukpisut
1
edited Feb 3 at 9:02
Jneven
953322
asked Feb 3 at 8:45
Ploen VoraprukpisutPloen Voraprukpisut
1
edited Feb 3 at 9:02
Jneven
953322
edited Feb 3 at 9:02
Jneven
953322
edited Feb 3 at 9:02
Jneven
953322
953322
asked Feb 3 at 8:45
Ploen VoraprukpisutPloen Voraprukpisut
1
asked Feb 3 at 8:45
Ploen VoraprukpisutPloen Voraprukpisut
1
asked Feb 3 at 8:45
Ploen VoraprukpisutPloen Voraprukpisut
1
1
add a comment |
add a comment |
1 Answer
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$begingroup$
If you are familiar with the polar coordinates, you can easily get the answer.
First you find the intersection of the two surfaces, which is $$x^2+y^2=4$$ or in polar coordinates it is r=2.
The integral for the volume is $$ V= int _0^{2pi} int _0^2 (12-3r^2)rdr dtheta $$
Which results in $$V=24pi$$
answered Feb 3 at 9:56
Mohammad Riazi-KermaniMohammad Riazi-Kermani
42.1k42061
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$begingroup$
If you are familiar with the polar coordinates, you can easily get the answer.
First you find the intersection of the two surfaces, which is $$x^2+y^2=4$$ or in polar coordinates it is r=2.
The integral for the volume is $$ V= int _0^{2pi} int _0^2 (12-3r^2)rdr dtheta $$
Which results in $$V=24pi$$
answered Feb 3 at 9:56
Mohammad Riazi-KermaniMohammad Riazi-Kermani
42.1k42061
$endgroup$
add a comment |
$begingroup$
If you are familiar with the polar coordinates, you can easily get the answer.
First you find the intersection of the two surfaces, which is $$x^2+y^2=4$$ or in polar coordinates it is r=2.
The integral for the volume is $$ V= int _0^{2pi} int _0^2 (12-3r^2)rdr dtheta $$
Which results in $$V=24pi$$
answered Feb 3 at 9:56
Mohammad Riazi-KermaniMohammad Riazi-Kermani
42.1k42061
$endgroup$
add a comment |
$begingroup$
If you are familiar with the polar coordinates, you can easily get the answer.
First you find the intersection of the two surfaces, which is $$x^2+y^2=4$$ or in polar coordinates it is r=2.
The integral for the volume is $$ V= int _0^{2pi} int _0^2 (12-3r^2)rdr dtheta $$
Which results in $$V=24pi$$
answered Feb 3 at 9:56
Mohammad Riazi-KermaniMohammad Riazi-Kermani
42.1k42061
$endgroup$
If you are familiar with the polar coordinates, you can easily get the answer.
First you find the intersection of the two surfaces, which is $$x^2+y^2=4$$ or in polar coordinates it is r=2.
The integral for the volume is $$ V= int _0^{2pi} int _0^2 (12-3r^2)rdr dtheta $$
Which results in $$V=24pi$$
answered Feb 3 at 9:56
Mohammad Riazi-KermaniMohammad Riazi-Kermani
42.1k42061
answered Feb 3 at 9:56
Mohammad Riazi-KermaniMohammad Riazi-Kermani
42.1k42061
answered Feb 3 at 9:56
Mohammad Riazi-KermaniMohammad Riazi-Kermani
42.1k42061
answered Feb 3 at 9:56
Mohammad Riazi-KermaniMohammad Riazi-Kermani
42.1k42061
42.1k42061
add a comment |
add a comment |
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